Determinant of exterior power and symmetric power

Question

I am self studying the tensor algebra, symmetric algebra and exterior algebra.

The exterior algebra brings a new definition of determinant to me.

I am curious what if $f:V \to V$ is a linear endomorphism on $n$ dimensional vector space $V$also with two linear maps $$\Lambda^i(f):\Lambda^i(V) \to \Lambda^i(v)\text{ and } S^i(f):S^i(V) \to S^i(V), i \geq 1$$

then what is the determinant of these maps?

The determinant of $f$ is just the map $\Lambda^n(f)$. So the determinant of $\Lambda^i(f)$ is just the map $\Lambda^{\binom{n}{i}}(\Lambda^i(f))$.

Is there any way to find $\Lambda^{\binom{n}{i}}(\Lambda^i(f))$ without involving matrices? I am looking for a general way to solve it. However, I am frustrated in the first step expanding

$$\Lambda^{\binom{n}{i}}(\Lambda^i(f))(w_1 \wedge \cdots\wedge w_{\binom{n}{i}}).$$

Is there any mistake?

Also I have no idea on $\Lambda^{\binom{\binom{n+k-1}{k}}{i}}(S^i(f))$. Is is possible to find them without involving matrices?

Answer 1

Pass to the algebraic closure of the ground field if necessary. Then:

1. $\det(f)$ is the product $\prod_{i=1}^n \lambda_i$ of the eigenvalues of $f$ (with algebraic multiplicity).
2. The eigenvalues of $\wedge^k(f)$ are products of unordered $k$-tuples of distinct eigenvalues of $f$, and the eigenvalues of $S^k(f)$ are products of unordered $k$-tuples of eigenvalues (not necessarily distinct) of $f$ (exercise).

So when computing $\det(\wedge^k(f))$ every eigenvalue $\lambda_i$ of $f$ appears the same number of times. A total of $k {n \choose k}$ eigenvalues appear, so each eigenvalue appears $\frac{k}{n} {n \choose k} = {n-1 \choose k-1}$ times and hence

$$\boxed{ \det(\wedge^k(f)) = \det(f)^{ {n-1 \choose k-1} } }.$$

Similarly, when computing $\det(S^k(f))$ every eigenvalue also appears the same number of times. Here a total of $k {n+k-1 \choose k}$ eigenvalues appear, so each eigenvalue appears $\frac{k}{n} {n+k-1 \choose k} = {n+k-1 \choose k-1}$ times and hence

$$\boxed{ \det(S^k(f)) = \det(f)^{ {n+k-1 \choose k-1} } }.$$

Abstractly, by functoriality both $\det(\wedge^k(f))$ and $\det(S^k(f))$ are polynomial multiplicative homomorphisms $\text{End}(V) \to F$ ($F$ the underlying field); it's possible to show that any such homomorphism must be a non-negative integer power of the determinant, so it only remains to compute which power.