# If $\mathrm{M,N}$ are $3\times 2, 2 \times 3$ matrices such that $\mathrm{MN}=$ is given. Then $\mathrm{det(NM)}$ is?

If $$\mathrm{M,N}$$ are $$3\times 2, 2 \times 3$$ matrices such that $$\mathrm{MN}=\pmatrix{8& 2 & -2\\2& 5& 4\\-2& 4&5}$$, then $$\mathrm{det(NM)}$$ is?

($$\mathrm{NM}$$ is invertible.)

$$\mathrm{det(MN)}$$ must be (and is) zero. But how to find $$\mathrm{det(NM)}$$? Any hint?

• Interesting. We could potentially cook up different sets of $M,N$ that yield the same product $MN$, why would $\det N_1M_1 = \det N_2M_2$ even have to hold? Jun 12, 2020 at 17:52
• You mean, without giving $N,M$?
– lcv
Jun 12, 2020 at 17:56
• @AlvinLepik In general it does not: Since $MN$ is a $3 \times 3$ product of matrices of rank $\leq 2$, we have $\operatorname{rank}(MN) \leq 2$, and so $\det(MN) = 0$. But one can choose $M, N$ such that $\det(NM) \neq 0$. For example, take $M$ and $N$ to be matrices given by padding the $2 \times 2$ identity matrix with zero entries on the bottom and right, resp. Jun 12, 2020 at 18:06
• Just as an aside: The question is problem B6 on the 1969 Putnam Exam. Jun 12, 2020 at 18:46
• here is an answer found among Putnam archives. prase.cz/kalva/putnam/psoln/psol6912.html The question was also different. The point is that the minimal polynomial is $x^2 - 9x$ Jun 12, 2020 at 19:29

Hint If $$p \geq q$$ and $$M, N$$ are $$p \times q$$ and $$q \times p$$ matrices, respectively, then the characteristic polynomials of $$p, q$$ are related by $$p_{MN}(\lambda) = \lambda^{p - q} p_{NM}(\lambda) .$$

• here is an answer, evidently the original question was rather different.. prase.cz/kalva/putnam/psoln/psol6912.html where they were told to show the reverse product gave $9I$ Jun 12, 2020 at 19:20
• Travis, is there a name I might be able to look up for the theorem you are quoting? New one on me. Jun 12, 2020 at 21:00
• @Sudix That depends on whether your convention for $p_A(\lambda)$ is $\det (\lambda I - A)$ or its negative; I prefer the former convention, so that $p_A$ is monic even when the size $n$ of $A$ is odd (this comes at the cost of the constant term being $(-1)^n \det A$ rather than just $\det A$). If you use the latter convention, then like you say, $\lambda^{q - p}$ in the identity should be replaced with $(-\lambda)^{q - p}$. Jun 12, 2020 at 21:37
• Travis, thank you. I have H+J, in my (paperback) edition it is 1.3.20 on page 53. That's really nice. There is a proof from pages 53-54, then an alternate as question 9 on page 55 Jun 12, 2020 at 22:53
• @WillJagy You're welcome. The theorem number I mentioned is that in the second edition (2013, apparently), where the result appears on p. 65. By the way, the Jordan block statement does not always hold for the zero eigenvalue, and hence the analogous statement for minimal polynomials is false, i.e., one can pick $M, N$ (both square, even) such that $m_{MN} \neq m_{NM}$. Jun 12, 2020 at 23:09

One way to proceed, not knowing the relevant theorem( new one for me as well) is to guess that, being symmetric, this is a Gram matrix. Look for integer rectangles...

$$\left( \begin{array}{rr} 2 & 2 \\ 2 & -1 \\ 1 & -2 \end{array} \right) \left( \begin{array}{rrr} 2 & 2 & 1 \\ 2 & -1 & -2 \end{array} \right) = \left( \begin{array}{rrr} 8 & 2 & -2 \\ 2 & 5 & 4 \\ -2 & 4 & 5 \end{array} \right)$$

$$\left( \begin{array}{rrr} 2 & 2 & 1 \\ 2 & -1 & -2 \end{array} \right) \left( \begin{array}{rr} 2 & 2 \\ 2 & -1 \\ 1 & -2 \end{array} \right) = \left( \begin{array}{rr} 9 & 0 \\ 0 & 9 \end{array} \right)$$

Indeed, the first characteristic polynomial is $$x^3 - 18 x^2 + 81x$$ and the second is $$x^2 - 18 x + 81$$

• Nice but this doesn't show that the answer is independent from the concrete 'parametrization'.
– lcv
Jun 12, 2020 at 22:24
• @lcv sure. I would like to encourage the students to experiment a bit when they have no idea what is going on. Jun 12, 2020 at 23:08
• Agree. Always a good philosophy.
– lcv
Jun 12, 2020 at 23:45

In this particular case, you can calculate the value of $$NM$$ explicitly by solving for the right variables.

The order isn't all that sophisticated though.