Exercise 1.3.6

Question

Prove that $\operatorname{tr}(a A+b B)=a \operatorname{tr}(A)+b \operatorname{tr}(B)$ for any $A, B \in \mathrm{M}_{n 	imes n}(F)$.

Jephian C.-H. Lin · Accepted Answer

\begin{proof}
We have that $$\operatorname{tr}(aA+bB)=\sum_{i=1}^n{aA_{ii}+bB_{ii}}=a\sum_{i=1}^n{A_{ii}}+b\sum_{i=1}^n{B_{ii}}=a \operatorname{tr}(A)+b\operatorname{tr}(B).$$
\end{proof}

Exercise 1.3.6

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Exercise 1.3.6

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