Why linear maps act like matrix multiplication?

In Linear Algebra Done Right, it said

Suppose $$T \in \mathcal{L}(V,W)$$ and $$v \in V$$. Suppose $$v_1,...,v_n$$ is a basis of $$V$$ and $$w_1,...,w_m$$ is a basis of $$W$$. Then $$M(Tv) = M(T)M(v)$$

$$M(T)$$ is the m-by-n matrix whose entries $$A_{j,k}$$ are defined by $$Tv_k = A_{1,k}w_1 + ... + A_{m,k}w_m$$ suppose $$T \in \mathcal{L}(V,W)$$ and $$v_1,...,v_n$$ is a basis of $$V$$ and $$w_1,...,w_m$$ is a basis of $$W$$.

$$M(v)$$ is the matrix of vector $$v$$.

I generally follow the following proof:

Suppose $$v = c_1v_1 + ... + c_nv_n$$, where $$c_1,...,c_n \in \mathbb{F}$$. Thus $$Tv = c_1Tv_1 +...+c_nTv_n$$

Hence

$$\begin{equation} \begin{split} M(Tv) &= c_1M(Tv_1) + ...+ c_nM(Tv_n)\\ & = c_1M(T)_{.,1} +...+c_nM(T)_{.,n} \\ & = M(T)M(v) \end{split} \end{equation}$$

But I have questions on the meaning of the proof. The book said it means each m-by-n matrix $$A$$ induces a linear map from $$\mathbb{F}^{n,1}$$ to $$\mathbb{F}^{m,1}$$. The result can be used to think of every linear map as a matrix multiplication map after suitable relabeling via the isomorphisms given by $$M$$.

1. Is the shape of $$M(Tv)$$ m by 1, $$M(T)$$ m by n, and $$M(v)$$ n by 1?
2. What is meant by suitable relabeling via the isomorphisms given by $$M$$? Does it just mean $$M(T)$$ is a isomorphism linear map between $$M(v)$$ and $$M(Tv)$$?
• In your blockquote, you haven't told us what $M$ means. – Gerry Myerson Apr 7 at 7:02
• @GerryMyerson revised – JOHN Apr 7 at 7:13
• Sorry, I don't know what it means for $M(v)$ to be the matrix of the vector $v$. What's the matrix of the vector $(1,2,3)$? – Gerry Myerson Apr 7 at 7:15
• @GerryMyerson Knowing Axler, it will be the coordinate vector for the given basis (in this case, $v_1, \ldots, v_n$). For example, if the basis is $(0, 0, 1), (0, 1, 1), (1, 1, 1)$ in the space $\Bbb{R}^3$, then $M(1, 2, 3) = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$. – Theo Bendit Apr 7 at 7:17

In answer to your first question, yes to all three: $$v$$ is an element of the $$n$$-dimensional space $$V$$, so the coordinate vector with respect to the basis will be an $$n \times 1$$ column vector. Similarly, $$Tv \in W$$, which is an $$n$$-dimensional space, so $$M(Tv)$$ will be an $$m \times 1$$ column vector. Finally, $$M(T)$$ is built from transforming the $$n$$ basis vectors of the domain, forming each an $$m \times 1$$ coordinate column vector, which are put into an $$m \times n$$ matrix. The process of applying $$T$$ to a vector $$v \in V$$ is the top row of the diagram. However, there's a parallel process happening between $$\Bbb{F}^n$$ and $$\Bbb{F}^m$$, mirroring the same process.

The isomorphism being referred to are the double arrows, taking us between $$V$$ and $$\Bbb{F}^n$$ and $$W$$ and $$\Bbb{F}^m$$, by way of coordinate vectors. The coordinate vector map on $$V$$ is a linear map between $$V$$ and $$\Bbb{F}^n$$ that is invertible, making it an isomorphism (and similarly for $$W$$). That is, the two spaces are structurally identical, and anything we can do with one space, we can view it in the other.

In $$V$$, we have some abstract vectors, and an abstract linear transformation $$T$$ that maps vectors in $$V$$ to vectors in $$W$$. However, using this isomorphism, we can view $$V$$ slightly differently as $$\Bbb{F}^n$$, and similarly for $$W$$, which means $$T$$ boils down to a linear map from $$\Bbb{F}^n$$ to $$\Bbb{F}^m$$, which can be characterised as matrix multiplication. The matrix, in particular, is $$M(T)$$.

• Very nice graph! – JOHN Apr 7 at 9:35
• Upvote for the commutative paint diagram! – Jannik Pitt Apr 7 at 13:27
• it seems that T and M(T) is also isomorphic ? – JOHN Apr 7 at 22:32
• Isomorphisms are maps between vector spaces; only vector spaces can be isomorphic (at least, until you study category theory). The map $M$ (given fixed bases) is itself an isomorphism between the space of linear maps from $V$ to $W$ and the $m \times n$ matrices, but even then, it's not really correct to say that a map $T$ is "isomorphic" to a matrix $M(T)$. It would be more accurate (though not conventional) to describe the map $T$ as "similar" to the matrix $M(T)$, in a similar sense to similar matrices (bear in mind, $T$ is not a matrix). – Theo Bendit Apr 7 at 23:02
• @JOHN Simply put, vectors cannot be isomorphic. Maps cannot be isomorphic. Only spaces can be isomorphic. – Theo Bendit Apr 7 at 23:06
1. Yes, those would be the shapes of those vectors when represented as matrices. Given that we're multiply by vectors on the right.

2. There is a theorem that if $$V$$ is an $$n-$$dimensional vector space over a field $$F,$$ then $$V$$ is isomorphic to $$F^n.$$ Here the isomorphic mappings assign coordinates to our vectors and our linear transformation. It doesn't mean that $$M$$ is an isomorphism between $$M(v)$$ and $$M(Tv).$$ These are particular vectors. The map $$M$$ actually induces an isomorphism from $$V\to F^n$$, isomorphism from $$T\to F^{n\times m}$$, and an isomorphism from $$W\to F^m.$$

I actually like the way that this is done. The Author is telling you that you're representation of $$T$$ by a matrix depends on your choice of basis in $$F^n$$. A fact that is important to remember.