Similarity of Matrices

In the last few sections, the following has been discussed in detail:
Given a finite dimensional vector space $V$ of dimension $n,$ we fixed an ordered basis ${\cal B}.$ For any ${\mathbf v}\in V,$ we calculated the column vector $[{\mathbf v}]_{{\cal B}},$ to obtain the coordinates of ${\mathbf v}$ with respect to the ordered basis ${\cal B}.$ Also, for any linear transformation $T: V {\longrightarrow}V,$ we got an $n \times n$ matrix $T[{\cal B},{\cal B}],$ the matrix of $T$ with respect to the ordered basis ${\cal B}.$ That is, once an ordered basis of $V$ is fixed, every linear transformation is represented by a matrix with entries from the scalars.

In this section, we understand the matrix representation of $T$ in terms of different bases ${\cal B}_1$ and ${\cal B}_2$ of $V.$ That is, we relate the two $n \times n$ matrices $T[{\cal B}_1, {\cal B}_1]$ and $T[{\cal B}_2, {\cal B}_2].$ We start with the following important theorem. This theorem also enables us to understand WHY THE MATRIX PRODUCT IS DEFINED SOMEWHAT DIFFERENTLY.

THEOREM 4.4.1 (Composition of Linear Transformations)   Let $V,$ $\;W$ and $Z$ be finite dimensional vector spaces with ordered bases ${\cal B}_1, {\cal B}_2, {\cal B}_3,$ respectively. Also, let $T: V {\longrightarrow}W$ and $S: W {\longrightarrow} Z$ be linear transformations. Then the composition map $S \circ T: V {\longrightarrow}Z$ is a linear transformation and

$$(S \circ T) \;[{\cal B}_1, {\cal B}_3] = S[{\cal B}_2, {\cal B}_3] \;\; T[{\cal B}_1, {\cal B}_2].$$

Proof. Let ${\cal B}_1 = ({\mathbf u}_1, {\mathbf u}_2, \ldots, {\mathbf u}_n), \; {\cal B}_2 = ({\mathbf v}_1, {\mathbf v}_2, \ldots, {\mathbf v}_m)$ and ${\cal B}_3 = ({\mathbf w}_1, {\mathbf w}_2, \ldots, {\mathbf w}_p)$ be ordered bases of $V, W$ and $Z,$ respectively. Then

$$(S\circ T) \; [ {\cal B}_1, {\cal B}_3] = [ [S \circ T({\mathbf u... ...\mathbf u}_2)]_{{\cal B}_3}, \ldots, [S \circ T({\mathbf u}_n)]_{{\cal B}_3} ].$$

Now for $1 \leq t \leq n,$

$$(S \circ T)\; ({\mathbf u}_t) \!\! = S ( T({\mathbf u}_t) ) = S \biggl( \sum\limits_{j=1}^m (T[{\cal B... ...\biggr) = \sum\limits_{j=1}^m (T[{\cal B}_1, {\cal B}_2])_{jt} S({\mathbf v}_j)$$

$$= \sum\limits_{j=1}^m (T[{\cal B}_1, {\cal B}_2])_{jt} \sum\limits_{k=1}^p (S[{\cal B}_2, {\cal B}_3])_{kj} {\mathbf w}_k$$

$$= \sum\limits_{k=1}^p (\sum\limits_{j=1}^m (S[{\cal B}_2, {\cal B}_3])_{kj} (T[{\cal B}_1, {\cal B}_2])_{jt}) {\mathbf w}_k$$

$$= \sum\limits_{k=1}^p ( S[{\cal B}_2, {\cal B}_3] \;\; T[{\cal B}_1, {\cal B}_2])_{kt} {\mathbf w}_k.$$

So,

$$[(S\circ T)\;({\mathbf u}_t)]_{{\cal B}_3} = ( (S[{\cal B}_2,{\ca... ...1t}, \ldots, (S[{\cal B}_2,{\cal B}_3] \; T[{\cal B}_1, {\cal B}_2])_{pt})^t.$$

Hence,

$$(S \circ T)\; [{\cal B}_1,{\cal B}_3] = \bigl[ \; [(S \circ T)\; ... ...{{\cal B}_3} \bigr] = S[{\cal B}_2, {\cal B}_3] \;\; T[{\cal B}_1, {\cal B}_2].$$

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PROPOSITION 4.4.2   Let $V$ be a finite dimensional vector space and let $T, S: V {\longrightarrow}V$ be a linear transformations. Then

$$\nu(T) + \nu(S) \geq \nu(T\circ S) \geq \max \{ \nu(T), \nu(S) \}.$$

Proof. We first prove the second inequality.
Suppose that ${\mathbf v}\in {\cal N}(S).$ Then $\; T\circ S({\mathbf v}) = T(S({\mathbf v})) = T({\mathbf 0}) = {\mathbf 0}.$ So, ${\cal N}(S) \subset {\cal N}(T\circ S).$ Therefore, $\nu(S) \leq \nu(T \circ S).$

Suppose $\dim (V) = n.$ Then using the rank-nullity theorem, observe that

$$\nu(T\circ S) \geq \nu(T) \Longleftrightarrow n - \nu(T\circ S) \leq n- \nu(T) \Longleftrightarrow \rho (T\circ S) \leq \rho (T).$$

So, to complete the proof of the second inequality, we need to show that ${\cal R}(T \circ S) \subset {\cal R}(T).$ This is true as ${\cal R}(S) \subset V.$

We now prove the first inequality.
Let $k = \nu(S)$ and let $\{{\mathbf v}_1, {\mathbf v}_2, \ldots, {\mathbf v}_k\}$ be a basis of ${\cal N}(S).$ Clearly, $\{{\mathbf v}_1, {\mathbf v}_2, \ldots, {\mathbf v}_k\} \subset {\cal N}(T\circ S)$ as $T({\mathbf 0}) = {\mathbf 0}.$ We extend it to get a basis $\{{\mathbf v}_1, {\mathbf v}_2, \ldots, {\mathbf v}_k, {\mathbf u}_1, {\mathbf u}_2, \ldots, {\mathbf u}_{\ell}\}$ of ${\cal N}(T\circ S).$

Claim: The set $\{S({\mathbf u}_1), S({\mathbf u}_2), \ldots, S({\mathbf u}_{\ell})\}$ is linearly independent subset of ${\cal N}(T).$

As ${\mathbf u}_1, {\mathbf u}_2, \ldots, {\mathbf u}_{\ell} \in {\cal N}(T\circ S),$ the set $\{S({\mathbf u}_1), S({\mathbf u}_2), \ldots, S({\mathbf u}_{\ell})\}$ is a subset of ${\cal N}(T).$ Let if possible the given set be linearly dependent. Then there exist non-zero scalars $c_1, c_2, \ldots, c_{\ell}$ such that

$$c_1 S({\mathbf u}_1) + c_2 S({\mathbf u}_2) + \cdots + c_{\ell} S({\mathbf u}_{\ell}) = {\mathbf 0}.$$

So, the vector $\sum\limits_{i=1}^{\ell} c_i {\mathbf u}_i \in {\cal N}(S)$ and is a linear combination of the basis vectors ${\mathbf v}_1, {\mathbf v}_2, \ldots, {\mathbf v}_k$ of ${\cal N}(S).$ Therefore, there exist scalars ${\alpha}_1, {\alpha}_2, {\alpha}_k$ such that

$$\sum\limits_{i=1}^{\ell} c_i {\mathbf u}_i = \sum\limits_{i=1}^{k} {\alpha}_i {\mathbf v}_i.$$

Or equivalently

$$\sum\limits_{i=1}^{\ell} c_i {\mathbf u}_i + \sum\limits_{i=1}^{k} (-{\alpha}_i) {\mathbf v}_i = {\mathbf 0}.$$

That is, the ${\mathbf 0}$ vector is a non-trivial linear combination of the basis vectors ${\mathbf v}_1, {\mathbf v}_2, \ldots, {\mathbf v}_k, {\mathbf u}_1, {\mathbf u}_2, \ldots, {\mathbf u}_{\ell}$ of ${\cal N}(T\circ S).$ A contradiction.

Thus, the set $\{S({\mathbf u}_1), S({\mathbf u}_2), \ldots, S({\mathbf u}_{\ell})\}$ is a linearly independent subset of ${\cal N}(T)$ and so $\nu(T) \geq \ell.$ Hence,

$$\nu(T \circ S) = k + \ell \leq \nu(S) + \nu(T).$$

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Recall from Theorem 4.1.8 that if $T$ is an invertible linear Transformation, then $T^{-1} : V {\longrightarrow}V$ is a linear transformation defined by $T^{-1}({\mathbf u}) = {\mathbf v}$ whenever $T({\mathbf v}) = {\mathbf u}.$ We now state an important result about inverse of a linear transformation. The reader is required to supply the proof (use Theorem 4.4.1).

THEOREM 4.4.3 (Inverse of a Linear Transformation)   Let $V$ be a finite dimensional vector space with ordered bases ${\cal B}_1$ and ${\cal B}_2.$ Also let $T: V {\longrightarrow}V$ be an invertible linear transformation. Then the matrix of $T$ and $T^{-1}$ are related by

$$T[{\cal B}_1, {\cal B}_2]^{-1} = T^{-1}[{\cal B}_2, {\cal B}_1].$$

EXERCISE 4.4.4   For the linear transformations given below, find the matrix $T[{\cal B},{\cal B}].$

1. Let ${\cal B}=\bigl((1,1,1), (1,-1,1), (1,1,-1)\bigr)$ be an ordered basis of ${\mathbb{R}}^3.$ Define $T: {\mathbb{R}}^3 {\longrightarrow}{\mathbb{R}}^3$ by $T(1,1,1) = (1,-1,1),$ $\; T(1,-1,1)=(1,1,-1),$ and $\; T(1,1,-1)=(1,1,1).$ Is $T$ an invertible linear transformation? Give reasons.
2. Let ${\cal B}=\bigl(1, x, x^2, x^3)\bigr)$ be an ordered basis of ${\cal P}_3({\mathbb{R}}).$ Define $T: {\cal P}_3({\mathbb{R}}) {\longrightarrow}{\cal P}_3({\mathbb{R}})$ by
   $$T(1) = 1, T(x) =1+x, T(x^2)= (1+x)^2, {\mbox{ and }} T(x^3)=(1+x)^3.$$
   Prove that $T$ is an invertible linear transformation. Also, find $T^{-1}[{\cal B},{\cal B}].$

Let $V$ be a vector space with $\dim (V) = n.$ Let ${\cal B}_1 = ({\mathbf u}_1, {\mathbf u}_2, \ldots, {\mathbf u}_n)$ and ${\cal B}_2 = ({\mathbf v}_1, {\mathbf v}_2, \ldots, {\mathbf v}_n \}$ be two ordered bases of $V.$ Recall from Definition 4.1.5 that $I: V {\longrightarrow}V$ is the identity linear transformation defined by $I ({\mathbf x}) = {\mathbf x}$ for every ${\mathbf x}\in V.$ Suppose ${\mathbf x}\in V$ with $[{\mathbf x}]_{{\cal B}_1} = (\alpha_1, \alpha_2, \ldots, \alpha_n)^t$ and $[{\mathbf x}]_{{\cal B}_2} = (\beta_1, \beta_2, \ldots, \beta_n)^t.$

We now express each vector in ${\cal B}_2$ as a linear combination of the vectors from ${\cal B}_1.$ Since ${\mathbf v}_i \in V,$ for $1 \leq i \leq n,$ and ${\cal B}_1$ is a basis of $V,$ we can find scalars $a_{ij}, 1\leq i, j \leq n$ such that

$${\mathbf v}_i = I ({\mathbf v}_i) = \sum\limits_{j=1}^n a_{ji} {\mathbf u}_j \;\; {\mbox{ for all}} \;\; i, 1 \leq i \leq n.$$

Hence, $[I({\mathbf v}_i)]_{{\cal B}_1} = [{\mathbf v}_i]_{{\cal B}_1} = ( a_{1i}, a_{2i}, \cdots, a_{ni} )^t$ and

$$I[{\cal B}_2, {\cal B}_1] = [ [I({\mathbf v}_1)]_{{\cal B}_1}, [I({\mathbf v}_2)]_{{\cal B}_1}, \ldots, [I({\mathbf v}_n)]_{{\cal B}_1} ]$$

$$= \begin{bmatrix}a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{... ...& \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{bmatrix}.$$

Thus, we have proved the following result.

THEOREM 4.4.5 (Change of Basis Theorem)   Let $V$ be a finite dimensional vector space with ordered bases ${\cal B}_1 = ({\mathbf u}_1, {\mathbf u}_2, \ldots, {\mathbf u}_n \}$ and ${\cal B}_2 = ({\mathbf v}_1, {\mathbf v}_2, \ldots, {\mathbf v}_n \}.$ Suppose ${\mathbf x}\in V$ with $[{\mathbf x}]_{{\cal B}_1} = (\alpha_1, \alpha_2, \ldots, \alpha_n)^t$ and $[{\mathbf x}]_{{\cal B}_2} = (\beta_1, \beta_2, \ldots, \beta_n)^t.$ Then

$$[{\mathbf x}]_{{\cal B}_1} = I[{\cal B}_2, {\cal B}_1] \; \; [{\mathbf x}]_{{\cal B}_2}.$$

Equivalently,

$$\begin{bmatrix}\alpha_1 \\ \alpha_2\\ \vdots \\ \alpha_n \end{b... ...rix} \; \begin{bmatrix}\beta_1 \\ \beta_2 \\ \vdots \\ \beta_n \end{bmatrix}.$$

Note: Observe that the identity linear transformation $I: V {\longrightarrow}V$ defined by $I ({\mathbf x}) = {\mathbf x}$ for every ${\mathbf x}\in V$ is invertible and

$$I[{\cal B}_2, {\cal B}_1]^{-1} = I^{-1}[{\cal B}_1, {\cal B}_2]= I[{\cal B}_1, {\cal B}_2].$$

Therefore, we also have

$$[{\mathbf x}]_{{\cal B}_2} = I[{\cal B}_1, {\cal B}_2] \; [{\mathbf x}]_{{\cal B}_1}.$$

Let $V$ be a finite dimensional vector space and let ${\cal B}_1$ and ${\cal B}_2$ be two ordered bases of $V.$ Let $T: V {\longrightarrow}V$ be a linear transformation. We are now in a position to relate the two matrices $T[{\cal B}_1, {\cal B}_1]$ and $T[{\cal B}_2, {\cal B}_2].$

THEOREM 4.4.6   Let $V$ be a finite dimensional vector space and let ${\cal B}_1 = ({\mathbf u}_1, {\mathbf u}_2, \ldots, {\mathbf u}_n)$ and ${\cal B}_2 = ({\mathbf v}_1, {\mathbf v}_2, \ldots, {\mathbf v}_n)$ be two ordered bases of $V.$ Let $T: V {\longrightarrow}V$ be a linear transformation with $B= T[{\cal B}_1, {\cal B}_1]$ and $C = T[{\cal B}_2, {\cal B}_2]$ as matrix representations of $T$ in bases ${\cal B}_1$ and ${\cal B}_2.$

Also, let $A = [a_{ij}] = I[{\cal B}_2, {\cal B}_1],$ be the matrix of the identity linear transformation with respect to the bases ${\cal B}_1$ and ${\cal B}_2.$ Then $B A = A C.$ Equivalently $B = A C A^{-1}.$

Proof. For any ${\mathbf x}\in V$ , we represent $[T({\mathbf x})]_{{\cal B}_2}$ in two ways. Using Theorem 4.2.1, the first expression is

$$[T({\mathbf x})]_{{\cal B}_2} = T[{\cal B}_2, {\cal B}_2] \; [{\mathbf x}]_{{\cal B}_2}.$$ (4.4.1)

Using Theorem 4.4.5, the other expression is

$$[T({\mathbf x})]_{{\cal B}_2} = I[{\cal B}_1, {\cal B}_2] \; [T({\mathbf x})]_{{\cal B}_1}$$

$$= I[{\cal B}_1, {\cal B}_2] \; T[{\cal B}_1, {\cal B}_1] \; [{\mathbf x}]_{{\cal B}_1}$$

$$= I[{\cal B}_1, {\cal B}_2] \; T[{\cal B}_1, {\cal B}_1] \; I[{\cal B}_2, {\cal B}_1] \; [{\mathbf x}]_{{\cal B}_2}.$$ (4.4.2)

Hence, using (4.4.1) and (4.4.2), we see that for every ${\mathbf x}\in V,$

$$I[{\cal B}_1, {\cal B}_2] \; T[{\cal B}_1, {\cal B}_1] \; I[{\cal... ...hbf x}]_{{\cal B}_2} =T[{\cal B}_2, {\cal B}_2] \; [{\mathbf x}]_{{\cal B}_2}.$$

Since the result is true for all ${\mathbf x}\in V,$ we get

$$I[{\cal B}_1, {\cal B}_2] \; T[{\cal B}_1, {\cal B}_1] \; I[{\cal B}_2, {\cal B}_1] = T[{\cal B}_2,{\cal B}_2].$$ (4.4.3)

That is, $A^{-1} B A = C$ or equivalently $A C A^{-1} = B.$ height6pt width 6pt depth 0pt

Another Proof:

Let $B= [b_{ij}]$ and $C = [c_{ij}].$ Then for $1 \leq i \leq n,$

$$T({\mathbf u}_i) = \sum\limits_{j=1}^n b_{ji} {\mathbf u}_j \; {\mbox{ and }} \; T({\mathbf v}_i) = \sum\limits_{j=1}^n c_{ji} {\mathbf v}_j.$$

So, for each $j, 1 \leq j \leq n,$

$$T({\mathbf v}_j) = T( I({\mathbf v}_j)) = T( \sum\limits_{k=1}^n a_{kj} {\mathbf u}_k) = \sum\limits_{k=1}^n a_{kj} T({\mathbf u}_k)$$

$$= \sum\limits_{k=1}^n a_{kj} (\sum\limits_{\ell=1}^n b_{\ell k} {\m... ...um\limits_{\ell=1}^n ( \sum\limits_{k=1}^n b_{\ell k} a_{kj} ) {\mathbf u}_\ell$$

and therefore,

$$[T({\mathbf v}_j)]_{{\cal B}_1} = \begin{bmatrix}\sum\limits_{k=1... ...matrix} = B \begin{bmatrix}a_{1j} \\ a_{2j} \\ \vdots \\ a_{nj} \end{bmatrix}.$$

Hence $T[{\cal B}_2,{\cal B}_1] = B A.$

Also, for each $j, 1 \leq j \leq n,$

$$T({\mathbf v}_j) = \sum\limits_{k=1}^n c_{kj} {\mathbf v}_k = \sum\limits_{k=1}^n c_... ...\sum\limits_{k=1}^n c_{kj} (\sum\limits_{\ell=1}^n a_{\ell k} {\mathbf u}_\ell)$$

$$= \sum\limits_{\ell=1}^n ( \sum\limits_{k=1}^n a_{\ell k} c_{kj} ) {\mathbf u}_\ell$$

and so

$$[T({\mathbf v}_j)]_{{\cal B}_1} = \begin{bmatrix}\sum\limits_{k=1... ...matrix} = A \begin{bmatrix}c_{1j} \\ c_{2j} \\ \vdots \\ c_{nj} \end{bmatrix}.$$

This gives us $T[{\cal B}_2,{\cal B}_1] = A C.$ We thus have $A C = T[{\cal B}_2,{\cal B}_1] = B A.$ height6pt width 6pt depth 0pt

Let $V$ be a vector space with $\dim(V) = n,$ and let $T: V {\longrightarrow}V$ be a linear transformation. Then for each ordered basis ${\cal B}$ of $V,$ we get an $n \times n$ matrix $T[{\cal B},{\cal B}].$ Also, we know that for any vector space we have infinite number of choices for an ordered basis. So, as we change an ordered basis, the matrix of the linear transformation changes. Theorem 4.4.6 tells us that all these matrices are related.

Now, let $A$ and $B$ be two $n \times n$ matrices such that $P^{-1} A P = B$ for some invertible matrix $P.$ Recall the linear transformation $T_A: {\mathbb{R}}^n {\longrightarrow}{\mathbb{R}}^n$ defined by $T_A({\mathbf x}) = A {\mathbf x}$ for all ${\mathbf x}\in {\mathbb{R}}^n.$ Then we have seen that if the standard basis of ${\mathbb{R}}^n$ is the ordered basis ${\cal B},$ then $A = T_A[{\cal B},{\cal B}].$ Since $P$ is an invertible matrix, its columns are linearly independent and hence we can take its columns as an ordered basis ${\cal B}_1.$ Then note that $B = T_A[{\cal B}_1, {\cal B}_1].$ The above observations lead to the following remark and the definition.

Remark 4.4.7   The identity (4.4.3) shows how the matrix representation of a linear transformation $T$ changes if the ordered basis used to compute the matrix representation is changed. Hence, the matrix $I[{\cal B}_1, {\cal B}_2]$ is called the ${\cal B}_1:{\cal B}_2$ change of basis matrix.

DEFINITION 4.4.8 (Similar Matrices)   Two square matrices $B$ and $C$ of the same order are said to be similar if there exists a non-singular matrix $P$ such that $B = P C P^{-1}$ or equivalently $B P = P C.$

Remark 4.4.9   Observe that if $A = T[{\cal B},{\cal B}]$ then

$$\{S^{-1} A S : S {\mbox{ is }} n \times n {\mbox{ invertible matrix }} \}$$

is the set of all matrices that are similar to the given matrix $A.$ Therefore, similar matrices are just different matrix representations of a single linear transformation.

EXAMPLE 4.4.10  

1. Consider ${\cal P}_2({\mathbb{R}}),$ with ordered bases
   $${\cal B}_1 = \bigl(1, 1+x, 1+x+x^2\bigr) \;\; {\mbox{ and }} \;\; {\cal B}_2 = \bigl( 1 + x - x^2, 1+ 2 x + x^2, 2 + x + x^2\bigr).$$
   Then
   $$[1+x - x^2]_{{\cal B}_1} = 0 \cdot 1 + 2 \cdot (1 + x) + (-1) \cdot (1+x+x^2) = (0,2, -1)^t,$$
   $$[1+2x+x^2]_{{\cal B}_1} = (-1) \cdot 1 + 1 \cdot (1+x) + 1 \cdot (1+x+x^2) = (-1,1,1)^t, \; {\mbox{ and}}$$
   $$[2+x+x^2]_{{\cal B}_1} = 1 \cdot 1 + 0 \cdot (1+x) + 1 \cdot (1+x+x^2) =(1,0,1)^t.$$
   Therefore,
   

   $$I[{\cal B}_2, {\cal B}_1] = [ [I(1 + x - x^2)]_{{\cal B}_1}, [I(1 + 2x + x^2)]_{{\cal B}_1}, [I(2 + x + x^2)]_{{\cal B}_1} ]$$

   $$= [ [1 + x - x^2]_{{\cal B}_1}, [1+ 2x + x^2]_{{\cal B}_1}, [2 + x + x^2]_{{\cal B}_1} ]$$

   $$= \begin{bmatrix}0 & -1 & 1 \\ 2 & 1 & 0 \\ -1 & 1 & 1 \end{bmatrix}.$$

   
   Find the matrices $T[{\cal B}_1, {\cal B}_1]$ and $T[{\cal B}_2, {\cal B}_2].$ Also verify that
   

   $$T[{\cal B}_2,{\cal B}_2] = I[{\cal B}_1, {\cal B}_2] \; T[{\cal B}_1, {\cal B}_1] \; I[{\cal B}_2, {\cal B}_1]$$

   $$= I^{-1}[{\cal B}_2, {\cal B}_1] \; T[{\cal B}_1, {\cal B}_1] \; I[{\cal B}_2, {\cal B}_1].$$

   
   
2. Consider two bases ${\cal B}_1= \bigl((1,0,0), (1,1,0), (1,1,1)\bigr)$ and ${\cal B}_2 = \bigl((1,1,-1), (1,2,1), (2,1,1) \bigr)$ of ${\mathbb{R}}^3.$ Suppose $T: {\mathbb{R}}^3 {\longrightarrow}{\mathbb{R}}^3$ is a linear transformation defined by
   $$T((x,y,z)) = (x + y, x+ y + 2 z, y-z).$$
   Then
   $$T[{\cal B}_1, {\cal B}_1] = \begin{bmatrix}0 & 0 & -2 \\ 1 & 1 &... ...}-{4/5} & 1 & {8/5} \\ -{2/5} & 2 & {9/5} \\ {8/5} & 0 & -{1/5} \end{bmatrix}.$$
   Find $I[{\cal B}_1, {\cal B}_2]$ and verify,
   $$I[{\cal B}_1, {\cal B}_2] \; T[{\cal B}_1, {\cal B}_1] \; I[{\cal B}_2, {\cal B}_1] = T[{\cal B}_2,{\cal B}_2].$$
   Check that,
   $$T[{\cal B}_1, {\cal B}_1] \; I[{\cal B}_2, {\cal B}_1] = I[{\cal ... ... B}_2] = \begin{bmatrix}2 & -2 & -2 \\ -2 & 4 & 5 \\ 2 & 1 & 0 \end{bmatrix}.$$

EXERCISE 4.4.11  

1. Let $V$ be an $n$ -dimensional vector space and let $T: V {\longrightarrow}V$ be a linear transformation. Suppose $T$ has the property that $T^{n-1} \neq {\mathbf 0}$ but $T^n = {\mathbf 0}.$
   1. Then prove that there exists a vector ${\mathbf u}\in V$ such that the set
      $$\{{\mathbf u}, T({\mathbf u}), \ldots, T^{n-1}({\mathbf u})\}$$
      is a basis of $V.$
   2. Let ${\cal B}= ({\mathbf u}, T({\mathbf u}), \ldots, T^{n-1}({\mathbf u})).$ Then prove that
      $$T[{\cal B}, {\cal B}] = \begin{bmatrix}0 & 0 & 0 & \cdots & 0 \\ ... ...ots & & \ddots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 & 0 \\ \end{bmatrix}.$$

   3. Let $A$ be an $n \times n$ matrix with the property that $A^{n-1} \neq {\mathbf 0}$ but $A^n = {\mathbf 0}.$ Then prove that $A$ is similar to the matrix given above.
2. Let $T: {\mathbb{R}}^3 {\longrightarrow}{\mathbb{R}}^3$ be a linear transformation given by
   $$T((x,y,z)) = (x + y + 2z, x - y - 3z, 2x + 3y + z).$$
   Let ${\cal B}$ be the standard basis and ${\cal B}_1 = \bigl( (1,1,1), (1,-1,1), (1,1,2) \bigr)$ be another ordered basis.
   1. Find the matrices $T[{\cal B}, {\cal B}]$ and $T[{\cal B}_1, {\cal B}_1].$
   2. Find the matrix $P$ such that $P^{-1} T[{\cal B},{\cal B}] \; P = T[{\cal B}_1, {\cal B}_1].$
3. Let $T: {\mathbb{R}}^3 {\longrightarrow}{\mathbb{R}}^3$ be a linear transformation given by
   $$T((x,y,z)) = (x, x + y, x+y+z).$$
   Let ${\cal B}$ be the standard basis and ${\cal B}_1= \bigl((1,0,0), (1,1,0), (1,1,1)\bigr)$ be another ordered basis.
   1. Find the matrices $T[{\cal B}, {\cal B}]$ and $T[{\cal B}_1, {\cal B}_1].$
   2. Find the matrix $P$ such that $P^{-1} T[{\cal B},{\cal B}] \; P = T[{\cal B}_1, {\cal B}_1].$
4. Let ${\cal B}_1 = \bigl( (1,2,0), (1, 3, 2), (0, 1, 3) \bigr)$ and ${\cal B}_2 = \bigl( (1,2,1), (0,1,2), (1, 4, 6) \bigr)$ be two ordered bases of ${\mathbb{R}}^3.$
   1. Find the change of basis matrix $P$ from ${\cal B}_1$ to ${\cal B}_2.$
   2. Find the change of basis matrix $Q$ from ${\cal B}_2$ to ${\cal B}_1.$
   3. Verify that $P Q = I = Q P.$
   4. Find the change of basis matrix from the standard basis of ${\mathbb{R}}^3$ to ${\cal B}_1.$ What do you notice?