Determinant of a linear transform between two different vector spaces with the same dimension

Mathematics Asked by lyrin on November 28, 2020

Let $T:V to W$ be a linear transform where $V$ and $W$ have the same dimension. If $V = W$, the determinant of $T$ is independent of the choice of basis. It measures the $T$’s dilation ratio of the volume spaned by the basis vectors and the ratio is indepedent of the basis because the two space have the same basis. However, if we use different bases for $V$ and $W$ (although the two spaces are isomorphic anyway), the determinant may depend on the choice of the bases. Is that right? In this case, what kind of meaning does the determinant have?

2 Answers

(Rewritten - the original post went off into unnecessary territory, but I had leave and didn't have time to prune it appropriately then.)

Determinants are only defined on transformations of a finite-dimensional vector space into itself. They are not defined for transformations into another vector space, not even if that vector space is isomorphic to the first.

Most commonly, determinants are first defined on square matrices. To define them on transformations from an $n$-dimensional vector space $V$ over the field $Bbb F$ (if you don't understand that, just consider $Bbb F$ to be either $Bbb R$ or $Bbb C$, depending on whether your vector space is real or complex), you first find a basis for $V$. By that basis, you can identify every vector $v in V$ with an ordered $n$-tuple $(v_1, v_2, ..., v_n) in Bbb F^n$. This identification defines an invertible linear transformation $R : V to Bbb F^n$. If $T : V to V$ is a linear transformation, then $RTR^{-1}=Rcirc T circ R^{-1}$ defines a linear transformation from $Bbb F^n to Bbb F^n$. I.e., $RTR^{-1}$ is an $n times n$ matrix. As such its determinant is already defined. So we define $det T := det(RTR^{-1})$.

Now extrinsically, that definition depends on $R$ - which itself depends on the basis originally chosen. But it can be shown that the exact same value for $det T$ is obtained for any basis, so in fact $det T$ does not depend on the basis.

But one must be careful here: we make use of $R$ twice. Once to convert from $Bbb F^n$ to $V$ and then convert back from $V$ to $Bbb F^n$. Suppose we decided to choose different bases for those two conversions. We could try $RTS^{-1}$ where $S$ is defined by some other basis. But while it can be shown that $det STS^{-1} = det RTR^{-1}$, in general $det RTS^{-1} ne det RTR^{-1}$. For example, if the vectors for the basis defining $R$ happen to be $2$ times the vectors in the basis defining $S$, then $det RTS^{-1} = 2^ndet RTR^{-1}$. So the magic that allows you to get the same value of $det T$ regardless of the basis chosen requires the bases used for both transformations to be the same.

However, if $T : V to W$ with $V ne W$, then even when $V$ and $W$ are of the same dimension and therefore both isomorphic to $Bbb F^n$, you cannot choose the same bases for transforming $Bbb F^n to V$ and for transforming $W to Bbb F^n$. If $V$ and $W$ had the same basis, they would necessarily be the same vector space. Because we have no way of relating the two transformations $S : V to Bbb F^n$ and $R : W to Bbb F^n$, we cannot define a single $det T$ in this case. If $S'$ and $R'$ are tranformations defined by different bases, then in general, $det RTS^{-1} ne det R'TS'^{-1}$

So you can only talk about the determinant of a transformation $T : V to V$. The determinant of a transformation $T : V to W$ is not well-defined, even when $V$ and $W$ have the same dimension.

Correct answer by Paul Sinclair on November 28, 2020

You can in fact define the determinant for maps between different (non-canonically isomorphic) vector spaces. Let $V,W$ be vector space of dimension $n$. The determinant of a linear map $T:Vto W$ is the induced map on top wedge $mathrm{det}(T):bigwedge^n(V)tobigwedge^n(W);v_1wedgecdotswedge v_n mapsto Tv_1wedgecdotswedge Tv_n$. This function takes values in the vector space $mathrm{Hom}(bigwedge^n(V),bigwedge^n(W))$ which has dimension 1. When we have chosen an isomorphism of $Vcong W$ we get an isomorphism $mathrm{Hom}(bigwedge^n(V),bigwedge^n(W)) cong mathrm{End}(bigwedge^n(V)) =mathbb{F}$ which then gives the determinant as you may recognise it. Without this choice of isomorphism we cannot find a canonical isomorphism with $mathbb{F}$ but we can still say intelligent things about our determinant function. For example it is zero if the rank of $T$ is less than $n$.

Answered by Callum on November 28, 2020

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