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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.

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

1 Answer 1

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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, $\det f = \det M = \det M_1 \times \det M_2 = \det f_1 \times \det f_2$.

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