# Direct sum and determinant

Consider the vector space $$V = V_1 \oplus V_2$$, where $$V_1,V_2$$ are subspaces. Let $$f_i\in\mathcal{L}(V_i)$$ be a linear map for $$i=1,2$$, and define $$f\in\mathcal{L}(V)$$ by $$f(v)= f_1(v_1)+f_2(v_2)$$, where $$v=v_1+v_2$$ and $$v_i\in V_i$$, for $$i=1,2.$$
Prove that $$\det(f) = \det(f_1)\cdot\det(f_2)$$. I have no idea where to start, any ideas are welcome.

• I don't think your statement is true. Say $f_i = \mathrm{Id}_{V_i}$. Then $f = f_1 + f_2 = \mathrm{Id}_V$ and the determinants are all $1$ Jun 3, 2020 at 12:56
• Try to see how the matrix of $f$ looks like when you choose a suitable basis. Consdider some concrete examples. Jun 3, 2020 at 12:57
• @DIdier_ I think he meant product instead of sum. Jun 3, 2020 at 12:57
• Determinant isn't a linear form (convince yourself by applying LaPlace). This will be true, e.g. $$\begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{bmatrix}a&0\\c&d\end{bmatrix}+\begin{bmatrix}0&b\\c&d\end{bmatrix}$$ Jun 3, 2020 at 13:00
• Btw, avoid no-clue questions, those aren't suitable. You don't set a good example to new contributors. Jun 3, 2020 at 13:01

I think you are talking about finite dimensional vector spaces $$V_1$$ and $$V_2$$ (so that one can talk about determinant). Choose $$(e_1,\ldots,e_n)$$ and $$(f_1,\ldots, f_m)$$ be basis of $$V_1$$ and $$V_2$$. Then $$(e_1,\ldots,e_n,f_1,\ldots,f_m)$$ is a basis of $$V=V_1 \oplus V_2$$.
Take $$f_i \in \mathrm{End}(V_i)$$, and define $$f = f_1 \oplus f_2$$. Then $$f \in \mathrm{End}(V)$$. If $$M_i$$ is the matrix of $$f_i$$ in the basis $$(e_j)$$ (or $$(f_k)$$), then the matrix of $$f$$ in the basis $$(e_1,\ldots,e_n,f_1,\ldots,f_m)$$ is $$M=\begin{pmatrix} M_1 & 0 \\ 0 & M_2\end{pmatrix}$$.
Thus, $$\det f = \det M = \det M_1 \times \det M_2 = \det f_1 \times \det f_2$$.