# Find determinant and trace of product of non square matrices

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.

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

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 $$5$$.

• Excellent solution,congratulations ! Commented Dec 29, 2018 at 14:22

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 $$\det(CB)=6$$.