# Change of Basis Matrix: Prove $[T]_{\gamma^{'}}^{\beta^{'}} = P^{-1}[T]_{\gamma}^{\beta} Q$

Let $$T : V → W$$ be a linear transformation from a finite-dimensional vector space $$V$$ to a finite-dimensional vector space $$W$$. Let $$\beta$$ and $$\beta^{'}$$ be ordered bases for $$V$$ and let $$\gamma$$ and $$\gamma^{'}$$ be ordered bases for $$W$$.

Prove $$[T]_{\gamma^{'}}^{\beta^{'}} = P^{-1}[T]_{\gamma}^{\beta} Q$$

where $$Q$$ is the matrix that changes $$\beta^{'}$$-coordinates to $$\beta$$-coordinates and $$P$$ is the matrix that changes $$\gamma^{'}$$-coordinates to $$\gamma$$-coordinates.

I'm having a hard time keeping straight the notation of change of basis matrices and what exactly they represent. I know I can write $$Q$$ as $$[I_V]_{\beta}^{\beta^{'}}$$ (simlarly for P) and I think that might help, but I'm not sure where to start! Any help is appreciated.

• My answer here may help. – Theo Bendit Oct 9 '18 at 0:41

Both $$P,Q$$ are matrices from identity transformations, so $$Q=[I_V]_\beta^{\beta'}$$ and

\begin{align*}P^{-1}=([I_W]_\gamma^{\gamma'})^{-1}=[I_W^{-1}]_{\gamma'}^\gamma=[I_W]_{\gamma'}^\gamma\end{align*}

since the inverse of $$I_W$$ is itself. Now

\begin{align*}P^{-1}[T]_{\gamma}^{\beta} Q&=[I_W]_{\gamma'}^\gamma[T]_\gamma^\beta[I_V]_\beta^{\beta'}\\ &=[I_WT]_{\gamma'}^{\beta}[I_V]_\beta^{\beta'}\\ &=[T]_{\gamma'}^\beta[I_V]_\beta^{\beta'}\\ &=[TI_V]_{\gamma'}^{\beta'}\\ &=[T]_{\gamma'}^{\beta'}\end{align*}

since the definition of matrix multiplication means composition of (linear-)transformations.

In case you don't know why, here is bonus about matrix multiplication (But notice that the notation I used is different from your, for $$T:V_\beta\to W_\gamma$$ I written $$[T]_\beta^\gamma$$):

Let $$T:\mathsf{V}\to\mathsf{W}$$ a linear transformation and $$\beta=\{v_1,v_2\}, \gamma=\{w_1,w_2\}$$ are bases of $$\mathsf{V}, \mathsf{W}$$ respectively. The value of interest is

$$T(v).$$

Let $$v=xv_1+yv_2$$, then

\begin{align} T(v)&=T(xv_1+yv_2)\\ &=xT(v_1)+yT(v_2). \end{align}

No matter what value of $$v$$ is, $$T(v_1),T(v_2)$$ are needed, the notation can be simplified. Let

$$T(v_1)=aw_1+bw_2,\\ T(v_2)=cw_1+dw_2,$$

represent $$T(v_1), T(v_2)$$ in columns

$$\begin{array}{ll} T(v_1) & T(v_2)\\ aw_1 & cw_1\\ {+} & {+}\\ bw_2 & dw_2\\ \end{array}$$

Put $$w_1, w_2$$ on the left side as a note and omit the plus signs

$$\begin{array}{lll} & T(v_1) & T(v_2) \\ w_1 & a & c \\ w_2 & b & d \\ \end{array}$$

Since $$T(v)=xT(v_1)+yT(v_2)$$

$$\begin{array}{} & x & y \\ & T(v_1)\ \ \ +& T(v_2)\ \ = & T(v) \\ w_1 & a & c & e \\ w_2 & b & d & f \\ \end{array}$$

An $$\color{blue}{operation}$$ can be defined such that

$$e=\color{blue}{x}a+\color{blue}{y}c\\ f=\color{blue}{x}b+\color{blue}{y}d$$

that is

$$\begin{bmatrix}e\\f\end{bmatrix} {=} \begin{bmatrix}a & c\\b & d\end{bmatrix} \color{blue}{oper.} \begin{bmatrix}\color{blue}{x}\\\color{blue}{y}\end{bmatrix},$$

The order $$w_1, w_2$$ are listed is associated to this notation, so the idea of ordered basis is required to denote the linear transformation matrix

$$\large[T]_\beta^\gamma$$

which here the lower-script $$\beta$$ means the matrix will work as a transformation when you multiply it with

$$\large[v]_\beta$$

the coordinate vector relative to $$\beta$$, and then you will get the output $$w$$ as coordinate vector relative to $$\gamma$$. We like this operation, so the $$\color{blue}{operation}$$ is defined s.t.

$$\large[T(v)]_\gamma = [T]_\beta^{\gamma} \color{blue}{\Large\cdot} [v]_\beta$$

Now open your book, find the definition of $$\color{blue}{matrix\ multiplication}$$ again and appreciate it.

--

Since for a linear transformation matrix $$\large[U]_\alpha^\beta$$, we can decompose it into column vectors from left to right, each as a coordinate vector, the composition of $$\large[T]_{\beta}^{\gamma}$$ and $$\large[U]_{\alpha}^{\beta}$$ now is

\begin{align*} \large[T]_\beta^\gamma[U]_\alpha^\beta &=\large[T]_\beta^\gamma[U(a_1)]_\beta \Bigg| [T]_\beta^\gamma[U(a_2)]_\beta\Bigg|\dots\Bigg|[T]_\beta^\gamma[U(a_n)]_\beta\\ &=\large[T(U(a_1))]_\gamma\Bigg|[T(U(a_2))]_\gamma\Bigg|\dots\Bigg|[T(U(a_n))]_\gamma\\ &=\large[TU(a_1)]_\gamma\Bigg|[TU(a_2)]_\gamma\Bigg|\dots\Bigg|[TU(a_n)]_\gamma\\ &=\large[TU]_\alpha^\gamma \end{align*}

(For those vertical bars I meant augmentation of column-vector(s))