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Let $A = \begin{pmatrix} 2&-2 & 3\\ 0&0&3 \\ 0&0&3 \\ \end{pmatrix}$. $B$ is a $3\times 2$ matrix and $C$ is a $2 \times 3$ matrix such that $BC=A$. Find $\det(CB)$ and $\operatorname{Tr}(CB)$.

I read this thread but i have not learned about characteristic polynomials yet and the answer that does not use those is tedious.

I also read this but in my case it is the other way around, i know the determinant which equals $0$, but i do not know how to find the "other one".

Also, I am not that familiar with spaces and dimensions in linear algebra so i would prefer a solution that does not involve deep knowledge in this area.

Any tips ?

Semiclassical
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Pete42
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  • The trace side of this is actually simple: So long as $AB$ is a square matrix, then $\operatorname{Tr}(AB)=\operatorname{Tr}(BA)$. No other properties of $A,B$ are needed. – Semiclassical Feb 07 '21 at 18:49

2 Answers2

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Since $A$ has rank $2$, $B$ has full column rank. It follows that if $Av=BCv=0$ for some vector $v$, we must have $Cv=0$.

Now let $v=(1,1,0)^T$. Then $Av=0$. Hence $Cv=0$. Let $u=(0,1,0)^T$. Then $u^Tv=1$. Therefore \begin{aligned} \det(CB) &=\det\pmatrix{CB&0\\ \ast&1}\\ &=\det\left[\pmatrix{C\\ u^T}\pmatrix{B&v}\right]\\ &=\det\left[\pmatrix{B&v}\pmatrix{C\\ u^T}\right]\\ &=\det(A+vu^T)\\ &=\det\pmatrix{2&-1&3\\ 0&1&3\\ 0&0&3}=6. \end{aligned} Finally, as the trace of a product of matrices is invariant under cyclic permutation of the multiplicands, we have $\operatorname{tr}(CB)=\operatorname{tr}(BC)=\operatorname{tr}(A)=5$.

user1551
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A practical approach is to simply find suitable matrices $B$ and $C$ for which $BC = A$. If the question is well formulated, then it suffices to compute the trace and determinant of $CB$ for our particular choice of matrices $C$ and $B$.

To that end: we can take $B$ to be the matrix $$ B = \pmatrix{1 & 1\\0&1\\0&1} $$ because each column of $A$ is a multiple of one of these columns. With that in mind, we can take $$ C = \pmatrix{2&-2&0\\0&0&3}. $$ Verify that we indeed have $BC = A$. With that, we now compute $$ CB = \pmatrix{2&-2&0\\0&0&3}\pmatrix{1&1\\0&1\\0&1} = \pmatrix{2&0\\0&3}. $$ It is easy to compute the determinant $2 \cdot 3 = 6$ and the trace $2 + 3 = 5$.

Ben Grossmann
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