No determinant of a homomorphism even when $\dim V = \dim W$?

Question

Why is there no determinant of a homomorphism $f: V \to W$, even when $\dim V = \dim W$?

I don't know how the concepts in linear algebra are related to each other and I am new to it.

Thanks

Answer 1

Because the determinant should be independent of basis. Intuitively, the determinant measures how volumes are scaled under a linear transformation. The important thing here is that it's only about scaling, not about absolute volumes. The determinant only compares volumes before and after transforming. It says: Whatever volume the set $X\subset V$ has, the set $f(X)$ has $\vert\det f\vert$ times that volume. We don't actually have a natural way to give $X$ a volume. We could choose some basis and say that the parallelepiped spanned by it has volume $1$. But we could also choose a completely different basis and say that the parallelepiped spanned by that basis has volume 1. In reality, we do so by switching between cubic feet, cubic meters, liters, etc.: We choose orthogonal vectors of length 1 ft., 1m or 10cm which span 3d space, and then say that the parallelepiped spanned by them has volume 1 (cubic foot, cubic meter or liter). If we stay within the same vector space, this is not a problem: The determinant doesn't care which basis we choose, as long as we stick with it, since it only measures how volumes are scaled, not their absolute values. But this breaks if we have two different vector spaces. Now we have to specify two bases to compare volumes, and how volume changes depends on our choices. For instance, if, say, the unit cube wrt. some basis in $V$ is mapped to the parallelepiped spanned by the basis $B_W=\{v_1,\dots,v_n\}$ of $W$ and we measure how volume changes here, we I will get something different when measuring wrt. this same basis than when we choose the basis $B'_W=\{2v_1,\dots,2v_n\}$, by a factor of $2^{-n}$. So a determinant could no longer be independent of basis, which is bad. So we just don't define it in this case.

Answer 2

Generally you can only take the determinant of matrices. The reason why this definition carries over to endomorphisms is because this way it turns out that it doesn't depend on the basis: If you choose a different basis of your vector space, that affects both domain and codomain of your endomorphism, or put differently, as matrices you'd have to basis-transform before and after applying your endomorphism, which cancels after taking the determinant. If you have two different vector spaces, the representing matrix depends on both bases, and changing one of them would change the determinant.

We can see this using the following construction: if you have an isomorphism $\phi: V\to W$, then you could choose a basis $(v_1,\dots,v_n)$ on $V$ and $(\phi(v_1),\dots,\phi(v_n))$ on $W$. Then, written in these bases, $\phi$ would just correspond to the identity matrix, so it would have determinant 1. But the same would be true for $2\phi$, but then the determinant wouldn't be multilinear.

One way in which you could achieve what you want to do is to fix an isomorphism $\psi$ between $V,W$. If you now have a homomorphism $f: V\to W$, then $\psi^{-1}\circ f: V\to V$ is an endomorphism and has a determinant.

Answer 3

Simply knowing that the dimensions are equal merely tells you that there is an isomorphism between the two vector spaces. However, there are, typically, infinitely many such isomorphisms and there is no canonical one. Now, what is the determinant? Algebraically, it is given by a formula (whatever is your favourite one). Geometrically, it is measuring volume. In particular, the determinant of $n$ vectors in $\mathbb R^n$ measures the (signed) area of the parallelepiped determined by these vectors (in the given order). However, volume is not something that exists in a vector space without having fixed a frame of reference. So, to speak of volume we must first choose a basis (usually in $\mathbb R^n$ this would be the standard basis) and decide that the parallelepiped they form has volume one (thus effectively normalising the determinant) and measuring volumes of other parallelepipeds relative to it.

So, what is going on here is that once you speak of the determinant in $V$ you really implicitly chose a basis for $V$. Now, if you have another vector space $W$ and you wish to speak of determinants there, then you also chose a basis for $W$. A linear transformation $T\colon V\to W$ need not care at all about these chosen bases. So, if you measure the volume deformation borne by $T$ (this is what the determinant of $T$ is --- the factor by which it deforms volumes of parallelepipeds) and you don't make sure $T$ respects the chosen bases, then you will really be comparing apples and oranges. If, however, $T$ does preserve the bases, then the whole situation is isomorphic to $T$ being an endomorphism on a single vector space $V$ with just a single chosen basis in it to give meaning to determinants. All is well, and the usual notion of the determinant of an endomorphism is restored.

Answer 4

Without attempting to discredit any one of the very instructive answers already given, let me just say that

It is indeed possible to define a determinant for all linear transformations between vector spaces of the same finite dimension.

In order to do this one may choose, once and for all, a basis for each and every vector space. Then, given a linear transformation $f:V\to W$, one may define $\text{det}(f)$ to be the determinant of the matrix representing $f$ relative to the chosen bases of $V$ and $W$.

Whether or not such a concept is useful is another question altogether. Vector spaces often occur in applications without any preferred basis, or perhaps with one which might not be the one we have chosen in advance, so the above definition of determinant is likely to be totally useless!