9.4. Multiplicities and Diagonalization

The Eigenvalue Decomposition¶

Recall that if $A$ is an $n \times n$ matrix, then $A$ has $n$ eigenvalues, it’s just that some of the eigenvalues may be complex (leading to complex-valued eigenvectors), and some eigenvalues may be repeated. These $n$ eigenvalues are all solutions to the characteristic polynomial of $A$ ,

p(\lambda) = \text{det}(A - \lambda I)

The corresponding eigenvectors are the solutions to the system of equations $(A - \lambda I) \vec v = \vec 0$ .

Suppose $A$ has $n$ linearly independent eigenvectors $\vec v_1, \vec v_2, \cdots, \vec v_n$ , with eigenvalues $\lambda_1, \lambda_2, \cdots, \lambda_n$ . What I’m about to propose next will seem a little arbitrary, but bear with me – you’ll see the power of this idea shortly. What happens if we multiply $A$ by a matrix, say $V$ , whose columns are the eigenvectors of $A$ ?

\begin{align*} AV &= A \begin{bmatrix} | & | && | \\ \vec v_1 & \vec v_2 & \cdots & \vec v_n \\ | & | && | \end{bmatrix} \\ &= \begin{bmatrix} | & | && | \\ A\vec v_1 & A\vec v_2 & \cdots & A\vec v_n \\ | & | && | \end{bmatrix} \\ &= \begin{bmatrix} | & | && | \\ \lambda_1 \vec v_1 & \lambda_2 \vec v_2 & \cdots & \lambda_n \vec v_n \\ | & | && | \end{bmatrix} \end{align*}

$AV$ is a matrix whose columns are eigenvectors of $A$ , each scaled by the corresponding eigenvalue! Is there another way to write the last line above?

\begin{align*} AV &= \begin{bmatrix} | & | && | \\ \lambda_1 \vec v_1 & \lambda_2 \vec v_2 & \cdots & \lambda_n \vec v_n \\ | & | && | \end{bmatrix} \\ &= \begin{bmatrix} | & | && | \\ \vec v_1 & \vec v_2 & \cdots & \vec v_n \\ | & | && | \end{bmatrix} \begin{bmatrix} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_n \end{bmatrix} \\ &= V \Lambda \end{align*}

Here, $\mathcal{\Lambda}$ (the capitalized Greek letter for $\mathbf{\lambda}$ ) is a diagonal matrix of eigenvalues in the same order as the eigenvectors in $V$ .

AV = V \Lambda

We can take this a step further. If $V$ is invertible – which it is here, since we assumed $A$ has $n$ linearly independent eigenvectors – then we can multiply both sides of the equation above by $V^{-1}$ on the right to get

A = V \Lambda V^{-1}

The existence of this decomposition is contingent on $V$ being invertible, which happens when $A$ has $n$ linearly independent eigenvectors. If $A$ doesn’t have “enough” eigenvectors, then $V$ wouldn’t be invertible, and we wouldn’t be able to decompose $A$ in this way.

Diagonalizable Matrices¶

Definition: Diagonalizable Matrix

A matrix $A$ is diagonalizable if it has an eigenvalue decomposition

A = V \Lambda V^{-1}

using the same definitions of $V$ and $\mathcal{\Lambda}$ above. $A$ is diagonalizable if and only if it has $n$ linearly independent eigenvectors (which allows for $V$ to be invertible). Otherwise, $A$ is not diagonalizable, and is sometimes called defective.

Often, $A$ is defined to be diagonalizable if and only if it can be written in the form $A = PDP^{-1}$ where $P$ is some invertible matrix and $D$ is some diagonal matrix. This is a general definition of diagonalizability. But, the only matrices that can be written in $A = PDP^{-1}$ have eigenvalue decompositions and vice versa. The eigenvalue decomposition tells you how to actually diagonalize $A$ .

A First Example¶

Let’s start with a familiar example, $A = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}$ , which was the first example we saw in Chapter 9.1. $A$ has eigenvalues $\lambda_1 = 3$ and $\lambda_2 = -1$ , and corresponding eigenvectors $\vec v_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$ and $\vec v_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}$ . These eigenvectors are linearly independent, so $A$ is diagonalizable, and we can write

V = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}, \qquad \Lambda = \begin{bmatrix} 3 & 0 \\ 0 & -1 \end{bmatrix}

which tells us that

V \Lambda V^{-1} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 3 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 1/2 & 1/2 \\ 1/2 & -1/2 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} = A

Let’s check with numpy.

V = np.array([[1, 1], 
              [1, -1]])

Lambda = np.diag([3, -1]) # New tool.
V_inv = np.linalg.inv(V)
V @ Lambda @ V_inv # The same as A!

array([[1., 2.],
       [2., 1.]])

Is this decomposition unique? No, because we could have chosen a different set of eigenvectors, or wrote them in a different order (in which case $\mathcal{\Lambda}$ would be different). We’ll keep the eigenvectors in the same order, but instead let’s consider

V = \begin{bmatrix} 2 & -5 \\ 2 & 5 \end{bmatrix}

which is also a valid eigenvector matrix for $A$ . (Remember that any scalar multiple of an eigenvector is still an eigenvector!)

It is still true that $A = V \Lambda V^{-1}$ . This may seem a little unbelievable (I wasn’t convinced at first), but remember that $V^{-1}$ “undoes” any changes in scaling we introduce to $V$ .

V \Lambda V^{-1} = \begin{bmatrix} 2 & -5 \\ 2 & 5 \end{bmatrix} \begin{bmatrix} 3 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 1/4 & 1/4 \\ -1/10 & 1/10 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} = A

V = np.array([[2, -5], 
              [2, 5]])
              
Lambda = np.diag([3, -1])
V_inv = np.linalg.inv(V)
V @ Lambda @ V_inv # Same as above!

array([[1., 2.],
       [2., 1.]])

Activity 1¶

Activity 1

Suppose $\vec v_1$ , $\vec v_2$ , and $\vec v_3$ are eigenvectors of $A$ with eigenvalues $\lambda_1$ , $\lambda_2$ , and $\lambda_3$ , respectively, and all three of $\lambda_1$ , $\lambda_2$ , and $\lambda_3$ are distinct.

Prove that $\vec v_1$ and $\vec v_2$ are linearly independent.
Prove that $\vec v_1$ , $\vec v_2$ , and $\vec v_3$ are linearly independent.

Solution

Part 1 We’ll start by assuming that $A \vec v_1 = \lambda_1 \vec v_1$ , $A \vec v_2 = \lambda_2 \vec v_2$ , and $\lambda_1 \neq \lambda_2$ .

If $\vec v_1$ and $\vec v_2$ are linearly independent, it means that the only solution to the equation

c_1 \vec v_1 + c_2 \vec v_2 = \vec 0

is $c_1 = c_2 = 0$ . (Otherwise, there are non-zero scalars $c_1$ and $c_2$ that make the equation true.)

Let’s start with $c_1 \vec v_1 + c_2 \vec v_2 = \vec 0$ and do two things to it:

Multiply both sides (on the left) by $A$ , yielding

\begin{align*} A(c_1 \vec v_1 + c_2 \vec v_2) &= A \vec 0 \\ c_1 A \vec v_1 + c_2 A \vec v_2 &= \vec 0 \\ c_1 \lambda_1 \vec v_1 + c_2 \lambda_2 \vec v_2 &= \vec 0 \end{align*}

Multiply both sides by $\lambda_1$ , yielding

\lambda_1 (c_1 \vec v_1 + c_2 \vec v_2) = \lambda_1 \vec 0 \implies c_1 \lambda_1 \vec v_1 + c_2 \lambda_1 \vec v_2 = \vec 0

Now, we have two similar equations:

\begin{align*} c_1 \lambda_1 \vec v_1 + c_2 \lambda_2 \vec v_2 &= \vec 0 \\ c_1 \lambda_1 \vec v_1 + c_2 \lambda_1 \vec v_2 &= \vec 0 \end{align*}

Subtracting the second equation from the first, we get

c_2 (\lambda_2 - \lambda_1) \vec v_2 = \vec 0

Since $\lambda_1$ and $\lambda_2$ are distinct, $\lambda_2 - \lambda_1 \neq 0$ , so we must have $c_2 = 0$ .

But if $c_2 = 0$ , then $c_1 = 0$ also! We can see this by substituting $c_2 = 0$ into the first equation we started working with.

c_1 \vec v_1 + c_2 \vec v_2 = \vec 0 \implies c_1 \vec v_1 = \vec 0 \implies c_1 = 0

So, the only solution to $c_1 \vec v_1 + c_2 \vec v_2 = \vec 0$ is $c_1 = c_2 = 0$ . This means that $\vec v_1$ and $\vec v_2$ are linearly independent.

Part 2

Now we need to show that if $\vec v_1$ , $\vec v_2$ , and $\vec v_3$ are eigenvectors of $A$ with distinct eigenvalues ( $\lambda_1$ , $\lambda_2$ , and $\lambda_3$ , respectively), then they are linearly independent. We need to show that the only solution to

c_1 \vec v_1 + c_2 \vec v_2 + c_3 \vec v_3 = \vec 0

is $c_1 = c_2 = c_3 = 0$ . The general idea is the same as in Part 1: we need to use the fact that $\lambda_1$ , $\lambda_2$ , and $\lambda_3$ are different.

Instead of multiplying $c_1 \vec v_1 + c_2 \vec v_2 + c_3 \vec v_3 = \vec 0$ by $A$ and then $\lambda_1$ and then subtracting the two equations, here’s an idea: why don’t we multiply by $A - \lambda_1 I$ and see what happens?

Doing so gives

\begin{align*} (A - \lambda_1 I)(c_1 \vec v_1 + c_2 \vec v_2 + c_3 \vec v_3) &= (A - \lambda_1 I) \vec 0 \\ c_1 (A - \lambda_1 I) \vec v_1 + c_2 (A - \lambda_1 I) \vec v_2 + c_3 (A - \lambda_1 I) \vec v_3 &= \vec 0 \\ c_1 (\lambda_1 \vec v_1 - \lambda_1 \vec v_1) + c_2 (\lambda_2 \vec v_2 - \lambda_1 \vec v_2) + c_3 (\lambda_3 \vec v_3 - \lambda_1 \vec v_3) &= \vec 0 \\ c_2 (\lambda_2 - \lambda_1) \vec v_2 + c_3 (\lambda_3 - \lambda_1) \vec v_3 &= \vec 0 \end{align*}

Now, what if we take this last equation and multiply both sides by $A - \lambda_2 I$ ?

\begin{align*} (A - \lambda_2 I)(c_2 (\lambda_2 - \lambda_1) \vec v_2 + c_3 (\lambda_3 - \lambda_1) \vec v_3) &= (A - \lambda_2 I) \vec 0 \\ c_2 (\lambda_2 - \lambda_1)(A - \lambda_2 I) \vec v_2 + c_3 (\lambda_3 - \lambda_1)(A - \lambda_2 I) \vec v_3 &= \vec 0 \\ c_2 (\lambda_2 - \lambda_1)(\lambda_2 \vec v_2 - \lambda_2 \vec v_2) + c_3 (\lambda_3 - \lambda_1)(\lambda_3 \vec v_3 - \lambda_2 \vec v_3) &= \vec 0 \\ c_3 (\lambda_3 - \lambda_1)(\lambda_3 - \lambda_2) \vec v_3 &= \vec 0 \end{align*}

Note that each multiplication by $A - \lambda_i I$ “killed” one of the eigenvectors. What we’re left with is

c_3 (\lambda_3 - \lambda_1)(\lambda_3 - \lambda_2) \vec v_3 = \vec 0

But since $\lambda_1$ , $\lambda_2$ , and $\lambda_3$ are distinct, $\lambda_3 - \lambda_1$ and $\lambda_3 - \lambda_2$ are non-zero, so we must have $c_3 = 0$ .

Substituting $c_3 = 0$ into the equation involving $c_2$ and $c_3$ only gives

c_2 (\lambda_2 - \lambda_1) \vec v_2 = \vec 0 \implies c_2 = 0

And finally, substituting $c_2 = c_3 = 0$ into $c_1 \vec v_1 + c_2 \vec v_2 + c_3 \vec v_3 = \vec 0$ gives

c_1 \vec v_1 + c_2 \vec v_2 + c_3 \vec v_3 = \vec 0 \implies c_1 \vec v_1 = \vec 0 \implies c_1 = 0

So, the only solution to $c_1 \vec v_1 + c_2 \vec v_2 + c_3 \vec v_3 = \vec 0$ is $c_1 = c_2 = c_3 = 0$ . This means that $\vec v_1$ , $\vec v_2$ , and $\vec v_3$ are linearly independent! This argument generalizes to any number of eigenvectors, as long as they all correspond to distinct eigenvalues.

Key Applications¶

Rightfully, you might be asking what the point of this is. There are (at least) two main uses of the eigenvalue decomposition.

Application 1: Matrix Powers¶

The first is that it makes it easy to compute powers of $A$ , which we know is a useful concept in understanding the long-run behavior of a Markov chain.

Suppose $A = V \Lambda V^{-1}$ . Then,

A^2 = (V \Lambda V^{-1})(V \Lambda V^{-1}) = V \Lambda (V^{-1}V) \Lambda V^{-1} = V \Lambda^2 V^{-1}

What does this say about $A^3$ ?

A^3 = (V \Lambda V^{-1})(V \Lambda V^{-1})(V \Lambda V^{-1}) = V \Lambda (V^{-1}V) \Lambda (V^{-1}V) \Lambda V^{-1} = V \Lambda^3 V^{-1}

In general, if $k$ is a positive integer,

\boxed{A^k = V \Lambda^k V^{-1}}

So, to compute $A^k$ , we don’t need to multiply $k$ matrices together (which would be a computational nightmare for large $k$ ). Instead, all we need to do is compute $V \Lambda^k V^{-1}$ . And remember, $\mathcal{\Lambda}$ is a diagonal matrix, so computing $\Lambda^k$ is easy: we just raise the diagonal entries to the power in question.

For example, $A^{10} = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}^{10}$ is

\begin{align*} A^{10} &= V \Lambda^{10} V^{-1} \\ &= \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 3^{10} & 0 \\ 0 & (-1)^{10} \end{bmatrix} \begin{bmatrix} 1/2 & 1/2 \\ 1/2 & -1/2 \end{bmatrix} \\ &= \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} 3^{10}/2 & 3^{10}/2 \\1 /2 & -1/2 \end{bmatrix} \\ &= \begin{bmatrix} (3^{10} + 1)/2 & (3^{10} - 1)/2 \\ (3^{10} - 1)/2 & (3^{10} + 1)/2 \end{bmatrix} \end{align*}

Pretty neat!

np.linalg.matrix_power(A, 10)

array([[29525., 29524.],
       [29524., 29525.]])

Application 2: Understanding Linear Transformations¶

Remember that if $A$ is an $n \times n$ matrix, then $f(\vec x) = A \vec x$ is a linear transformation from $\mathbb{R}^n$ to $\mathbb{R}^n$ . But, if $A = V \Lambda V^{-1}$ , then

f(\vec x) = A \vec x = V \Lambda V^{-1} \vec x

This allows us to understand the effect of $f$ on $\vec x$ in three stages. Remember that $V$ ’s columns are the eigenvectors of $A$ , so the act of multiplying a vector $\vec y$ by $V$ is equivalent to taking a linear combination of $V$ ’s columns ( $A$ ’s eigenvectors) using the weights in $\vec y$ .

V \vec y = \begin{bmatrix} | & | & \cdots & | \\ \vec v_1 & \vec v_2 & \cdots & \vec v_n \\ | & | & \cdots & | \end{bmatrix} \begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_n \end{bmatrix} = y_1 \vec v_1 + y_2 \vec v_2 + \cdots + y_n \vec v_n

For example, $V \begin{bmatrix} 3 \\ -4 \end{bmatrix}$ says take 3 of the first eigenvector, and -4 of the second eigenvector, and add them together to get a new vector. If $V \begin{bmatrix} 3 \\ -4 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$ , this says that taking 3 of the first eigenvector and -4 of the second eigenvector is the same as taking 1 of the first standard basis vector, $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and 2 of the second standard basis vector, $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ .

Intuitively, think of $V$ as a matrix that takes in “amounts of each eigenvector” and outputs “amounts of each standard basis vector”, where the standard basis vectors are the columns of $I$ .

So, if

V: \text{eigenvector amounts} \to \text{standard basis vector amounts}

then $V^{-1}$ does the opposite, and maps

V^{-1}: \text{standard basis vector weights} \to \text{eigenvector amounts}

i.e. multiplying $V^{-1}$ by $\vec x$ expresses $\vec x$ as a linear combination of the eigenvectors of $A$ . If $\vec z$ is the output of $V^{-1} \vec x$ , then $\vec x = V \vec z$ , meaning $\vec z$ contains the amounts of each eigenvector needed to produce $\vec x$ .

(This is non-standard notation, since $V$ is not a function and $\text{eigenvector amounts}$ and $\text{standard basis vector amounts}$ are not sets but concepts, but I hope this helps convey the role of $V$ and $V^{-1}$ in this context.)

Taking a step back, recall

f(\vec x) = A \vec x = V \Lambda V^{-1} \vec x

What this is saying is $f$ does three things:

First, it takes $\vec x$ and expresses it as a linear combination of the eigenvectors of $A$ , which is $V^{-1} \vec x$ . (This contains the “amounts of each eigenvector” needed to produce $\vec x$ .)
Then, it takes that linear combination $V^{-1} \vec x$ and scales each eigenvector by its corresponding eigenvalue, i.e. it scales or stretches each eigenvector by a different amount. Remember that diagonal matrices only scale, they don’t do anything else!
Finally, it takes the resulting scaled vector $\Lambda V^{-1} \vec x$ and expresses it as a linear combination of the standard basis vectors, i.e. it combines the correct amounts of the stretched eigenvectors to get the final result.

\underbrace{\text{express in eigenvector basis}}_{\text{multiply by } V^{-1}} \to \underbrace{\text{scale eigenvectors}}_{\text{multiply by } \Lambda} \to \underbrace{\text{express in standard basis}}_{\text{multiply by } V}

This is better understood visually, as you’ll see in Chapter 9.5: Symmetric Matrices and the Spectral Theorem.

Examples¶

Example: Non-Diagonalizable Matrices¶

Most matrices we’ve seen in Chapter 9.1 and here so far in this section have been diagonalizable. I’ve tried to shield you from the messy world of non-diagonalizable matrices until now, but we’re ready to dive deeper.

Consider $A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ . The characteristic polynomial of $A$ is

p(\lambda) = \text{det}(A - \lambda I) = \text{det} \begin{bmatrix} 1 - \lambda & 1 \\ 0 & 1 - \lambda \end{bmatrix} = (1 - \lambda)^2

which has a double root of $\lambda = 1$ . So, $A$ has a single eigenvalue $\lambda = 1$ . What are all of $A$ ’s eigenvectors? They’re solutions to $A \vec v = 1 \vec v$ , i.e. $(A - I) \vec v = \vec 0$ . Let’s look at the null space of $A - I$ :

A - I = \begin{bmatrix} 1 - 1 & 1 \\ 0 & 1 - 1 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}

The null space of $A - I$ is spanned by the single vector $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ . So, $A$ has a single line of eigenvectors, spanned by $\vec v_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ .

So, does $A$ have an eigenvalue decomposition? No, because we don’t have enough eigenvectors to form $V$ . If we try and form $V$ by using $\vec v_1$ in the first column and a column of zeros in the second column, we’d have

V = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}

but this is not an invertible matrix, so $V$ is not invertible, and we can’t write $A = V \Lambda V^{-1}$ . $A$ is not diagonalizable! This means that it’s harder to interpret $A$ as a linear transformation through the lens of eigenvectors, and it’s harder to understand the long-run behavior of $A^k \vec x$ for an arbitrary $\vec x$ .

Let me emphasize what I mean by $A$ needing to have $n$ linearly independent eigenvectors. Above, we found that $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ is an eigenvector of $A$ , which means that any scalar multiple of $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ is also an eigenvector of $A$ . But, the matrix

\begin{bmatrix} 1 & 2 \\ 0 & 0 \end{bmatrix}

is not invertible, meaning it can’t be the $V$ in $A = V \Lambda V^{-1}$ .

Example: The Identity Matrix¶

The $2 \times 2$ identity matrix $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ has the characteristic polynomial $p(\lambda) = (\lambda - 1)^2$ . So, $I$ has a single eigenvalue $\lambda = 1$ , just like the last example.

But, $\lambda = 1$ corresponds to two different eigenvector directions! I can pick any two linearly independent vectors in $\mathbb{R}^2$ and they’ll both be eigenvectors for $I$ . For example, let $\vec v_1 = \begin{bmatrix} 1 \\ -3 \end{bmatrix}$ and $\vec v_2 = \begin{bmatrix} 11 \\ 98 \end{bmatrix}$ , meaning

V = \begin{bmatrix} 1 & 11 \\ -3 & 98 \end{bmatrix}

The matrix $\mathcal{\Lambda}$ is just $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ which is the same as $I$ itself. This means that as long as $V$ is invertible (i.e. as long as I pick two linearly independent vectors in $\mathbb{R}^2$ ),

V \Lambda V^{-1} = V I V^{-1} = V V^{-1} = I

which means that indeed, $I$ is diagonalizable. (Any matrix that is diagonal to begin with is diagonalizable: in $A = PDP^{-1}$ , $P$ is just the identity matrix.)

If you’d rather look at this example through the lens of solving systems of equations, an eigenvector $\vec v = \begin{bmatrix} a \\ b \end{bmatrix}$ of $I$ with eigenvalue $\lambda = 1$ satisfies

I \vec v = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} a \\ b \end{bmatrix} = 1 \begin{bmatrix} a \\ b \end{bmatrix}

But, as a system this just says $a = a$ and $b = b$ , which is true for any $\vec v$ . Think of $a$ and $b$ both as independent variables; the set of possible $\vec v$ ’s, then, is two dimensional (and is all of $\mathbb{R}^2$ ).

Example: Non-Invertible Matrices¶

Let $A = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}$ .

$A$ is not invertible (column 2 is double column 1), so it has an eigenvalue of $\lambda_1 = 0$ , which corresponds to the eigenvector $\vec v_1 = \begin{bmatrix} -2 \\ 1 \end{bmatrix}$ .
It also has an eigenvalue of $\lambda_2 = 5$ , corresponding to the eigenvector $\vec v_2 = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$ . (Remember, the quick way to spot this eigenvalue is to remember that the sum of the eigenvalues of $A$ is equal to the trace of $A$ , which is $1 + 4 = 5$ ; since 0 is an eigenvalue, the other eigenvalue must be 5.)

Does $A$ have an eigenvalue decomposition? Let’s see if anything goes wrong when we try and construct the eigenvector matrix $V$ and the diagonal matrix $\mathcal{\Lambda}$ and multiplying $V \Lambda V^{-1}$ .

V = \begin{bmatrix} -2 & 1 \\ 1 & 2 \end{bmatrix}, \qquad \Lambda = \begin{bmatrix} 0 & 0 \\ 0 & 5 \end{bmatrix}

V \Lambda V^{-1} = \begin{bmatrix} -2 & 1 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 0 & 0 \\ 0 & 5 \end{bmatrix} \begin{bmatrix} -2/5 & 1/5 \\ 1/5 & 2/5 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}

Everything worked out just fine! $A$ is diagonalizable, even though it’s not invertible.

Algebraic and Geometric Multiplicity¶

As we saw in an earlier example, the matrix $\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ does not have two linearly independent eigenvectors, and so is not diagonalizable. I’d like to dive deeper into identifying when a matrix is diagonalizable.

Activity 2¶

Activity 2

Suppose an $n \times n$ matrix $A$ has the characteristic polynomial

p(\lambda) = (\lambda + 1)^2 \lambda (\lambda - 1)^3 (\lambda - 4)^2 (\lambda - 5) (\lambda - 12)^2

What is $n$ (i.e. the number of rows/columns of $A$ )?
What is the determinant of $A$ ?
What are all of $A$ ’s eigenvalues and their algebraic multiplicities?

Solution

$n$ is equal to the sum of the exponents of the polynomial, which is $2 + 1 + 3 + 2 + 1 + 2 = 11$ . Remember that $p(\lambda)$ is a polynomial of degree $n$ .
The determinant of $A$ is the product of the eigenvalues. Notice that $p(\lambda)$ has a factor of $\lambda$ , i.e. $p(0) = 0$ , meaning 0 is an eigenvalue, so $\text{det}(A) = 0$ .
$A$ has eigenvalues:
- -1 with multiplicity $\text{AM}(-1) = 2$
- 0 with multiplicity $\text{AM}(0) = 1$
- 1 with multiplicity $\text{AM}(1) = 3$
- 4 with multiplicity $\text{AM}(4) = 2$
- 5 with multiplicity $\text{AM}(5) = 1$
- 12 with multiplicity $\text{AM}(12) = 2$
These algebraic multiplicities come from the exponents of the factors in $p(\lambda)$ .

The algebraic multiplicity of an eigenvalue, as the name suggests, is purely a property of the characteristic polynomial. Alone, they don’t tell us whether or not a matrix is diagonalizable. Instead, we’ll need to look at another form of multiplicity alongside the algebraic multiplicity.

Definition: Eigenspace and Geometric Multiplicity

The eigenspace of an eigenvalue $\mathbf{\lambda}_i$ is the set of all eigenvectors with an eigenvalue of $\mathbf{\lambda}_i$ . Equivalently:

It is the set of all vectors $\vec v$ such that $A \vec v = \lambda_i \vec v$ .
It is the subspace $\text{nullsp}(A - \lambda_i I)$ , though $\vec 0$ (which is in the null space of every matrix, including $A - \lambda_i I$ ) is not technically an eigenvector.

The geometric multiplicity of $\mathbf{\lambda}$ is the dimension of the eigenspace of $\mathbf{\lambda}$ .

\text{GM}(\lambda_i) = \text{dim}(\text{nullsp}(A - \lambda_i I))

Equivalently, the geometric multiplicity of $\lambda_i$ is the number of linearly independent eigenvectors corresponding to $\mathbf{\lambda}_i$ .

The geometric multiplicity of an eigenvalue can’t be determined from the characteristic polynomial alone – instead, it involves finding $\text{nullsp}(A - \lambda_i I)$ and finding its dimension, i.e. the number of linearly independent vectors needed to span it. But in general,

1 \leq \text{GM}(\lambda_i) \leq \text{AM}(\lambda_i)

Think of the algebraic multiplicity of an eigenvalue as the “potential” number of linearly independent eigenvectors for an eigenvalue, sort of like the number of slots we have for that eigenvalue. The geometric multiplicity, on the other hand, is the number of linearly independent eigenvectors we actually have for that eigenvalue. As we’ll see in the following examples, when the geometric multiplicity is less than the algebraic multiplicity for any eigenvalue, the matrix in question is not diagonalizable.

In each of the following matrices $A$ , we’ll

Find the algebraic multiplicity of each of $A$ ’s eigenvalues.
For each eigenvalue, find the geometric multiplicity and a basis for the eigenspace.
Conclude whether $A$ is diagonalizable.

As usual, attempt these examples on your own first before peeking at the solutions.

A First Example¶

A = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}

We’ll walk through this one together.

First, we’ll find the characteristic polynomial of $A$ :
$p(\lambda) = \text{det}(A - \lambda I) = \begin{vmatrix} 1 - \lambda & 1 & 0 \\ 0 & 1 - \lambda & 1 \\ 0 & 0 & 1 - \lambda \end{vmatrix} = (1 - \lambda)^3$
(You should practice quickly finding the characteristic polynomial of a $3 \times 3$ matrix; Chapter 9.2 has relevant examples.)
This tells us that $A$ has a single eigenvalue $\lambda = 1$ , with algebraic multiplicity $\boxed{\text{AM}(1) = 3}$ .
The geometric multiplicity of $\lambda = 1$ is $\text{dim}(\text{nullsp}(A - I))$ . Let’s look at $A - I$ :
$A - I = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$
$A-I$ has rank 2, so $\text{dim}(\text{nullsp}(A - I)) = 3 - 2 = 1$ (from the rank-nullity theorem), so the geometric multiplicity of $\lambda = 1$ is $\boxed{\text{GM}(1) = 1}$ .
The vector $\vec v_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$ is an eigenvector for $\lambda = 1$ , and so is any scalar multiple of $\vec v_1$ ; I found this by noticing that the first column of $A-I$ is all zeros. So,
$\boxed{\text{nullsp}(A - I) = \text{span} \left( \left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \right\} \right)}$
Since $\text{GM}(1) = 1 < \text{AM}(1) = 3$ , $A$ is not diagonalizable. $A$ only has one linearly independent eigenvector!

Now, it’s your turn.

Example: Adjacency Matrices¶

Consider this adjacency matrix of a Markov chain with two states. (This is the same adjacency matrix we introduced in Chapter 9.3.)

A = \begin{bmatrix} 0.8 & 0.3 \\ 0.2 & 0.7 \end{bmatrix}

Is $A$ diagonalizable? If so, find $V$ and $\Lambda$ such that $A = V \Lambda V^{-1}$ .

Solution

In Chapter 9.3, we found that $A$ has eigenvalues $\lambda_1 = 1$ and $\lambda_2 = 0.5$ , but we’ll find the characteristic polynomial again for clarity:
$\begin{align*} p(\lambda) &= \det(A - \lambda I) \\ &= \begin{vmatrix} 0.8 - \lambda & 0.3 \\ 0.2 & 0.7 - \lambda \end{vmatrix} \\ &= (0.8 - \lambda)(0.7 - \lambda) - (0.3)(0.2) \\ &= \lambda^2 - 1.5\lambda + 0.5 \\ &= (1 - \lambda)(0.5 - \lambda) \end{align*}$
So, the eigenvalues are $\lambda_1 = 1$ and $\lambda_2 = 0.5$ , each with algebraic multiplicity $\boxed{\text{AM}(1) = \text{AM}(0.5) = 1}$ .
The fact that $A$ has two distinct eigenvalues alone tells us that $A$ is diagonalizable, since each eigenvalue has exactly one independent eigenvector (which really means a line of eigenvectors, or a 1-dimensional eigenspace) and eigenvectors for different eigenvalues are always linearly independent. Let’s find those eigenvectors to be sure.
- For $\lambda_1 = 1$ , we solve $(A - I) \vec v = \vec 0$ :
  $A - I = \begin{bmatrix} -0.2 & 0.3 \\ 0.2 & -0.3 \end{bmatrix}$
  The null space of this matrix is spanned by $\begin{bmatrix} 3 \\ 2 \end{bmatrix}$ , so $\boxed{\text{nullsp}(A - I) = \text{span} \left( \left\{ \begin{bmatrix} 3 \\ 2 \end{bmatrix} \right\} \right)}$ and $\boxed{\text{GM}(1) = 1}$ .
- For $\lambda_2 = 0.5$ , we solve $(A - 0.5I) \vec v = \vec 0$ :
  $A - 0.5I = \begin{bmatrix} 0.3 & 0.3 \\ 0.2 & 0.2 \end{bmatrix}$
  The null space of this matrix is spanned by $\begin{bmatrix} -1 \\ 1 \end{bmatrix}$ , so $\boxed{\text{nullsp}(A - 0.5I) = \text{span} \left( \left\{ \begin{bmatrix} -1 \\ 1 \end{bmatrix} \right\} \right)}$ and $\boxed{\text{GM}(0.5) = 1}$ .
Since $A$ has two linearly independent eigenvectors, it is diagonalizable:
$A = V \Lambda V^{-1} = \begin{bmatrix} 3 & -1 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 0.5 \end{bmatrix} \left( \begin{bmatrix} 3 & -1 \\ 2 & 1 \end{bmatrix} \right)^{-1}$

Example: Another Diagonalizable Matrix¶

A = \begin{bmatrix} 2 & 1 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}

Is $A$ diagonalizable? If so, find $V$ and $\Lambda$ such that $A = V \Lambda V^{-1}$ .

Solution

The characteristic polynomial of $A$ is
$\begin{align*} p(\lambda) &= \det(A - \lambda I) \\ &= \begin{vmatrix} 2 - \lambda & 1 & 0 \\ 1 & 2 - \lambda & 0 \\ 0 & 0 & 3 - \lambda \end{vmatrix} \\ &= (2 - \lambda)\left((2 - \lambda)(3 - \lambda) - 0 \cdot 0 \right) - 1 \left( (1)(3 - \lambda) - 0 \cdot 0 \right) + 0 \\ &= (2 - \lambda)(2 - \lambda)(3 - \lambda) - (3 - \lambda) \\ &= (3 - \lambda)\left( (2 - \lambda)^2 - 1 \right) \\ &= (3 - \lambda)(\lambda^2 - 4\lambda + 3) \\ &= (3 - \lambda)^2(1 - \lambda) \end{align*}$
So, $A$ has eigenvalues $\lambda_1 = 3$ with $\boxed{\text{AM}(3) = 2}$ and $\lambda_2 = 1$ with $\boxed{\text{AM}(1) = 1}$ . There’s a quicker way we could have spotted $A$ ’s eigenvalues, discussed after the solutions box.
- For $\lambda_1 = 3$ , we solve $(A - 3I) \vec v = \vec 0$ :
  $A - 3I = \begin{bmatrix} -1 & 1 & 0 \\ 1 & -1 & 0 \\ 0 & 0 & 0 \end{bmatrix}$
  $\text{rank}(A - 3I) = 1$ (since all columns are multiples of $\begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix}$ ), so $\text{dim}(\text{nullsp}(A - 3I)) = 3 - 1 = 2$ , which means $\boxed{\text{GM}(3) = 2}$ . At this point, we can conclude that $A$ is diagonalizable! But, let’s find a basis for the eigenspace $\text{nullsp}(A - 3I)$ to be thorough.
  Two linearly independent vectors in $\text{nullsp}(A - 3I)$ are $\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$ , so $\boxed{\text{nullsp}(A - 3I) = \text{span} \left( \left\{ \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \right\} \right)}$ . This is just one possible basis for the eigenspace; it’s not the only one.
  What this says is any linear combination of $\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$ is also an eigenvector for $\lambda = 3$ . For example, $\begin{bmatrix} 2 & 1 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix} \begin{bmatrix} -1 \\ -1 \\ 13 \end{bmatrix} = 3 \begin{bmatrix} -1 \\ -1 \\ 13 \end{bmatrix}$ .
- For $\lambda_2 = 1$ , we solve $(A - I) \vec v = \vec 0$ :
  $A - I = \begin{bmatrix} 1 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix}$
  Since $\text{AM}(1) = 1$ , we know that there could only be one linearly independent eigenvector for $\lambda_2 = 1$ , so $\boxed{\text{GM}(1) = 1}$ too. One such eigenvector is $\begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}$ , so $\boxed{\text{nullsp}(A - I) = \text{span} \left( \left\{ \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} \right\} \right)}$ .
Since $A$ has three linearly independent eigenvectors, it is diagonalizable:
$A = V \Lambda V^{-1} = \begin{bmatrix} 1 & 0 & 1 \\ 1 & 0 & -1 \\ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 3 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix} \left( \begin{bmatrix} 1 & 0 & 1 \\ 1 & 0 & -1 \\ 0 & 1 & 0 \end{bmatrix} \right)^{-1}$

In the solutions above, I found $A$ ’s eigenvalues the traditional way, by solving for the roots of $A$ ’s characteristic polynomial. But there’s a quicker way to find $A$ ’s eigenvalues in this case that I want to highlight. $A$ is a block diagonal matrix:

A = \begin{bmatrix} {\color{3d81f6} 2} & {\color{3d81f6} 1} & 0 \\ {\color{3d81f6} 1} & {\color{3d81f6} 2} & 0 \\ 0 & 0 & {\color{orange} 3} \end{bmatrix} = \begin{bmatrix} {\color{3d81f6} A_1} & \vec 0_{2 \times 1} \\ \vec 0_{1 \times 2} & {\color{orange} A_2} \end{bmatrix}

In this case, $A_1 = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$ and $A_2 = \begin{bmatrix} 3 \end{bmatrix}$ . Think of this as us placing $A_1$ and $A_2$ on the diagonal of $A$ , and then filling the rest of $A$ with zeros.

In general, if $A$ is a block diagonal matrix, then the eigenvalues of $A$ are the eigenvalues of the blocks on the diagonal combined. To see why this is true, suppose you know that $\lambda$ is an eigenvalue of one of $A$ ’s diagonal blocks with a corresponding eigenvector of $\vec v$ . To get a corresponding eigenvector of $A$ , just extend $\vec v$ by placing 0’s in the positions of the other blocks. In other words, if $\vec v$ is an eigenvector of $A_1$ , then $\begin{bmatrix} \vec v \\ 0 \\ \vdots \\ 0 \end{bmatrix}$ is an eigenvector of $A$ .

Here, the eigenvalues of $A_1$ are 1 and 3, and the sole eigenvalue of $A_2$ is 3, so $A$ ’s eigenvalues are 1 and 3 (with algebraic multiplicity 2).

Example: Another Non-Diagonalizable Matrix¶

A = \begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}

Is $A$ diagonalizable? If so, find $V$ and $\Lambda$ such that $A = V \Lambda V^{-1}$ . Tip: Use the fact that $A$ is upper triangular to find its eigenvalues quickly. We first discussed this trick in Chapter 9.2.

Solution

$A$ is upper triangular, so its eigenvalues are the entries on the diagonal: $\lambda_1 = 2$ , $\lambda_2 = 2$ , $\lambda_3 = 3$ . So, $\boxed{\text{AM}(2) = 2}$ and $\boxed{\text{AM}(3) = 1}$ .
- For $\lambda_1 = 2$ , we solve $(A - 2I) \vec v = \vec 0$ :
$A - 2I = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$\text{rank}(A-2I) = 2$ , so $\text{dim}(\text{nullsp}(A-2I)) = 1$ , so $\boxed{\text{GM}(2) = 1}$ . At this point, we can conclude that $A$ is not diagonalizable, since $\text{GM}(2) < \text{AM}(2)$ . Noticing that the first column of $A - 2I$ is all zeros, $\vec v_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$ is an eigenvector, and
$\boxed{\text{nullsp}(A-2I) = \text{span} \left( \left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \right\} \right)}$
- For $\lambda_3 = 3$ , we solve $(A - 3I) \vec v = \vec 0$ :
$A - 3I = \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{bmatrix}$
Using similar logic, $\boxed{\text{GM}(3) = 1}$ and
$\boxed{\text{nullsp}(A-3I) = \text{span} \left( \left\{ \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \right\} \right)}$
$A$ only has two linearly independent eigenvectors, one for $\lambda = 2$ and one for $\lambda = 3$ , but is a $3 \times 3$ matrix, and hence its not diagonalizable. (Also, for $\lambda_1 = 2$ , the geometric multiplicity, 1, is less than the algebraic multiplicity, 2, but we need equality for all eigenvalues in order for $A$ to be diagonalizable.)

Example: Working Backwards¶

In the previous examples, we were given a matrix $A$ and then found its eigenvalues and eigenvectors. Let’s go in the other direction.

Suppose $A$ is a $3 \times 3$ matrix with:

eigenvalue $\lambda_1 = 3$ with eigenvector $\vec v_1 = \begin{bmatrix} 2 \\ 0 \\ 1 \end{bmatrix}$
eigenvalue $\lambda_2 = -2$ with eigenspace
$\text{span}\left(\left\{\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}\right\}\right)$

Find $A$ .

Solution

If we’re given eigenvalues and enough linearly independent eigenvectors, we can build $A$ directly from the decomposition $A = V \Lambda V^{-1}$ .

Since the eigenspace for $\lambda = -2$ is 2-dimensional, we can use $\begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}$ as two linearly independent eigenvectors for $\lambda = -2$ . So, one valid choice is

V = \begin{bmatrix} 2 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 2 \end{bmatrix}, \quad \Lambda = \begin{bmatrix} 3 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -2 \end{bmatrix}

where the columns of $V$ are the eigenvectors, and the diagonal entries of $\Lambda$ are the corresponding eigenvalues in the same order.

Now we compute

A = V \Lambda V^{-1} = \begin{bmatrix} 2 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 2 \end{bmatrix} \begin{bmatrix} 3 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -2 \end{bmatrix} \left(\begin{bmatrix} 2 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 2 \end{bmatrix}\right)^{-1} = \boxed{\begin{bmatrix} 2 & -4 & 2 \\ 0 & -2 & 0 \\ 2 & -2 & -1 \end{bmatrix}}

You can quickly verify that this works:

$A \begin{bmatrix} 2 \\ 0 \\ 1 \end{bmatrix} = 3 \begin{bmatrix} 2 \\ 0 \\ 1 \end{bmatrix}$
$A \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} = -2 \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}$
$A \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix} = -2 \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}$

The main takeaway is that if you know a full eigenbasis and the matching eigenvalues, you can reconstruct the original matrix by putting the eigenvectors into $V$ , the eigenvalues into $\Lambda$ , and multiplying $V \Lambda V^{-1}$ .

In Chapter 9.5, I want to build on our understanding of diagonalizability by focusing on a special class of matrices: symmetric matrices. It turns out that symmetric matrices are always diagonalizable, and beyond that, their corresponding eigenvectors obey a really nice property.

Activity 3¶

Activity 3

For each statement, determine whether it is true or false and provide a brief justification. (For false statements, provide a counterexample.)

True or False: If all of $A$ ’s eigenvalues are distinct, then $A$ is diagonalizable.
True or False: If all of $A$ ’s eigenvalues are positive, then $A$ is diagonalizable.
True or False: If $A$ is diagonalizable, then $A$ is invertible.
True or False: If $A$ is invertible, then $A$ is diagonalizable.

Solution

If all of $A$ ’s eigenvalues are distinct, then $A$ is diagonalizable. True. Remember that the minimum possible geometric multiplicity is 1, so if all eigenvalues are distinct, they each have an algebraic and geometric multiplicity of 1. Eigenvectors for different eigenvalues are always linearly independent; think of it this way, an eigenvector can only ever be paired with a single eigenvalue, so eigenvectors for different eigenvalues can’t be linearly dependent. Pairing these two facts, $A$ must have $n$ linearly independent eigenvectors, and so is diagonalizable.
If all of $A$ ’s eigenvalues are positive, then $A$ is diagonalizable. False. In the example above titled “Another Non-Diagonalizable Matrix”, $A$ had eigenvalues of 2, 2, and 3, all of which are positive, but $A$ was not diagonalizable.
If $A$ is diagonalizable, then $A$ is invertible. False. We saw an example of a diagonalizable matrix that was not invertible earlier in this section:
$\begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} = \underbrace{\begin{bmatrix} 2 & 1 \\ -1 & 2 \end{bmatrix}}_V \underbrace{\begin{bmatrix} 0 & 0 \\ 0 & 5 \end{bmatrix}}_{\Lambda} \underbrace{\left( \begin{bmatrix} 2 & 1 \\ -1 & 2 \end{bmatrix} \right)^{-1}}_{V^{-1}}$
If $A$ is invertible, then $A$ is diagonalizable. False. The matrix from the example “Another Non-Diagonalizable Matrix” above had eigenvalues of 2, 2, and 3, none of which are 0, meaning $A$ is invertible, but $A$ was not diagonalizable.