Let A$_{3×2}$ and B$_{2×3}$ be matrices such that their product $AB$ is $$AB=\begin{pmatrix} 8&2 & -2\\ 2&5&4 \\ -2&4&5 \\ \end{pmatrix}$$ And $BA$ is nonsingular

Find the determinant of $BA$.

I have no idea , how to solve this type of question. All I could notice is that $|AB| = 0$ and it's a symmetric matrix. I tried assuming a general matrix , but I get simply too many unknowns and very few equations.

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    $\begingroup$ Try Cauchy-Binet, or compute the characteristic polynomial... BA and AB have the same charpolys except AB has an extra zero in its polynomial for dimension reasons $\endgroup$ – user8675309 Feb 14 at 10:03
  • $\begingroup$ I dont know what Cauchy Binet is ... But I do have a vague idea of characteristic polynomials.... If you say so , that the property is true, how to decipher, which of the root of AB , is bot part of BA?? $\endgroup$ – RandomAspirant Feb 14 at 11:54
  • $\begingroup$ my read is: the point of this exercise is you are supposed to learn to relate the characteristic polynomials of AB and BA... I infer you are working in $\mathbb R$ here -- another choice is to compute $\text{trace}\big(AB\big)$ --which must be $= \text{trace}\big(BA\big)$ and compute $\text{trace}\big((AB)^2\big)$-- --which must be $ = \text{trace}\big((BA)^2\big)$-- and Newton's Identities give you the result $\endgroup$ – user8675309 Feb 14 at 18:59

A direct calculation is also possible in case one wants to find out the answer easily. If we denote the coefficients of $A$ by $a_i$ and the ones of $B$ by $b_j$, and the given matrix by $C$, then the matrix equation $AB=C$ is equivalent to equations in $a_i,b_j$. We can solve them case by case. The first equation is $a_1b_1 + a_2b_4 =8$. For $a_1=0$ we obtain $a_2\neq 0$ and $$ b_4=\frac{8}{a_2}, b_5= \frac{2}{a_2}, b_6=-\frac{2}{a_2}, a_4= \frac{a_2a_3b_3 - 4a_2}{2}, b_3= -\frac{1}{a_3(a_3b_2 - 9}, b_2= \frac{1}{(4a_3)(a_3b_1 + 18)}, a_6= -\frac{a_2a_5b_1 + 2a_2}{8}, a_5=a_3. $$ Then we obtain $$ BA=\begin{pmatrix}9 & 0 \cr 0 & 9\end{pmatrix}. $$ The other case $a_1\neq 0$ is similar. Note that $\det(AB)\neq \det(BA)$ in general, but $tr(AB)=tr(BA)$ is true in general. So the determinant is $81$.

  • $\begingroup$ Thanks a lot :) , but isn't there any general way to calculate it ??... That is , without taking assuming values. $\endgroup$ – RandomAspirant Feb 14 at 15:01
  • $\begingroup$ We don't assume values. We just do both cases, first assuming $a_1=0$ and then $a_1\neq 0$. This covers everything. The second case has the advantage that we can use the first equation $a_1b_1 + a_2b_4 =8$ to eliminate $b_1$. I have edited my post. $\endgroup$ – Dietrich Burde Feb 14 at 15:09

The characteristic polynomial of $AB$ is $$ p(x)=\left| \begin{array}{rrr} 8-x & 2 & -2 \\ 2 & 5-x & 4 \\ -2 & 4 & 5-x \end{array} \right|=\cdots =-x^3+18x^2-81x=-x(x-9)^2 $$ Cayley-Hamilton Theorem implies that $p(AB)=0$, i.e., $$ -(AB)^3+18(AB)^2-81AB=0 $$ In fact, since $AB$ is symmetric, it is diagonalisable and hence its minimal polynomial is $m(x)=x(x-9)$.

Hence $$ (AB)^2-9AB=0 \quad\Longrightarrow\quad B(AB)^2A-9BABA=0 \quad\Longrightarrow\quad (BA)^3-9(BA)^2=0. \quad\Longrightarrow\quad (BA)^2\big((BA)-9\big)=0. $$ Clearly, rank$(AB)=2$, and also rank$((AB)^2)=2$ (as $(AB)^2$ is also symmetric with eigenvalues $81,81,0$) and hence rank$(BA)=2$, and thus $BA$ is invertible and thus $$ (BA)^2\big((BA)-9\big)=0\quad\Longrightarrow\quad (BA)-9=0, $$ i.e., $$ BA=\left(\begin{array}{cc} 9&0 \\ 0&9\end{array}\right) $$ and hence det$(BA)=81$.


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