## Chapter 5

### Notes

Diagonalizable Matrices

Consider the diagonalizable matrix $A$

(1)Linear Transformation Matrix

The matrix for $T$ relative to the bases $\mathcal{B}$ and $\mathcal{C}$ is called $M$

(2)### Definitions

Eigen Value

A scalar $\lambda$ such that there exists a nonzero vector of the form $A\vec{x}=\lambda \vec{x}$.

Eigen Vector

A nonzero vector $\vec{x}$ such that $A\vec{x}=\lambda \vec{x}$.

Eigen Space

The eigen space of $A$ corresponding to the eigen value $\lambda$ is the null space of $(A-\lambda I)$.

Similar Matrices

Matrix $A$ is similar to matrix $B$ if there exists an invertible matrix $P$, such that $A=PBP^{-1}$.

Note: A matrix $A$ that row reduces to matrix $B$ does not imply that the two matrices are similar.

Diagonalizable

An $n \times n$ matrix $A$ is diagonalizable if $A$ is similar to some diagonal matrix such that

$\ \ \ \ A=PDP^{-1}$

where $P$ is an invertible matrix and $D$ is a diagonal matrix.

### Theorems

Theorem 1

The eigenvalues of a triangular matrix are the entries on its main diagonal.

Theorem 2

If $\vec{v}_1,...,\vec{v}_r$ are eigenvectors that correspond to distinct eigenvalues $\lambda_1,...,\lambda_r$ of an $n \times n$ matrix $A$, then the set $\{ \vec{v}_1,...,\vec{v}_r \}$ is linearly independent.

The Invertible Matrix Theorem (continued)

Let A be an nxn matrix. Then A is invertible if and only if:

s. The number 0 is not an eigenvalue of A. ($\lambda \neq 0$)

t. The determinant of A is not zero.

Theorem 3

Let A and B be nxn matricies

a. A is invertible if and only if $\det(A) \neq 0$

b. $\det(AB) = \det(A)\det(B)$

c. $\det(A)^T = \det(A)$

d. If A is triangular, then $\det(A)$ is the product of the the entries on the main diagonal of A.

e. A row replacement operation on A does not change the determinant. A row interchange changes the sign of the determinant. A row scaling also scales the determinant by the same factor.

Theorem 4

If $n \times n$ matrices A and B are similar, then they have the same characteristic polynomial and hence the same eigenvalues (with the same multiplicities).

Theorem 5

An $n \times n$ matrix $A$ is diagonalizable if and only if $A$ has $n$ linearly independent eigenvectors.

Furthermore, for a diagonalizable matrix $D$,

$\ \ \ \ A=PDP^{-1}$

if and only if the columns of $P$ are $n$ linearly independent eigenvectors of $A$. The diagonal entries of $D$ are then eigenvalues of $A$ that correspond, respectively, to the eigenvectors of $P$.

Theorem 6

An nxn matrix with n distinct eigenvalues is diagonalizable.

Theorem 7

Let A be an nxn matrix whose distinct eigenvalues are $\lambda_1, \dots, \lambda_{p}$.

a) For $1\leq K \leq p$, the dimension of the eigenspace for $\lambda_k$ is less than or equal to the multiplicity of the eigenvalue $\lambda_k$.

b) The matrix $A$ is diagonalizable if and only if the sum of the dimensions of the eigenspace equals n, and this happens if and only if (i) the characteristic polynomial factors completely into linear factors and (ii) the dimension of the eigenspace for each $\lambda_k$ equals the multiplicity of $\lambda_k$.

c) If $A$ is diagonalizable and $\beta_k$ is a bsis for the eigenspace corresponding to $\lambda_k$ for each k, then the total collection of vectors in the sets $\beta_1,\dots,\beta_p$ forms an eigenvector basis for $\mathbb{R}^{n}$.

Theorem 8

Suppose $A=PDP^{-1}$, where D is a diagonal nxn matrix. If $B$ is the basis for $R^{n}$ formed from the columns of P, then D is the $B$-matrix for the transformation x ->$Ax$

Theorem 9

Let $A$ be a real $2x2$ matrix with a complex eigenvalue $\lambda = a - bi$ where $(b \neq 0)$ and an associated eigenvector $v$ in $C^2$. Then $A=PDP^{-1}$, where $P = [Re (v) Im (v)]$ and $C = \begin{bmatrix}a&-b\\b&a \end{bmatrix}$

#### Homework Problems

**5.1**

25. Let $\lambda$ be an eigen value of an invertible matrix $A$. Show $\lambda^{-2}$ is an eigenvalue of $A^{-1}$.

We are given that $A\vec{x}=\lambda\vec{x}$. Notice the following manipulations, by the question, may hold,

$A\vec{x}\frac{1}{\lambda}=\vec{x}$

$\vec{x}\frac{1}{\lambda}=A^{-1}\vec{x}$

$A^{-1}\vec{x}=\frac{1}{\lambda}\vec{x}$

Clearly, this implies that $\frac{1}{\lambda}= \lambda^{-1}$ is an eigenvalue of $A^{-1}$, as the equation immediately above is what it means for it to be true.

**5.2**

5. To find the characteristic equation of $A$, do the following:

$\begin{bmatrix}8&4\\4&8 \end{bmatrix} - \begin{bmatrix}\lambda&0\\0&\lambda \end{bmatrix} = \begin{bmatrix}8-\lambda&4\\4&8-\lambda \end{bmatrix}$

$det(\begin{bmatrix}8-\lambda&4\\4&8-\lambda \end{bmatrix}) = \lambda^2 - 16\lambda + 58$

This gives us the characteristic equation. Factoring this reveals the the eigenvalues -

$(\lambda - 4)(\lambda - 12)$, which implies $\lambda = 4,12$

15. List the real eigenvalues, repeated according to their multiplicities, for matrix $\begin{bmatrix}5&5&0&2\\0&2&-3&6\\0&0&3&-2\\0&0&0&5\end{bmatrix}$.

Since the matrix is upper triangular, we know that the characteristic polynomial is the product of each diagonal entry minus lambda, so we have $(5-\lambda)(2-\lambda)(3-\lambda)(5-\lambda)$ which means our eigenvalues are $2,3,5,5$ (5 has multiplicity 2).

19. Let A be an $n \times n$ matrix, and suppose A has n real eigenvalues, $\lambda_1,...\lambda_n$, repeated according to multiplicities, so that

det (A-$\lambda$I)=($\lambda_1$-$\lambda$)($\lambda_2$-$\lambda$)…($\lambda_n$-$\lambda$)

Explain why det A is the product of the $n$ eigenvalues of A.

Solution:

Since the above equation holds for any $\lambda$ E R we can set $\lambda=0$

det (A-0I)=($\lambda_1$-0)($\lambda_2$-0)…($\lambda_n$-0)=$\lambda_1, \lambda_2,...\lambda_n$

**5.3**

27. Prove if $A$ is diagonalizable and invertible, so is $A^{-1}$.

$A$ is similar, so $A=PDP^{-1}$, where $D$ is diagonal. Notice that by construction, the vectors of $D$ are linearly independent, so the matrix is invertible. Consider $D^{-1}$, constructed by taking $D$ and flipping the non-zero elements over one in place in the matrix (so $x$ becomes $\frac{1}{x}$ and does not change position in the matrix). A simple check will show $D$ and $D^{-1}$ are inverses, and we have unofficially argued that $D^{-1}$ is also diagonal. And so, we can do the following manipulations:

$A=PDP^{-1}$

$I=A^{-1}PDP^{-1}$

$PD^{-1}P^{-1}=A^{-1}$

And so, by the explanation above, we have shown $A^{-1}$ is similar. There is no need to show it is invertible, as we already know that it exists ($A$).

**5.5**

5. Find the eigenvalues and a basis for $\begin{bmatrix}3&-2\\1&5 \end{bmatrix}$.

$A-\lambda I=\begin{bmatrix}3-\lambda &-2\\1&5-\lambda \end{bmatrix}$

$det(A-\lambda I)=\lambda ^2-8\lambda +15+2=\lambda ^2-8\lambda +17$

$\lambda=\frac{8\pm \sqrt{64-68}}{2}=4\pm i$

$\lambda =4+i: \begin{bmatrix}-1-i&-2\\1&1-i \end{bmatrix}=>x_2=2; x_1=1-i=>\vec{v_1}=\begin{bmatrix}1-i\\2 \end{bmatrix}$

$\lambda =4-i: \vec{v_2}=\begin{bmatrix}1+i\\2 \end{bmatrix}$

$basis=\{ \begin{bmatrix}1-i\\2 \end{bmatrix},\begin{bmatrix}1+i\\2 \end{bmatrix} \}$

9. Find associated $r$ and angle $\Phi$ for the matrix $A= \begin{bmatrix}0&2\\-2&0 \end{bmatrix}$.

Due to the format of the $A$, Theorem 9 lets us say that are eigenvalues are $\lambda = 0 \pm 2i$. So this implies that $r=\sqrt{0^2+2^2 } = 2$.

So then, using this $r$, $A= 2\begin{bmatrix}0&1\\-1&0 \end{bmatrix}$. To find the angle of rotation, we compare the elements in this matrix to the rotation matrix, $\begin{bmatrix}\cos(\Phi)&-\sin(\Phi)\\ \sin(\Phi)& \cos(\Phi) \end{bmatrix}$.

This implies $\cos(\Phi)=0$ and $\sin(\Phi)=-1$. This is true if $\Phi = -\frac{\pi}{2}$.