Understanding the concept of isomorphism between Hom(V,W) and $M_{m\times n}$

Question

I'd appreciate a clarification about the following issue. It's known that Hom(V,W) is isomorphic to $M_{m\times n}$ Correct me if I'm wrong, but as I get it, the meaning of the above statement is that every linear transformation from V to W is represented uniquely by an mxn matrix, and vice versa. However, I'm having a hard time understanding something. Since we are free to choose any bases for V and W, consequently we get different representation matrices. How does it not contradict the statement mentioning the isomorphism, according to which, as I get it (and probably not correctly), there is a unique respective matrix?

Thanks in advance!

Answer 1

The statement means there is an isomorphism $\phi\colon{\rm Hom}\,(V,W)\to M_{m\times n}$. (It does not preclude many isomorphisms.)

Let $B$ and $B'$ be bases for $V$ and $W$ respectively. Then there is *one" isomorphism corresponding to $B,B'$, let us denote it by $\phi_{B,B'}$.

It is : $\phi_{B,B'}(T) = A$ where $A$ is the matrix of $T\colon V\to W$ for the choice of bases $B$ and $B'$ on $V$ and $W$ respectively.

Answer 2

Important that we place all our key statements in yellow boxes.

The first statement says that there is an(at least one) isomorphism an between $\mbox{Hom} (V,W)$ and $M_{m \times n}$. Such an isomorphism would, by its injective nature, assign to every linear transformation, a unique matrix.

unique up to choice of isomorphism!

And the choice of isomorphism is not unique...

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The point, however, is that :

Every choice of bases on $V$ and $W$ do The fact that for every choice of basis on $V$ and $W$ isomorphism between the above two spaces, does not therefore contradict the first statement at all.

In fact, it strengthens it, since now we have a plethora of isomorphisms to choose from.

Therefore, if we have a linear transformation, then depending on which isomorphism between the two spaces we are choosing , we will get a different matrix representation for that linear transformation.

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As Batominovski points out in the comments, we note that the isomorphism between the spaces depends on the choice of basis. Is it possible to write down an isomorphism which does not make use of a basis on $V$ and $W$ i.e. does not require more additional information about $V$ and $W$ other than the fact that they are vector spaces? Turns out this is not the case : a very interesting point. However, replacing matrices with another space $V^* \otimes W$ does give such an isomorphism(which we refer to as "canonical", for the reason that it avoids using further information about $V$ and $W$, and is "natural" in some sense), which is why it is sometimes more desirable to speak of the above space rather than the space of matrices.