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.

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

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


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


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


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


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))


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