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## Similar and Diagonalizable Matrices

Definition. Let $$A$$ and $$B$$ be $$n\times n$$ matrices. $$A$$ is similar to $$B$$ if $$A=PBP^{-1}$$ for some invertible matrix $$P$$.

Remark. If $$A$$ is similar to $$B$$, then $$B$$ is similar to $$A$$ because $$B=P^{-1}A(P^{-1})^{-1}$$. So we simply say $$A$$ and $$B$$ are similar.

Example. Consider $$A=\left[\begin{array}{rr}6&-1\\2&3\end{array} \right],\; B=\left[\begin{array}{rr}5&0\\0&4\end{array} \right],\; C=\left[\begin{array}{rr}5&4\\0&0\end{array} \right]$$.
$$A$$ and $$B$$ are similar because $$A=PBP^{-1}$$ where $$P=\left[\begin{array}{rr}1&1\\1&2\end{array} \right]$$ and $$P^{-1}=\left[\begin{array}{rr}2&-1\\-1&1\end{array} \right]$$. It can be verified that there is no invertible matrix $$P$$ such that $$A=PCP^{-1}$$. So $$A$$ and $$C$$ are not similar.

Theorem. If $$n\times n$$ matrices $$A$$ and $$B$$ are similar, then they have the same characteristic polynomial and consequently the same eigenvalues, counting multiplicities.

Let $$A=PBP^{-1}$$ for some invertible matrix $$P$$. Then $\begin{eqnarray*} \det(\lambda I-A)&=&\det(\lambda I-PBP^{-1})\\ &=&\det(\lambda PP^{-1} -PBP^{-1})\\ &=&\det(P(\lambda I-B)P^{-1})\\ &=&\det P\det(\lambda I-B)\det(P^{-1})\\ &=&\det(\lambda I-B)\det P\det(P^{-1})\\ &=&\det(\lambda I-B)\det(PP^{-1})\\ &=&\det(\lambda I-B)\cdot 1. \end{eqnarray*}$

Remark.

1. In the preceding example $$A$$ and $$B$$ are similar and then they have the same eigenvalues. Since the eigenvalues of $$C$$ and $$B$$ are different, $$C$$ is not similar to $$B$$ and hence $$A$$.

2. If $$A$$ and $$B$$ are similar, they have the same eigenvalues. But the converse is not true.
For example, $$\left[\begin{array}{rr}1&0\\0&0\end{array}\right]$$ and $$\left[\begin{array}{rr}1&1\\0&0\end{array}\right]$$ are not similar, but have the same eigenvalues.

Theorem. Let $$A$$ be an $$n\times n$$ matrix with eigenvalue $$\lambda$$. Then the geometric multiplicity of $$\lambda$$ is less than or equal to the algebraic multiplicity of $$\lambda$$.

Let $$k$$ be the geometric multiplicity of $$\lambda$$. Suppose $$\overrightarrow{x_1},\ldots,\overrightarrow{x_k}$$ are $$k$$ linearly independent eigenvectors of $$A$$ corresponding to $$\lambda$$. Let $$P=[\overrightarrow{x_1}\cdots\overrightarrow{x_k}*\ldots *]$$ be an $$n \times n$$ invertible matrix. Then $$P^{-1}AP=\left[\begin{array}{c|c}\lambda I_k&*\\\hline 0&*\end{array}\right]$$ and $$\lambda$$ is an eigenvalue of $$P^{-1}AP$$ with algebraic multiplicity at least $$k$$. Since $$A$$ and $$P^{-1}AP$$, being similar, have the same eigenvalues, the algebraic multiplicity of $$\lambda$$ is at least $$k$$.

Definition. A square matrix $$A$$ is diagonalizable if $$A$$ is similar to a diagonal matrix, i.e., $$A=PDP^{-1}$$ for some invertible matrix $$P$$ and some diagonal matrix $$D$$.

Example. In the first example, $$A$$ is diagonalizable as $$A=PBP^{-1}$$ where $$B$$ is a diagonal matrix.

Theorem. Let $$A$$ be an $$n\times n$$ matrix. Then TFAE.

1. $$A$$ is diagonalizable,

2. There are $$n$$ linearly independent eigenvectors of $$A$$,

3. The sum of the geometric multiplicities of the distinct eigenvalues of $$A$$ is $$n$$, and

4. Geometric multiplicity and algebraic multiplicity are the same for all eigenvalues of $$A$$.

First note that (b), (c), and (d) are equivalent. So we prove (a)$$\Longleftrightarrow$$(b).

(a)$$\Longrightarrow$$(b) There is an invertible matrix $$P=[\overrightarrow{p_1},\ldots,\overrightarrow{p_n}]$$ such that $$A=P\mbox{diag}(\lambda_1,\ldots,\lambda_n)P^{-1}$$, i.e., $$AP=P\; \mbox{diag}(\lambda_1,\ldots,\lambda_n)$$. So $$A\overrightarrow{p_i}=\lambda_i\overrightarrow{p_i}$$ for $$i=1,\ldots,n$$. So $$\overrightarrow{p_i}$$ is an eigenvector of $$A$$ corresponding to the eigenvalue $$\lambda_i$$ for $$i=1,\ldots,n$$. Since $$P$$ is invertible, its columns $$\overrightarrow{p_1},\ldots,\overrightarrow{p_n}$$ are linearly independent by the IMT.

(b)$$\Longrightarrow$$(a) Suppose $$\overrightarrow{x_1},\ldots,\overrightarrow{x_n}$$ are $$n$$ linearly independent eigenvectors of $$A$$ corresponding to the eigenvalues $$\lambda_1,\ldots,\lambda_n$$ respectively. Then $$P=[\overrightarrow{x_1},\ldots,\overrightarrow{x_n}]$$ is invertible by the IMT. Since $$A\overrightarrow{x_i}=\lambda_i\overrightarrow{x_i}$$ for $$i=1,\ldots,n$$, we get $\begin{eqnarray*} [A\overrightarrow{x_1}\ldots,A\overrightarrow{x_n}]&=&[\lambda_1\overrightarrow{x_1},\ldots,\lambda_n\overrightarrow{x_n}]\\ A[\overrightarrow{x_1},\ldots,\overrightarrow{x_n}]&=&[\overrightarrow{x_1},\ldots,\overrightarrow{x_n}]\mbox{diag}(\lambda_1,\ldots,\lambda_n)\\ AP&=&PD, \end{eqnarray*}$ where $$D=\mbox{diag}(\lambda_1,\ldots,\lambda_n)$$. Thus $$A=PDP^{-1}$$.

Corollary. Let $$A$$ be an $$n\times n$$ matrix.

1. If $$A$$ has $$n$$ distinct eigenvalues, then $$A$$ is diagonalizable.

2. Suppose that $$A$$ has $$k$$ distinct eigenvalues $$\lambda_1,\ldots,\lambda_k$$ with eigenbases $$B_1,\ldots,B_k$$ respectively. Then $$A$$ is diagonalizable if and only if $$B_1\cup\cdots\cup B_k$$ is a basis for $$\mathbb R^n$$.

A formula for $$A^k$$: Suppose $$A$$ is diagonalizable and $$A=PDP^{-1}$$ for some diagonal matrix $$D$$. Then $A^k=PD^kP^{-1}.$ It is easy to see that $AA\cdots A=(PDP^{-1})(PDP^{-1})\cdots (PDP^{-1})=PDD\cdots DP^{-1}.$ Note that $$D^k$$ is obtained from $$D$$ by raising the power of each diagonal entry of $$D$$ to $$k$$.

Example. Let $$A=\left[\begin{array}{rrr} 2&0&0\\ 1&2&1\\ -1&0&1 \end{array}\right]$$.

1. Diagonalize $$A$$, if possible.

2. Find $$A^k$$, if $$A$$ is diagonalizable.

Solution. $$\det(\lambda I-A)=\left|\begin{array}{ccc} \lambda-2&0&0\\ -1&\lambda-2&-1\\ 1&0&\lambda-1 \end{array}\right|=(\lambda-1)(\lambda-2)^2=0\implies \lambda=1,2,2$$. Verify the following: $\begin{eqnarray*} \operatorname{NS}(A-1I)&=&\displaystyle\operatorname{Span}\left\{\left[\begin{array}{r}0\\-1\\1\end{array}\right]\right\}\\ \operatorname{NS}(A-2I)&=&\displaystyle\operatorname{Span}\left\{ \left[\begin{array}{r}0\\1\\0\end{array}\right], \left[\begin{array}{r}-1\\0\\1\end{array}\right] \right\} \end{eqnarray*}$ (a) Since $$3\times 3$$ matrix $$A$$ has $$3$$ linearly independent eigenvectors, $$A$$ is diagonalizable and $$A=PDP^{-1}$$ where $$D=\left[\begin{array}{rrr} 1&0&0\\0&2&0\\0&0&2\end{array}\right]$$ and $$P=\left[\begin{array}{rrr} 0&0&-1\\-1&1&0\\1&0&1\end{array}\right]$$. You may verify this by showing $$AP=PD=\left[\begin{array}{rrr} 0&0&-2\\-1&2&0\\1&0&2\end{array}\right]$$.

(b) Since $$A=PDP^{-1}$$, $\begin{eqnarray*} A^k&=&PD^kP^{-1}\\ &=&\left[\begin{array}{rrr} 0&0&-1\\-1&1&0\\1&0&1\end{array}\right] \left[\begin{array}{rrr} 1&0&0\\0&2&0\\0&0&2\end{array}\right]^k \left[\begin{array}{rrr} 1&0&1\\1&1&1\\-1&0&0\end{array}\right]\\ &=&\left[\begin{array}{rrr} 0&0&-1\\-1&1&0\\1&0&1\end{array}\right] \left[\begin{array}{rrr} 1&0&0\\0&2^k&0\\0&0&2^k\end{array}\right] \left[\begin{array}{rrr} 1&0&1\\1&1&1\\-1&0&0\end{array}\right]\\ &=&\left[\begin{array}{ccc} 2^k&0&0\\-1+2^k&2^k&-1+2^k\\1-2^k&0&1\end{array}\right]. \end{eqnarray*}$

An interesting fact: If $$A$$ and $$B$$ are diagonalizable and they have the same eigenvectors, then $$AB=BA.$$

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