Let $T$ be a linear operator on a finite dimensional vector space $V$ and let $W_{1} oplus cdots oplus W_{k}$ where $W_{i}$ are $T$-invariant subspaces of $V$. Prove that $det(T) = det(T_{w_{1}})det(T_{w_{1}})cdots det(T_{w_{k}})$

Prior to attempting to prove the question I have two relevant pieces of information to use:

Attempt:

Let $beta_{i}$ be an ordered basis for $W_{i}$ such that $beta = cup beta_{i}$ is a basis for $V$. As such we let $A = (T)_{beta}$ and $B_{i} = (T_{W_{i}})_{beta_{i}}$.

As a result of this $A = B_{1} oplus B_{2} cdots oplus B_{k}$. Observe $A$ is a diagonal matrix. As a result $det(A)$ is the product of the diagonal.

$$det(A) = prod_{i = 1}^{k}B_{i}$$

Here is where I’m having an issue in making the logical jump. What I worked out occurring after working from the other direction would be

$$text{missing steps} \ = prod_{i=1}^{k} det(B_{i}) \ = prod_{i=1}^{k} det(T_{W_{i}})$$

I know I can take the determinants of the individual $B_{i}$ because they are square matrices, but the determinant is not a linear operator so I can’t just do

$$det( prod_{i=1}^{k}(B_{i}) = prod_{i=1}^{k} det(B_{i})$$

As well it is not the case that $A = prod_{i=1}^{k}(B_{i}$, only the $det(A) = prod_{i=1}^{k}(B_{i})$.

So what is it that I’m missing to get the final result I’m desiring?