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Let $B \in M_{3,2}(\mathbb{R})$,$C \in M_{2,3}(\mathbb{R})$ so that $BC=\begin{pmatrix}2 & -2 & 3\\ 0 & 0 & 3\\ 0 & 0 & 3 \end{pmatrix}$. Find $\det(CB)$ and $Tr(CB)$.
My attempt : I tried to compute the powers of $BC$ (I actually hoped that $BC$ was idempotent) and I saw that $(BC)^n=\begin{pmatrix}2^n & -2^n & 3^n\\ 0 & 0 & 3^n\\ 0 & 0 & 3^n \end{pmatrix}$ which doesn't really help.

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  • $\begingroup$ @PeterMelech $\det(C)$ doesn't exist because are not squared matrix $\endgroup$ Commented Dec 29, 2018 at 14:08
  • $\begingroup$ Yes sorry, I noticed already $\endgroup$ Commented Dec 29, 2018 at 14:08
  • $\begingroup$ @Yanko the trace formula doesn't hold for any non square matrices even though in this case according to the answer key it does. $\endgroup$
    – Math Guy
    Commented Dec 29, 2018 at 14:13
  • $\begingroup$ @MathGuy Ok I believe you, anyway I post an answer without using this. $\endgroup$
    – Yanko
    Commented Dec 29, 2018 at 14:18

2 Answers 2

6
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You need to work with this:

The trace of a matrix is the sum of the eigenvalues and the determinant is the product.

It is not hard to see that the eigenvalues of $BC$ are $2,3,0$. Let $v_1,v_2$ be the eigenvectors correspond to $2,3$ (you will see why we can't use the eigenvector for $0$ soon).

Consider the matrix $CB$ and look at the vectors $Cv_1,Cv_2$. Then $$CB(Cv_i) = C(BCv_i)=\lambda_i Cv_i$$ therefore since $Cv_i\not = 0$ (otherwise $BCv_i=0$) we have that $2,3$ are eigenvalues of $CB$ which means that the determinant is $6$ and the trace is $5$.

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    $\begingroup$ Excellent solution,congratulations ! $\endgroup$
    – Math Guy
    Commented Dec 29, 2018 at 14:22
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I have just come across this old question and I will post my simpler solution to it.
From the Cayley-Hamilton theorem for $CB$ we have that:
$(CB)^2-\operatorname{Tr}(CB)\cdot CB+\det(CB)\cdot I_2=O_2$
Now left multiply by $B$ and then right multiply by $C$ this relation and get that
$(BC)^3-\operatorname{Tr}(CB) \cdot (BC)^2 +\det(CB) \cdot BC =O_3$.
Now simply substitute $BC$'s powers into this equation to get that $\operatorname{Tr}(CB)=5$ and $\det(CB)=6$.

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