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4.H Diagonalizable matrices

A diagonal matrix is the simplest kind of matrix. Here are some facts that justify such a strong statement: For diagonal matrices, (1) the eigenvalues are the entries along the diagonal, (2) the standard basis vectors are eigenvectors, (3) the determinant is the product of diagonal entries, (4) the rank is the number of non-zero entries on the diagonal, and (5) all diagonal matrices commute with each other (that is, $AB=BA$ , if $A$ and $B$ are diagonal). In some statistical applications, diagonal matrices correspond to uncorrelated variables, which is the easiest situation to study. A diagonal adjacency matrix in graph theory corresponds to a completely disconnected graph.

If $T:V\to V$ is a linear transformation, it is desirable to choose a basis $\mathcal{B}$ of $V$ for which the matrix ${}_{\mathcal{B}}[T]_{\mathcal{B}}$ is diagonal. This is not always possible (hence the reason for most of Chapter 6), but when is it possible, we will call $T$ diagonalizable.

Theorem 4.52 shows how the matrix of a linear transformation changes when we change the basis. We would like to think of the resulting matrices as being closely related to each other in some way; that is the purpose of the following terminology.

Definition 4.53:

Let $A,B\in\operatorname{M}_{{n}}({F})$ . We say that $A$ is similar to $B$ if there exists an invertible matrix $P\in\operatorname{M}_{{n}}({F})$ such that $A=P^{-1}BP.$

A matrix $B$ is called diagonalizable if there exists a $P$ such that $P^{-1}BP$ is a diagonal matrix.

[Aside: In fact, any invertible matrix can be thought of as a change of basis matrix for an appropriate choice of bases; so if two matrices are similar to each other, then they can always be visualized as representing the same linear transformation, but with a different choice of basis.]

An important special case of Theorem 4.52 is when one of the bases consists of eigenvectors, as has been the case in several of the examples we have already seen. We summarize this case as follows, and omit the proof (compare with Theorem 4.11):

Theorem 4.54.

Let $A\in\operatorname{M}_{{n}}({F})$ be a square matrix. Let $\mathcal{C}=(\vec{e_{1}},\cdots,\vec{e_{n}})$ be the standard basis of $F^{n}$ .

i.

If $\mathcal{B}$ is a basis of eigenvectors of $A$ , and $P:={{}_{\mathcal{C}}[}\operatorname{Id}\nolimits]_{\mathcal{B}}$ , then $P^{-1}AP$ is diagonal.
ii.

If $P^{-1}AP$ is diagonal, then $P\vec{e_{1}},\cdots,P\vec{e_{n}}$ is a basis of eigenvectors of $A$ .

Note that $P\vec{e_{i}}$ are the column vectors of the matrix $P$ . The matrix $P^{-1}AP$ in part (i) is called a diagonalization of $A$ .

Example 4.55.

Let $A=\begin{bmatrix}2&3&-3\\ 4&1&-1\\ 4&2&-2\end{bmatrix}\in\operatorname{M}_{{3}}({{\mathbb{R}}}).$ Find a basis $\mathcal{B}$ of eigenvectors of $A$ . Verify Theorem 4.54(i) in this case.

Solution: Firstly, one can check that the eigenvalues of $A$ are $\lambda=0,-1,$ and 2, and the eigenspaces are as follows:

$V_{0}=\operatorname{span}\{\begin{bmatrix}0\\ 1\\ 1\end{bmatrix}\}$
$V_{-1}=\operatorname{span}\{\begin{bmatrix}1\\ -3\\ -2\end{bmatrix}\}$
$V_{2}=\operatorname{span}\{\begin{bmatrix}1\\ 2\\ 2\end{bmatrix}\}$

Therefore, we can define a basis consisting of eigenvectors:

\mathcal{B}:=((0,1,1),(1,-3,-2),(1,2,2)).

To verify Theorem 4.54(i), first

P={{}_{\mathcal{C}}[}\operatorname{Id}\nolimits]_{\mathcal{B}}=\begin{bmatrix}% 0&1&1\\ 1&-3&2\\ 1&-2&2\end{bmatrix},

, then compute

P^{-1}=\begin{bmatrix}-2&-4&5\\ 0&-1&1\\ 1&1&-1\end{bmatrix}.

Then verify the matrix product is a diagonal matrix:

P^{-1}AP=\begin{bmatrix}0&0&0\\ 0&-1&0\\ 0&0&2\end{bmatrix}.

Example 4.56.

i.

Let $A=\begin{bmatrix}0&-1\\ 1&0\end{bmatrix}\in\operatorname{M}_{{2}}({{\mathbb{C}}})$ , and prove that $\mathcal{B}=((i,1),(-i,1))$ is a basis of eigenvectors, and hence find a diagonalization of $A$ .

Solution: One checks that these basis vectors are eigenvectors as follows:
1. $\begin{bmatrix}0&-1\\ 1&0\end{bmatrix}\begin{bmatrix}i\\ 1\end{bmatrix}=i\begin{bmatrix}i\\ 1\end{bmatrix}$
2. $\begin{bmatrix}0&-1\\ 1&0\end{bmatrix}\begin{bmatrix}-i\\ 1\end{bmatrix}=-i\begin{bmatrix}-i\\ 1\end{bmatrix}$
So let $\mathcal{C}$ be the standard basis, and then we have
1. $P:={{}_{\mathcal{C}}[}\operatorname{Id}\nolimits]_{\mathcal{B}}=\begin{bmatrix% }i&-i\\ 1&1\end{bmatrix}$
2. $P^{-1}=\frac{1}{2i}\begin{bmatrix}1&i\\ -1&i\end{bmatrix}$
By Theorem 4.54, a diagonalization is

$P^{-1}AP=\frac{1}{2i}\begin{bmatrix}1&i\\ -1&i\end{bmatrix}\begin{bmatrix}0&-1\\ 1&0\end{bmatrix}\begin{bmatrix}i&-i\\ 1&1\end{bmatrix}=\frac{1}{2i}\begin{bmatrix}-2&0\\ 0&2\end{bmatrix}=\begin{bmatrix}i&0\\ 0&-i\end{bmatrix}.$
ii.

Let $A=\begin{bmatrix}0&-1\\ 1&0\end{bmatrix}\in\operatorname{M}_{{2}}({{\mathbb{R}}})$ . Prove $A$ is not diagonalizable ( $F={\mathbb{R}}$ ).

Solution: The matrix $A$ has no real eigenvalues, and therefore it has no eigenvectors in ${\mathbb{R}}^{2}$ . So by Theorem 4.54(ii), $P^{-1}AP$ can never be diagonal, and therefore $A$ is not diagonalizable (when $F={\mathbb{R}}$ ).

Exercise 4.57:

Using the basis $\mathcal{B}=((3,-1),(-2,1))$ , and $P$ from Exercise 4.51,

i.

Verify that $\mathcal{B}$ consists of eigenvectors of $A:=\begin{bmatrix}-5&-18\\ 3&10\end{bmatrix}$ .
ii.

Verify, by matrix multiplication, that $D:=P^{-1}AP$ is a diagonal matrix.
iii.

Verify, by matrix multiplication, that $A=PDP^{-1}$ .

[End of Exercise]

Exercise 4.58:

Prove that similar matrices always have the same eigenvalues.

[End of Exercise]