If $T$ is a linear map and $\mathcal M(T)$ is its matrix, what exactly does the multiplication $\mathcal M(T)v$ mean?

Question

If $T \in \mathcal L(V,W)$ and $\mathcal M(T)$ is its matrix with bases $(v_1,\dots,v_n),(w_1,\dots,w_m)$, then what does the multiplication $\mathcal M(T) v$ mean, with $v \in V$? Does this equal $T(v)$? Why or why not?

Thanks in advance.

Answer 1

When you write $\mathcal M(T)$ you are expressing $T$ in relation to a basis of $V$ and a basis of $W$.

In this case, from your notations I can infer $V$ is $n$ dimensional and $W$ is $m$ dimensional, so $\mathcal M(T)$ is a $m\times n$ matrix.

An object of $V$ can be represented in relation to a base as a vector, specifically a vector of length $n$, so a $n\times 1$ matrix. With this form, you can perform the multiplication $\mathcal M(T)v$ since it is a matrix-vector multiplication with the right dimensions.

It is equivalent to compute $Tv$ where $v$ is not represented wrt a basis and it will return an abstract element of $W$, but in the first case, $\mathcal M(T)v$ willl yeld a $m\times 1$ vector, that is an element of $W$ represented wrt a basis.

Answer 2

Let me give you an example in the case $V=W$. A plane given by its normal vector $n$ is defined by $\langle n,x\rangle=0$. The reflection $T$ on this plane takes vector $v$ to $$T(v)=v-2\frac{\langle v,n\rangle}{\|n\|^2}n.$$ That is a coordinate-free representation of $T$.

Now let $V=W=\mathbb R^2$ and $n=\begin{pmatrix}3\\4\end{pmatrix}$. You might represent $T$ by the standard basis, in which case you’ll come up with a somewhat unpleasant matrix $$\frac{1}{25}\begin{pmatrix}7&-24\\-24&7\end{pmatrix}.$$

If you choose $\left\{\begin{pmatrix}-4\\3\end{pmatrix},\begin{pmatrix}3\\4\end{pmatrix}\right\}$ instead, the representation would be $$\begin{pmatrix}1&0\\0&-1\end{pmatrix}.$$