What is the actual definition of "matrix representation in another basis"?

Question

A common type of question in linear algebra textbooks are these:

Let $A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}$ and consider a basis $B = \left\{b_1 = \begin{bmatrix} b_{11} \\ b_{12} \end{bmatrix}, b_2 = \begin{bmatrix} b_{21} \\ b_{22} \end{bmatrix}\right\}$

Find the matrix representation of A in basis

So what is the definition of: Matrix representation of A in basis ?

After some digging, I found an unsatisfactory definition:

The matrix representation of $A \in \mathbb{R}^{n \times n}$ in basis $B$ is the matrix $\widetilde A = [\widetilde a_{ji}]$ such that,

\begin{equation} A b_i = \sum\limits_{j = 1}^n {\widetilde a_{ji}} b_i, \forall i = 1, \ldots, n \end{equation}

There seems to be little or no motivation at all behind this definition. For instance, why am I multiplying $A$ by a basis vector $b_i$ and not any arbitrary vector? Certainly, any $Ax$, $x \in \mathbb{R}^n$, could be represented in the basis $B$, no?

Can someone provide a good definition to Matrix representation of A in basis and I would appreciate it if a reference is included.

Answer 1

Given a matrix $A$, its representation in a basis $V = (v_1,\ldots,v_n)$ is a matrix $B$ that takes $V$-coordinates of any element $v$ to the $V$ coordinates of $Av$. In terms of notations let's say

$v = a_1v_1 + \ldots + a_nv_n = \begin{bmatrix}v_1&\ldots&v_n\end{bmatrix} \begin{bmatrix}a_1\\ \vdots \\ a_n\end{bmatrix}$

Then $[a_1,\ldots, a_n]'$ is what you call the coordinates of $v$ with respect to the basis $V$ which I called $V$-coordinates. Now let's say

$Av =b_1v_1 + \ldots + b_nv_n $

Then matrix $B$ must send $[a_1,\ldots, a_n]'$ to $[b_1,\ldots, b_n]'$. In notations it means

$B \begin{bmatrix}a_1\\ \vdots \\ a_n\end{bmatrix} = \begin{bmatrix}b_1\\ \vdots \\ b_n\end{bmatrix}$

A compact way to put it is as follows

$Av = A \begin{bmatrix}v_1&\ldots&v_n\end{bmatrix} \begin{bmatrix}a_1\\ \vdots \\ a_n\end{bmatrix} = b_1v_1 + \ldots + b_nv_n = \begin{bmatrix}v_1&\ldots&v_n\end{bmatrix} \begin{bmatrix}b_1\\ \vdots \\ b_n\end{bmatrix} = \begin{bmatrix}v_1&\ldots&v_n\end{bmatrix} B\begin{bmatrix}a_1\\ \vdots \\ a_n\end{bmatrix} $

Thus ultimately

$A \begin{bmatrix}v_1&\ldots&v_n\end{bmatrix} \begin{bmatrix}a_1\\ \vdots \\ a_n\end{bmatrix} = \begin{bmatrix}v_1&\ldots&v_n\end{bmatrix} B\begin{bmatrix}a_1\\ \vdots \\ a_n\end{bmatrix} $

For all $[a_1,\ldots, a_n]'.$ This in particular will give you

$A \begin{bmatrix}v_1&\ldots&v_n\end{bmatrix} = \begin{bmatrix}v_1&\ldots&v_n\end{bmatrix} B $