Chapter 5: SimilarityContent check — Linear Algebra, Chapter 5: Similarity

One way to proceed is left to right.

$$\begin{array}{c}PT{P}^{-1}=\left(\begin{array}{cc}4& 2\\ -3& 2\end{array}\right)\left(\begin{array}{cc}1& 3\\ -2& -6\end{array}\right)\left(\begin{array}{cc}2/14& -2/14\\ 3/14& 4/14\end{array}\right)\\ =\left(\begin{array}{cc}0& 0\\ -7& -21\end{array}\right)\left(\begin{array}{cc}2/14& -2/14\\ 3/14& 4/14\end{array}\right)=\left(\begin{array}{cc}0& 0\\ -11/2& -5\end{array}\right)\end{array}$$

Because the matrix $(2)$ is $1\times 1$⁠, the matrices $P$ and $P^{-1}$ are also $1\times 1$ and so where $P=(p)$ the inverse is $P^{-1}=(1/p)$⁠. Thus $P(2)P^{-1}=(p)(2)(1/p)=(2)$⁠.

Yes: recall that we can bring scalar multiples out of a matrix $P(cI)P^{-1}=cPI{P}^{-1}=cI$⁠. By the way, the zero and identity matrices are the special cases $c=0$ and $c=1$⁠.

No, as this example shows.

$$\left(\begin{array}{cc}1& -2\\ -1& 1\end{array}\right)\left(\begin{array}{cc}-1& 0\\ 0& -3\end{array}\right)\left(\begin{array}{cc}-1& -2\\ -1& -1\end{array}\right)=\left(\begin{array}{cc}-5& -4\\ 2& 1\end{array}\right)$$

$$\begin{array}{ccc}{\displaystyle {ℂ}_{\text{wrt}B}^3}& \underset{\underset{}{T}}{\overset{\stackrel{t}{}}{\to }}& {\displaystyle {ℂ}_{\text{wrt}B}^3}\\ \begin{array}{c}\mathrm{id}\end{array}\downarrow & {\displaystyle }& \begin{array}{c}\mathrm{id}\end{array}\downarrow \\ {\displaystyle {ℂ}_{\text{wrt}D}^3}& \underset{\underset{}{\hat{T}}}{\overset{\stackrel{t}{}}{\to }}& {\displaystyle {ℂ}_{\text{wrt}D}^3}\end{array}$$

For each element of the starting basis $B$ find the effect of the transformation

$$\left(\begin{array}{c}1\\ 2\\ 3\end{array}\right)\stackrel{t}{⟼}\left(\begin{array}{c}-2\\ 3\\ 4\end{array}\right)\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)\stackrel{t}{⟼}\left(\begin{array}{c}0\\ 0\\ 2\end{array}\right)\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)\stackrel{t}{⟼}\left(\begin{array}{c}-1\\ 1\\ 0\end{array}\right)$$

and represented those outputs with respect to the ending basis $B$

$${\mathrm{Rep}}_B(\left(\begin{array}{c}-2\\ 3\\ 4\end{array}\right))=\left(\begin{array}{c}-2\\ 7\\ 10\end{array}\right){\mathrm{Rep}}_B(\left(\begin{array}{c}0\\ 0\\ 2\end{array}\right))=\left(\begin{array}{c}0\\ 0\\ 2\end{array}\right){\mathrm{Rep}}_B(\left(\begin{array}{c}-1\\ 1\\ 0\end{array}\right))=\left(\begin{array}{c}-1\\ 3\\ 3\end{array}\right)$$

to get the matrix.

$$T={\mathrm{Rep}}_{B,B}(t)=\left(\begin{array}{ccc}-2& 0& -1\\ 7& 0& 3\\ 10& 2& 3\end{array}\right)$$

Find the effect of the transformation on the elements of $D$

$$\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)\stackrel{t}{⟼}\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right)\stackrel{t}{⟼}\left(\begin{array}{c}1\\ 0\\ 2\end{array}\right)\left(\begin{array}{c}1\\ 0\\ 1\end{array}\right)\stackrel{t}{⟼}\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)$$

and represented those with respect to the ending basis $D$

$${\mathrm{Rep}}_D(\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right))=\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right){\mathrm{Rep}}_D(\left(\begin{array}{c}1\\ 0\\ 2\end{array}\right))=\left(\begin{array}{c}-1\\ 0\\ 2\end{array}\right){\mathrm{Rep}}_D(\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right))=\left(\begin{array}{c}-1\\ 1\\ 0\end{array}\right)$$

to get the matrix.

$$\hat{T}={\mathrm{Rep}}_{D,D}(t)=\left(\begin{array}{ccc}1& -1& -1\\ 0& 0& 1\\ 0& 2& 0\end{array}\right)$$

To go down on the right we need ${\mathrm{Rep}}_{B,D}(\mathrm{id})$ so we first compute the effect of the identity map on each element of $B$⁠, which is no effect, and then represent the results with respect to $D$⁠.

$${\mathrm{Rep}}_D(\left(\begin{array}{c}1\\ 2\\ 3\end{array}\right))=\left(\begin{array}{c}-4\\ 2\\ 3\end{array}\right){\mathrm{Rep}}_D(\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right))=\left(\begin{array}{c}-1\\ 1\\ 0\end{array}\right){\mathrm{Rep}}_D(\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right))=\left(\begin{array}{c}-1\\ 0\\ 1\end{array}\right)$$

So this is $P$⁠.

$$P=\left(\begin{array}{ccc}-4& -1& -1\\ 2& 1& 0\\ 3& 0& 1\end{array}\right)$$

For the other matrix ${\mathrm{Rep}}_{D,B}(\mathrm{id})$ we can either find it directly, as we just have with $P$⁠, or we can do the usual calculation of a matrix inverse.

$$P^{-1}=\left(\begin{array}{ccc}1& 1& 1\\ -2& -1& -2\\ -3& -3& -2\end{array}\right)$$

Because we describe $t$ with the members of $B$⁠, finding the matrix representation is easy:

$${\mathrm{Rep}}_B(t(x^2))={\left(\begin{array}{c}0\\ 1\\ 1\end{array}\right)}_B{\mathrm{Rep}}_B(t(x))={\left(\begin{array}{c}1\\ 0\\ -1\end{array}\right)}_B{\mathrm{Rep}}_B(t(1))={\left(\begin{array}{c}0\\ 0\\ 3\end{array}\right)}_B$$

gives this.

$${\mathrm{Rep}}_{B,B}(t)\left(\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 1& -1& 3\end{array}\right)$$

We will find $t(1)$⁠, $t(1+x)$⁠, and $t(1+x+x^2$⁠, to find how each is represented with respect to $D$⁠. We are given that $t(1)=3$⁠, and the other two are easy to see: $t(1+x)=x^2+2$ and $t(1+x+x^2)=x^2+x+3$⁠. By eye, we get the representation of each vector

$${\mathrm{Rep}}_D(t(1))={\left(\begin{array}{c}3\\ 0\\ 0\end{array}\right)}_D{\mathrm{Rep}}_D(t(1+x))={\left(\begin{array}{c}2\\ -1\\ 1\end{array}\right)}_D{\mathrm{Rep}}_D(t(1+x+x^2))={\left(\begin{array}{c}2\\ 0\\ 1\end{array}\right)}_D$$

and thus the representation of the map.

$${\mathrm{Rep}}_{D,D}(t)=\left(\begin{array}{ccc}3& 2& 2\\ 0& -1& 0\\ 0& 1& 1\end{array}\right)$$

The diagram

$$\begin{array}{ccc}{\displaystyle V_{\text{wrt}B}}& \underset{\underset{}{T}}{\overset{\stackrel{t}{}}{\to }}& {\displaystyle V_{\text{wrt}B}}\\ \begin{array}{c}\mathrm{id}\end{array}\downarrow \begin{array}{c}P\end{array}& {\displaystyle }& \begin{array}{c}\mathrm{id}\end{array}\downarrow \begin{array}{c}P\end{array}\\ {\displaystyle V_{\text{wrt}D}}& \underset{\underset{}{\hat{T}}}{\overset{\stackrel{t}{}}{\to }}& {\displaystyle V_{\text{wrt}D}}\end{array}$$

shows that these are $P={\mathrm{Rep}}_{B,D}(\mathrm{id})$ and $P^{-1}={\mathrm{Rep}}_{D,B}(\mathrm{id})$⁠.

$$P=\left(\begin{array}{ccc}0& -1& 1\\ -1& 1& 0\\ 1& 0& 0\end{array}\right)P^{-1}=\left(\begin{array}{ccc}0& 0& 1\\ 0& 1& 1\\ 1& 1& 1\end{array}\right)$$

$$\begin{array}{ccc}{\displaystyle {ℂ}_{\text{wrt}B}^2}& \underset{\underset{}{T}}{\overset{\stackrel{t}{}}{\to }}& {\displaystyle {ℂ}_{\text{wrt}B}^2}\\ \begin{array}{c}\mathrm{id}\end{array}\downarrow & {\displaystyle }& \begin{array}{c}\mathrm{id}\end{array}\downarrow \\ {\displaystyle {ℂ}_{\text{wrt}D}^2}& \underset{\underset{}{\hat{T}}}{\overset{\stackrel{t}{}}{\to }}& {\displaystyle {ℂ}_{\text{wrt}D}^2}\end{array}$$

For the right side we find the effect of the identity map, which is no effect,

$$\left(\begin{array}{c}1\\ 0\end{array}\right)\stackrel{\mathrm{id}}{⟼}\left(\begin{array}{c}1\\ 0\end{array}\right)\left(\begin{array}{c}1\\ 1\end{array}\right)\stackrel{\mathrm{id}}{⟼}\left(\begin{array}{c}1\\ 1\end{array}\right)$$

and represent those with respect to $D$

$${\mathrm{Rep}}_D(\left(\begin{array}{c}1\\ 0\end{array}\right))=\left(\begin{array}{c}1/2\\ 0\end{array}\right){\mathrm{Rep}}_D(\left(\begin{array}{c}1\\ 1\end{array}\right))=\left(\begin{array}{c}1/2\\ -1/2\end{array}\right)$$

so we have this.

$$P={\mathrm{Rep}}_{B,D}(\mathrm{id})=\left(\begin{array}{cc}1/2& 1/2\\ 0& -1/2\end{array}\right)$$

For the matrix on the left we can either compute it directly, as in the prior paragraph, or we can take the inverse.

$$P^{-1}={\mathrm{Rep}}_{D,B}(\mathrm{id})=\frac{1}{(-1/4)}\cdot \left(\begin{array}{cc}-1/2& -1/2\\ 0& 1/2\end{array}\right)=\left(\begin{array}{cc}2& 2\\ 0& -2\end{array}\right)$$

As with the prior item we can either compute it directly from the definition or compute it using matrix operations.

$$PT{P}^{-1}=\left(\begin{array}{cc}2& 2\\ 0& -2\end{array}\right)\left(\begin{array}{cc}1& -1\\ 2& 1\end{array}\right)\left(\begin{array}{cc}2& 2\\ 0& -2\end{array}\right)=\left(\begin{array}{cc}3& 3\\ -2& -1\end{array}\right)$$

One possible choice of the bases is

$$B=⟨\left(\begin{array}{c}1\\ 2\end{array}\right),\left(\begin{array}{c}-1\\ 1\end{array}\right)⟩D={ℰ}_2=⟨\left(\begin{array}{c}1\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 1\end{array}\right)⟩$$

(this $B$ comes from the map description). To find the matrix $\hat{T}={\mathrm{Rep}}_{B,B}(t)$⁠, solve the relations

$$c_1\left(\begin{array}{c}1\\ 2\end{array}\right)+c_2\left(\begin{array}{c}-1\\ 1\end{array}\right)=\left(\begin{array}{c}3\\ 0\end{array}\right){\hat{c}}_1\left(\begin{array}{c}1\\ 2\end{array}\right)+{\hat{c}}_2\left(\begin{array}{c}-1\\ 1\end{array}\right)=\left(\begin{array}{c}-1\\ 2\end{array}\right)$$

to get $c_1=1$⁠, $c_2=-2$⁠, ${\hat{c}}_1=1/3$ and ${\hat{c}}_2=4/3$⁠.

$${\mathrm{Rep}}_{B,B}(t)=\left(\begin{array}{cc}1& 1/3\\ -2& 4/3\end{array}\right)$$

Finding ${\mathrm{Rep}}_{D,D}(t)$ involves a bit more computation. We first find $t({\overrightarrow{e}}_1)$⁠. The relation

$$c_1\left(\begin{array}{c}1\\ 2\end{array}\right)+c_2\left(\begin{array}{c}-1\\ 1\end{array}\right)=\left(\begin{array}{c}1\\ 0\end{array}\right)$$

gives $c_1=1/3$ and $c_2=-2/3$⁠, and so

$${\mathrm{Rep}}_B({\overrightarrow{e}}_1)={\left(\begin{array}{c}1/3\\ -2/3\end{array}\right)}_B$$

making

$${\mathrm{Rep}}_B(t({\overrightarrow{e}}_1))={\left(\begin{array}{cc}1& 1/3\\ -2& 4/3\end{array}\right)}_{B,B}{\left(\begin{array}{c}1/3\\ -2/3\end{array}\right)}_B={\left(\begin{array}{c}1/9\\ -14/9\end{array}\right)}_B$$

and hence $t$ acts on the first basis vector ${\overrightarrow{e}}_1$ in this way.

$$t({\overrightarrow{e}}_1)=(1/9)\cdot \left(\begin{array}{c}1\\ 2\end{array}\right)-(14/9)\cdot \left(\begin{array}{c}-1\\ 1\end{array}\right)=\left(\begin{array}{c}5/3\\ -4/3\end{array}\right)$$

The computation for $t({\overrightarrow{e}}_2)$ is similar. The relation

$$c_1\left(\begin{array}{c}1\\ 2\end{array}\right)+c_2\left(\begin{array}{c}-1\\ 1\end{array}\right)=\left(\begin{array}{c}0\\ 1\end{array}\right)$$

gives $c_1=1/3$ and $c_2=1/3$⁠, so

$${\mathrm{Rep}}_B({\overrightarrow{e}}_1)={\left(\begin{array}{c}1/3\\ 1/3\end{array}\right)}_B$$

making

$${\mathrm{Rep}}_B(t({\overrightarrow{e}}_1))={\left(\begin{array}{cc}1& 1/3\\ -2& 4/3\end{array}\right)}_{B,B}{\left(\begin{array}{c}1/3\\ 1/3\end{array}\right)}_B={\left(\begin{array}{c}4/9\\ -2/9\end{array}\right)}_B$$

and hence $t$ acts on the second basis vector ${\overrightarrow{e}}_2$ in this way.

$$t({\overrightarrow{e}}_2)=(4/9)\cdot \left(\begin{array}{c}1\\ 2\end{array}\right)-(2/9)\cdot \left(\begin{array}{c}-1\\ 1\end{array}\right)=\left(\begin{array}{c}2/3\\ 2/3\end{array}\right)$$

Therefore

$${\mathrm{Rep}}_{D,D}(t)=\left(\begin{array}{cc}5/3& 2/3\\ -4/3& 2/3\end{array}\right)$$

and so this matrix.

$$P={\mathrm{Rep}}_{B,D}(\mathrm{id})=\left(\begin{array}{cc}1& -1\\ 2& 1\end{array}\right)$$

and this one change the bases.

$$P^{-1}=({\mathrm{Rep}}_{B,D}(\mathrm{id}){)}^{-1}={\left(\begin{array}{cc}1& -1\\ 2& 1\end{array}\right)}^{-1}=\left(\begin{array}{cc}1/3& 1/3\\ -2/3& 1/3\end{array}\right)$$

The check of these computations is routine.

$$\left(\begin{array}{cc}1& -1\\ 2& 1\end{array}\right)\left(\begin{array}{cc}1& 1/3\\ -2& 4/3\end{array}\right)\left(\begin{array}{cc}1/3& 1/3\\ -2/3& 1/3\end{array}\right)=\left(\begin{array}{cc}5/3& 2/3\\ -4/3& 2/3\end{array}\right)$$

Gauss’s Method shows that the first matrix represents maps of rank two while the second matrix represents maps of rank three.

The only representation of a zero map is a zero matrix, no matter what the pair of bases ${\mathrm{Rep}}_{B,D}(z)=Z$⁠, and so in particular for any single basis $B$ we have ${\mathrm{Rep}}_{B,B}(z)=Z$⁠. The case of the identity is slightly different: the only representation of the identity map, with respect to any $B,B$⁠, is the identity ${\mathrm{Rep}}_{B,B}(\mathrm{id})=I$⁠. (Remark: of course, we have seen examples where $B\ne D$ and ${\mathrm{Rep}}_{B,D}(\mathrm{id})\ne I$ — in fact, we have seen that any nonsingular matrix is a representation of the identity map with respect to some $B,D$⁠.)

No. If $A=PB{P}^{-1}$ then $A^2=$ $(PB{P}^{-1})(PB{P}^{-1})=$ $P{B}^2{P}^{-1}$⁠.

Matrix similarity is a special case of matrix equivalence (if matrices are similar then they are matrix equivalent) and matrix equivalence preserves nonsingularity.

A matrix is similar to itself; take $P$ to be the identity matrix: $P=IP{I}^{-1}=IPI$⁠.

If $\hat{T}$ is similar to $T$ then $\hat{T}=PT{P}^{-1}$ and so $P^{-1}\hat{T}P=T$⁠. Rewrite $T=(P^{-1})\hat{T}{(P^{-1})}^{-1}$ to conclude that $T$ is similar to $\hat{T}$⁠.

For transitivity, if $T$ is similar to $S$ and $S$ is similar to $U$ then $T=PS{P}^{-1}$ and $S=QU{Q}^{-1}$⁠. Then $T=PQU{Q}^{-1}{P}^{-1}=(PQ)U{(PQ)}^{-1}$⁠, showing that $T$ is similar to $U$⁠.

Let $f_x$ and $f_y$ be the reflection maps (sometimes called ‘flip’s). For any bases $B$ and $D$⁠, the matrices ${\mathrm{Rep}}_{B,B}(f_x)$ and ${\mathrm{Rep}}_{D,D}(f_y)$ are similar. First note that

$$S={\mathrm{Rep}}_{{ℰ}_2,{ℰ}_2}(f_x)=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)T={\mathrm{Rep}}_{{ℰ}_2,{ℰ}_2}(f_y)=\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)$$

are similar because the second matrix is the representation of $f_x$ with respect to the basis $A=⟨{\overrightarrow{e}}_2,{\overrightarrow{e}}_1⟩$⁠:

$$\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)=P\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)P^{-1}$$

where $P={\mathrm{Rep}}_{A,{ℰ}_2}(\mathrm{id})$⁠.

$$\begin{array}{ccc}{\displaystyle {ℂ}_{\text{wrt}A}^2}& \underset{\underset{}{T}}{\overset{\stackrel{f_x}{}}{\to }}& {\displaystyle V{ℂ}_{\text{wrt}A}^2}\\ \begin{array}{c}\mathrm{id}\end{array}\downarrow \begin{array}{c}P\end{array}& {\displaystyle }& \begin{array}{c}\mathrm{id}\end{array}\downarrow \begin{array}{c}P\end{array}\\ {\displaystyle {ℂ}_{\text{wrt}{ℰ}_2}^2}& \underset{\underset{}{S}}{\overset{\stackrel{f_x}{}}{\to }}& {\displaystyle {ℂ}_{\text{wrt}{ℰ}_2}^2}\end{array}$$

Now the conclusion follows from the transitivity part of Exercise 1.15.

We can also finish without relying on that exercise. Write ${\mathrm{Rep}}_{B,B}(f_x)=$ $QT{Q}^{-1}=$ $Q{\mathrm{Rep}}_{{ℰ}_2,{ℰ}_2}(f_x)Q^{-1}$ and ${\mathrm{Rep}}_{D,D}(f_y)=$ $RS{R}^{-1}=$ $R{\mathrm{Rep}}_{{ℰ}_2,{ℰ}_2}(f_y)R^{-1}$⁠. By the equation in the first paragraph, the first of these two is ${\mathrm{Rep}}_{B,B}(f_x)=$ $QP{\mathrm{Rep}}_{{ℰ}_2,{ℰ}_2}(f_y)P^{-1}{Q}^{-1}$⁠. Rewriting the second of these two as $R^{-1}\cdot {\mathrm{Rep}}_{D,D}(f_y)\cdot R=$ ${\mathrm{Rep}}_{{ℰ}_2,{ℰ}_2}(f_y)$ and substituting gives the desired relationship

$$\begin{array}{c}{\mathrm{Rep}}_{B,B}(f_x)=QP{\mathrm{Rep}}_{{ℰ}_2,{ℰ}_2}(f_y)P^{-1}{Q}^{-1}\\ =QP{R}^{-1}\cdot {\mathrm{Rep}}_{D,D}(f_y)\cdot R{P}^{-1}{Q}^{-1}=(QP{R}^{-1})\cdot {\mathrm{Rep}}_{D,D}(f_y)\cdot {(QP{R}^{-1})}^{-1}\end{array}$$

Thus the matrices ${\mathrm{Rep}}_{B,B}(f_x)$ and ${\mathrm{Rep}}_{D,D}(f_y)$ are similar.

We must show that if two matrices are similar then they have the same rank and the same determinant.

First, we’ve shown that the rank of a map equals the rank of any matrix representing that map, so the rank must be the same for all representations. (Said another way, rank is a property of the map since it is the dimension of the range space.)

As for determinants, $\left|PS{P}^{-1}\right|=$ $\left|P\right|\cdot \left|S\right|\cdot \left|P^{-1}\right|=$ $\left|P\right|\cdot \left|S\right|\cdot {\left|P\right|}^{-1}=$ $\left|S\right|$⁠.

The converse of the statement does not hold. For one thing, there are matrices with the same determinant that are not similar. As an example, consider a nonzero matrix with a determinant of zero. It is not similar to the zero matrix because the zero matrix is similar only to itself, but they have they same determinant. The same thinking gives an example for rank.

The matrix equivalence class containing all $n\times n$ rank zero matrices contains only a single matrix, the zero matrix. Therefore it has as a subset only one similarity class.

In contrast, the matrix equivalence class of $1\times 1$ matrices of rank one consists of those $1\times 1$ matrices $(k)$ where $k\ne 0$⁠. For any basis $B$⁠, the representation of multiplication by the scalar $k$ is ${\mathrm{Rep}}_{B,B}(t_k)=(k)$⁠, so each such matrix is alone in its similarity class. So this is a case where a matrix equivalence class splits into infinitely many similarity classes.

Yes, these are similar

$$\left(\begin{array}{cc}1& 0\\ 0& 3\end{array}\right)\left(\begin{array}{cc}3& 0\\ 0& 1\end{array}\right)$$

since, where the first matrix is ${\mathrm{Rep}}_{B,B}(t)$ for $B=⟨{\overrightarrow{\beta }}_1,{\overrightarrow{\beta }}_2⟩$⁠, the second matrix is ${\mathrm{Rep}}_{D,D}(t)$ for $D=⟨{\overrightarrow{\beta }}_2,{\overrightarrow{\beta }}_1⟩$⁠.

The $k$⁠-th powers are similar because, where each matrix represents the map $t$⁠, the $k$⁠-th powers represent $t^k$⁠, the composition of $k$⁠-many $t$⁠’s. (For instance, if $T=reptB,B$ then $T^2={\mathrm{Rep}}_{B,B}(t\circ t)$⁠.)

Restated more computationally, if $T=PS{P}^{-1}$ then $T^2=$ $(PS{P}^{-1})(PS{P}^{-1})=$ $P{S}^2{P}^{-1}$⁠. Induction extends that to all powers.

For the $k\le 0$ case, suppose that $S$ is invertible and that $T=PS{P}^{-1}$⁠. Note that $T$ is invertible: $T^{-1}={(PS{P}^{-1})}^{-1}=P{S}^{-1}{P}^{-1}$⁠, and that same equation shows that $T^{-1}$ is similar to $S^{-1}$⁠. Other negative powers are now given by the first paragraph.

In conceptual terms, both represent $p(t)$ for some transformation $t$⁠. In computational terms, we have this.

$$\begin{array}{cc}\hfill p(T)& =c_n{(PS{P}^{-1})}^n+\cdots +c_1(PS{P}^{-1})+c_{0}I\hfill \\ \hfill & =c_{n}P{S}^n{P}^{-1}+\cdots +c_{1}PS{P}^{-1}+c_{0}I\hfill \\ \hfill & =P{c}_n{S}^n{P}^{-1}+\cdots +P{c}_{1}S{P}^{-1}+P{c}_0{P}^{-1}\hfill \\ \hfill & =P(c_n{S}^n+\cdots +c_{1}S+c_0)P^{-1}\hfill \end{array}$$

There are two equivalence classes, (i) the class of rank zero matrices, of which there is one: ${𝒞}_1=\{(0)\}$⁠, and (2) the class of rank one matrices, of which there are infinitely many: ${𝒞}_2=\{(k)|k\ne 0\}$⁠.

Each $1\times 1$ matrix is alone in its similarity class. That’s because any transformation of a one-dimensional space is multiplication by a scalar $t_k:V\to V$ given by $\overrightarrow{v}\mapsto k\cdot \overrightarrow{v}$⁠. Thus, for any basis $B=⟨\overrightarrow{\beta }⟩$⁠, the matrix representing a transformation $t_k$ with respect to $B,B$ is $({\mathrm{Rep}}_B(t_k(\overrightarrow{\beta })))=(k)$⁠.

So, contained in the matrix equivalence class ${𝒞}_1$ is (obviously) the single similarity class consisting of the matrix $(0)$⁠. And, contained in the matrix equivalence class ${𝒞}_2$ are the infinitely many, one-member-each, similarity classes consisting of $(k)$ for $k\ne 0$⁠.

No. Here is an example that has two pairs, each of two similar matrices:

$$\left(\begin{array}{cc}1& -1\\ 1& 2\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 3\end{array}\right)\left(\begin{array}{cc}2/3& 1/3\\ -1/3& 1/3\end{array}\right)=\left(\begin{array}{cc}5/3& -2/3\\ -4/3& 7/3\end{array}\right)$$

and

$$\left(\begin{array}{cc}1& -2\\ -1& 1\end{array}\right)\left(\begin{array}{cc}-1& 0\\ 0& -3\end{array}\right)\left(\begin{array}{cc}-1& -2\\ -1& -1\end{array}\right)=\left(\begin{array}{cc}-5& -4\\ 2& 1\end{array}\right)$$

(this example is not entirely arbitrary because the center matrices on the two left sides add to the zero matrix). Note that the sums of these similar matrices are not similar

$$\left(\begin{array}{cc}1& 0\\ 0& 3\end{array}\right)+\left(\begin{array}{cc}-1& 0\\ 0& -3\end{array}\right)=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)\left(\begin{array}{cc}5/3& -2/3\\ -4/3& 7/3\end{array}\right)+\left(\begin{array}{cc}-5& -4\\ 2& 1\end{array}\right)\ne \left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$$

since the zero matrix is similar only to itself.

If $N=P(T-\lambda I)P^{-1}$ then $N=PT{P}^{-1}-P(\lambda I)P^{-1}$⁠. The diagonal matrix $\lambda I$ commutes with anything, so $P(\lambda I)P^{-1}=P{P}^{-1}(\lambda I)=\lambda I$⁠. Thus $N=PT{P}^{-1}-\lambda I$ and consequently $N+\lambda I=PT{P}^{-1}$⁠. (So not only are they similar, in fact they are similar via the same $P$⁠.)

Because we chose the basis vectors arbitrarily, many different answers are possible. However, here is one way to go; to diagonalize

$$T=\left(\begin{array}{cc}4& -2\\ 1& 1\end{array}\right)$$

take it as the representation of a transformation with respect to the standard basis $T={\mathrm{Rep}}_{{ℰ}_2,{ℰ}_2}(t)$ and look for $B=⟨{\overrightarrow{\beta }}_1,{\overrightarrow{\beta }}_2⟩$ such that

$${\mathrm{Rep}}_{B,B}(t)=\left(\begin{array}{cc}{\lambda }_1& 0\\ 0& {\lambda }_2\end{array}\right)$$

that is, such that $t({\overrightarrow{\beta }}_1)={\lambda }_1$ and $t({\overrightarrow{\beta }}_2)={\lambda }_2$⁠.

$$\left(\begin{array}{cc}4& -2\\ 1& 1\end{array}\right){\overrightarrow{\beta }}_1={\lambda }_1\cdot {\overrightarrow{\beta }}_1\left(\begin{array}{cc}4& -2\\ 1& 1\end{array}\right){\overrightarrow{\beta }}_2={\lambda }_2\cdot {\overrightarrow{\beta }}_2$$

We are looking for scalars $x$ such that this equation

$$\left(\begin{array}{cc}4& -2\\ 1& 1\end{array}\right)\left(\begin{array}{c}{b}_1\\ b_2\end{array}\right)=x\cdot \left(\begin{array}{c}{b}_1\\ b_2\end{array}\right)$$

has solutions $b_1$ and $b_2$⁠, which are not both zero. Rewrite that as a linear system

$$\begin{array}{ccccc}\hfill (4-x)\cdot b_1& +& \hfill -2\cdot b_2& =& \hfill 0\\ \hfill 1\cdot b_1& +& \hfill (1-x)\cdot b_2& =& \hfill 0\end{array}$$

If $x=4$ then the first equation gives that $b_2=0$⁠, and then the second equation gives that $b_1=0$⁠. We have disallowed the case where both $b$⁠’s are zero so we can assume that $x\ne 4$⁠.

$$\stackrel{(-1/(4-x)){\rho }_1+{\rho }_2}{⟶}\begin{array}{ccccc}\hfill (4-x)\cdot b_1& +& \hfill -2\cdot b_2& =& \hfill 0\\ \hfill & & \hfill ((x^2-5x+6)/(4-x))\cdot b_2& =& \hfill 0\end{array}$$

Consider the bottom equation. If $b_2=0$ then the first equation gives $b_1=0$ or $x=4$⁠. The $b_1=b_2=0$ case is not allowed. The other possibility for the bottom equation is that the numerator of the fraction $x^2-5x+6=(x-2)(x-3)$ is zero. The $x=2$ case gives a first equation of $2b_1-2b_2=0$⁠, and so associated with $x=2$ we have vectors whose first and second components are equal:

$${\overrightarrow{\beta }}_1=\left(\begin{array}{c}1\\ 1\end{array}\right)\text{(so }\left(\begin{array}{cc}4& -2\\ 1& 1\end{array}\right)\left(\begin{array}{c}1\\ 1\end{array}\right)=2\cdot \left(\begin{array}{c}1\\ 1\end{array}\right)\text{, and }{\lambda }_1=2\text{).}$$

If $x=3$ then the first equation is $b_1-2b_2=0$ and so the associated vectors are those whose first component is twice their second:

$${\overrightarrow{\beta }}_2=\left(\begin{array}{c}2\\ 1\end{array}\right)\text{(so }\left(\begin{array}{cc}4& -2\\ 1& 1\end{array}\right)\left(\begin{array}{c}2\\ 1\end{array}\right)=3\cdot \left(\begin{array}{c}2\\ 1\end{array}\right)\text{, and so }{\lambda }_2=3\text{).}$$

This picture

$$\begin{array}{ccc}{\displaystyle {ℂ}_{\text{wrt}{ℰ}_2}^2}& \underset{\underset{}{T}}{\overset{\stackrel{t}{}}{\to }}& {\displaystyle {ℂ}_{\text{wrt}{ℰ}_2}^2}\\ \begin{array}{c}\mathrm{id}\end{array}\downarrow & {\displaystyle }& \begin{array}{c}\mathrm{id}\end{array}\downarrow \\ {\displaystyle {ℂ}_{\text{wrt}B}^2}& \underset{\underset{}{D}}{\overset{\stackrel{t}{}}{\to }}& {\displaystyle {ℂ}_{\text{wrt}B}^2}\end{array}$$

shows how to get the diagonalization.

$$\left(\begin{array}{cc}2& 0\\ 0& 3\end{array}\right)={\left(\begin{array}{cc}1& 2\\ 1& 1\end{array}\right)}^{-1}\left(\begin{array}{cc}4& -2\\ 1& 1\end{array}\right)\left(\begin{array}{cc}1& 2\\ 1& 1\end{array}\right)$$

Comment. This equation matches the $T=PS{P}^{-1}$ definition under this renaming.

$$T=\left(\begin{array}{cc}2& 0\\ 0& 3\end{array}\right)P={\left(\begin{array}{cc}1& 2\\ 1& 1\end{array}\right)}^{-1}{P}^{-1}=\left(\begin{array}{cc}1& 2\\ 1& 1\end{array}\right)S=\left(\begin{array}{cc}4& -2\\ 1& 1\end{array}\right)$$

We want a basis where the member vectors satisfy

$$\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)\left(\begin{array}{c}{b}_1\\ b_2\end{array}\right)=x\cdot \left(\begin{array}{c}{b}_1\\ b_2\end{array}\right)$$

so consider the resulting linear system.

$$\begin{array}{ccccc}\hfill b_1& +& \hfill b_2& =& \hfill x\cdot b_1\\ \hfill & & \hfill 0& =& \hfill x\cdot b_2\end{array}$$

If $x=0$ then $b_1+b_2=0$⁠, so $b_1=-b_2$⁠. That is, associated with this case are basis elements where the entries have opposite signs. (The entries must be nonzero since the zero vector cannot be a member of a basis.)

If $x\ne 0$ then $b_2=0$ and substituting that into the top equation gives $b_1=x\cdot b_1$⁠. So in this case, either $b_1=0$ or $x=1$⁠. But since $b_2=0$ in this case, $b_1=0$ is impossible because the zero vector is not in any basis. Thus $x=1$ and the system becomes this.

$$\begin{array}{ccccc}\hfill b_1& +& \hfill b_2& =& \hfill 1\cdot b_1\\ \hfill & & \hfill 0& =& \hfill 1\cdot b_2\end{array}$$

So this case is associated with basis elements where $b_2=0$ and $b_1$ can have any nonzero value.

Therefore, a suitable basis is this.

$$B=⟨\left(\begin{array}{c}1\\ -1\end{array}\right),\left(\begin{array}{c}1\\ 0\end{array}\right)⟩$$

With that basis, to get the diagonalization, draw the similarity diagram.

$$\begin{array}{ccc}{\displaystyle {ℂ}_{\text{wrt}{ℰ}_2}^2}& \underset{\underset{}{T}}{\overset{\stackrel{t}{}}{\to }}& {\displaystyle {ℂ}_{\text{wrt}{ℰ}_2}^2}\\ \begin{array}{c}\mathrm{id}\end{array}\downarrow & {\displaystyle }& \begin{array}{c}\mathrm{id}\end{array}\downarrow \\ {\displaystyle {ℂ}_{\text{wrt}B}^2}& \underset{\underset{}{D}}{\overset{\stackrel{t}{}}{\to }}& {\displaystyle {ℂ}_{\text{wrt}B}^2}\end{array}$$

In this diagram,

$$T=\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)$$

and we will calculate the matrix $D$ using the basis $B$ and traversing the lower-left to upper-left, to upper-right, and to lower-right path.

We first want to go from the lower-left to the upper-left. Observe that because we have chosen the basis for the top of the diagram to be the standard basis ${ℰ}_2$⁠, the matrix representing that map is easy.

$${\mathrm{Rep}}_{B,{ℰ}_2}(\mathrm{id})=\left(\begin{array}{cc}1& 1\\ -1& 0\end{array}\right)$$

The calculation for the entire path does indeed give a diagonal result (of course, the lower-left to upper-left portion appears on the right side of the product of the three matrices).

$$\begin{array}{cc}\hfill D& ={\left(\begin{array}{cc}1& 1\\ -1& 0\end{array}\right)}^{-1}\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)\left(\begin{array}{cc}1& 1\\ -1& 0\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{cc}0& -1\\ 1& 1\end{array}\right)\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)\left(\begin{array}{cc}1& 1\\ -1& 0\end{array}\right)=\left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right)\hfill \end{array}$$