# linear algebra – Prove that \$det(T) = det(T_{w_{1}})det(T_{w_{1}})cdots det(T_{w_{k}})\$ where \$W_{i}\$ are \$T\$-invariant subspaces of \$V\$ – Trouble at last step.

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?