Section 4.5 Similarity and Diagonalization

Definition: If $A$ and $B$ are $n\times n$ matrices, then $A$ is similar to $B$ if there is an invertible matrix $P$ such that $P^{-1}AP=B$ or $A=PB{P}^{-1}$. We write $A\sim B$.

Remark: We can write $Q=P^{-1}$ because $P$ is invertible then we have $QA{Q}^{-1}=B$, i.e. $Q^{-1}BQ=A$ and $B$ is similar to $A$. Changing $A$ into $P^{-1}AP=B$ is called a similarity transformation.

 

Example1: If $A$ is similar to $B$ and either $A$ or $B$ is diagonalizable, show that the other is also diagonalizable.

 

 

Exercise 1: Show that if $A$ is similar to $B$ then $\text{det}A=\text{det}B$.

 

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

Remark: 1. Unfortunately, the converse of the theorem is not true. For example $A=\left[\begin{array}{cc}1& 2\\ 0& 3\end{array}\right]$ and $B=\left[\begin{array}{cc}1& 0\\ 0& 3\end{array}\right]$ have the same characteristic polynomial but they are not similar.

2. Two matrices that are row equivalent do not mean they are similar to each other. For example $B=EA$ where $E$ is just elementary matrix, and it does not mean $A$ is similar to $B$. Therefore, we cannot use row reduction to get the eigenvalues.

 

Theorem: If $A$ and $B$ are similar $n\times n$ matrices, then $A$ and $B$ have the same determinant, rank, characteristic polynomial, and eigenvalues.

 

Remark: 1. A square matrix $A$ is diagonalizable then there exists an invertible matrix $P$ such that $P^{-1}AP=D$ is a diagonal matrix, that is $A$ is similar to a diagonal matrix $D$.

2.The set of all solutions of $(A-\lambda I)\overrightarrow{x}=0$ is just the null space of the matrix $A-\lambda I$. So this set is a subspace of ${\mathbb{R}}^n$ and is called the eigenspace of $A$ corresponding to $\lambda$. The eigenspace consists of the zero vector and all the eigenvectors corresponding to $\lambda$.

3. If $\overrightarrow{x}$ is in the eigenspace of some eigenvalue $\lambda$, then $A\overrightarrow{x}$ is just a scalar of $\overrightarrow{x}$. This is saying the transformation defined by $A$ transforms the eigenspace into scalar multiple of itself.

 

Theorem: If $\overrightarrow{v_1},...,\overrightarrow{v_p}$ are eigenvectors that correspond to distinct eigenvalues ${\lambda }_1,...,{\lambda }_p$ of an matrix $A$, then the set $\{\overrightarrow{v_1},...,\overrightarrow{v_p}\}$ is linearly independent.

 

 

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

 

Remark: $A$ is diagonalizable if and only if there are enough eigenvectors to form a basis of ${\mathbb{R}}^n$. We call such a basis an eigenvector basis of ${\mathbb{R}}^n$.

 

 

Theorem: Let $A$ be a matrix whose distinct eigenvalues are ${\lambda }_1,....,{\lambda }_p$.

a. For $1\le k\le p$, 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 eigenspaces 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 $B_k$ is a basis for the eigenspace corresponding to ${\lambda }_k$
for each $k$, then the total collection of vectors in the sets $B_1,...,B_p$ forms an eigenvector basis for ${\mathbb{R}}^n$.

 

 

Example 2: Diagonalize the following matrix, if possible.

$A=\left[\begin{array}{cccc}1& 1& 4& 0\\ 0& 1& 1& 6\\ 0& 0& -2& 1\\ 0& 0& 0& -2\end{array}\right]$

 

 

Exercise 2: Diagonalize the following matrix, if possible.

$A=\left[\begin{array}{cccc}3& 0& 0& 0\\ 1& -1& 0& 0\\ 0& 1& 2& 0\\ 1& 0& 2& -1\end{array}\right]$

 

Example 3: Diagonalize the following matrix, if possible.

$A=\left[\begin{array}{ccc}3& 0& 6\\ 0& -3& 0\\ 5& 0& 2\end{array}\right]$

 

 

Exercise 3: Diagonalize the following matrix, if possible.

$A=\left[\begin{array}{ccc}4& 0& 0\\ 0& 2& 2\\ 0& 3& 1\end{array}\right]$

 

GroupWork 1: True or False. All matrices are $n\times n$ matrices.

a. $A$ is diagonalizable with eigenvalues ${\lambda }_1,...,{\lambda }_n$ then $\text{det}A={\lambda }_1\cdots {\lambda }_n$.

b. If ${\mathbb{R}}^n$ has a basis of eigenvectors of $A$, then $A$ is diagonalizable.

c. $A$ is diagonalizable if and only if $A$ has $n$ eigenvalues, counting multiplicities.

d. If $A$ is invertible, then $A$ is diagonalizable.

e. An eigenspace of $A$ is a null space of a certain matrix.

 

GroupWork 2: $A$ is a $6\times 6$ matrix with two eigenvalues. One eigenspace is 4-dimensional, and the other eigenspace is 2-dimensional. Is $A$ diagonalizable? Why?

 

GroupWork 3: If $A$ is an $n\times n$ matrix with $n$ distinct eigenvalues, show $A$ is diagonalizable.

 

GroupWork 4: Show that if $A$ is diagonalizable then $A$ is similar to $A^T$.

 

GroupWork 5: True or False. All matrices are $n\times n$ matrices. Justify each answer.

a. $A$ is diagonalizable if $A$ has $n$ eigenvectors.

b. $A$ is diagonalizable if $A$ has $n$ distinct eigenvectors.

c. $A\sim B$, then $\text{det}(A-xI)=\text{det}(B-xI)$.

d. $A$ is diagonalizable, then $A$ is invertible.

 

 

GroupWork 6: Let $A$ be an $3\times 3$ matrix with 2 eigenvalues. Each eigenspace is one-dimensional. Is $A$ diagonalizable?
Why?

 

GroupWork7: $A$ is a $5\times 5$ matrix with $3$ eigenvalues. Two of the eigenspaces are 2-dimensional. Is it possible that $A$ is not diagonalizable?