# Showing that $\det A=\det B\cdot\det C$ when $B,C$ are the restrictions of $A$ onto a subspace

I am a bit unsure about one approach that is mentioned to prove this determinant result.

Here is the quote from Pages 100-101 of Finite-Dimensional Vector Spaces by Halmos:

Here is another useful fact about determinants. If $$\mathcal{M}$$ is a subspace invariant under $$A$$, if $$B$$ is the transformation $$A$$ considered on $$\mathcal{M}$$ only, and if $$C$$ is the quotient transformation $$A/\mathcal{M}$$, then

$$\det{A}=\det{B}\cdot\det{C}$$

This multiplicative relation holds if, in particular, $$A$$ is the direct sum of two transformations $$B$$ and $$C$$. The proof can be based directly on the definition of determinants, or, alternatively, on the expansion obtained in the preceding paragraph.

What I am confused about is how you can use the definition of determinants to conclude this result.

In this book, the determinant of a linear transformation $$A$$ is defined as the scalar $$\delta$$ such that $$(\delta w)(x_{1},\cdots,x_{n})=w(A(x_{1}),\cdots,A(x_{n}))$$ for all alternating $$n$$-linear forms $$w$$, where $$V$$ is an $$n$$-dimensional vector space.

It is then shown that by fixing a coordinate system (or basis) and letting $$\alpha_{ij}$$ be the entries of the matrix of the linear transformation under the coordinate system, the determinant of the linear transformation $$A$$ in that coordinate system is:

$$\det{A}=\sum_{\pi}(\text{sgn}\,\pi)\alpha_{\pi{(1)},1}\cdots\alpha_{\pi{(n)},n}$$

where the summation goes over all permutations in $$\mathcal{S}_{n}$$.

I have been able to use the expression involving the coordinates to show this result, but I am not sure about how this would be done directly from the definition. I have tried looking at defining other alternating forms and using their product to show this, but I was not able to make much use of that approach.

Are there any suggestions for proving this result directly from the definition?

Edit: I would like to add that part of my confusion may be from the fact that $$A$$, $$B$$ and $$C$$ are all linear transformations on different vector spaces and I am not sure how the definition can be used in this situation.

Let $$d = \dim(\mathcal M)$$. Let $$v_1, \dots, v_n \in V$$ such that $$v_1, \ldots, v_d \in \mathcal M$$ is a basis of $$\mathcal M$$ and let $$\omega$$ be a non-zero alternating form on $$V$$. Then

$$\mathcal M \ni (w_1, \ldots, w_d) \mapsto \omega(w_1, \ldots, w_d, v_{d+1}, \ldots, v_n)$$

is an alternating form on $$\mathcal M$$ and

$$V \ni (w_{d+1}, \ldots, w_n) \mapsto \omega(v_1, \dots, v_d, w_{d+1}, \ldots, w_n)$$

is (or induces) an alternating form on $$V/\mathcal M$$. Therefore

$$\begin{eqnarray} \det(A) \, \omega(v_1, \ldots, v_n) & = & \omega(Av_1, \ldots, Av_n) \\ &=& \det(B) \, \omega(v_1, \ldots, v_d, Av_{d+1}, \ldots, Av_n) \\ &=& \det(B) \det(A/\mathcal M) \, \omega(v_1, \ldots, v_n). \end{eqnarray}$$

Since this holds for all $$v_1, \ldots, v_n$$ (with $$v_1, \ldots, v_d \in \mathcal M$$ a basis) and $$\omega$$ is non-zero it follows that $$\det(A) = \det(B) \det(A/\mathcal M).$$