Question

Let N=[3 -4 0
4 3 0
0 0 1]

If M is any 3*3 real matrix, then trace(NMN^T) is equal to
a) trace(M)
b) 2 trace(N) + trace(M)
c) trace(N)^2 . trace(M)
d) trace(N)^2 + trace(M)

Answer 1

Answer on Question #38450, Math, Matrix

There must be some mistake in the task. Plugging in any real matrix $M$, and explicitly evaluating trace does not match any of $a)$, $b)$, $c)$, $d)$ answers.

$$N = \left( \begin{array}{ccc} 3 & -4 & 0 \\ 4 & 3 & 0 \\ 0 & 0 & 1 \end{array} \right) .$$

For example, let $M = \begin{pmatrix} -1 & 0 & 2 \\ 1 & 2 & 3 \\ 1 & 1 & 1 \end{pmatrix}$ , so $N M N^T = \begin{pmatrix} 11 & -52 & -6 \\ -27 & 14 & 17 \\ -1 & 7 & 1 \end{pmatrix}$ and $tr(NMN^T) = 26$ . Then,

a) $trM = 2$

b) $2tr(N) + tr(M) = 2\cdot 7 + 2 = 16$

c) $(tr(N))^2\cdot tr(M) = 49\cdot 2 = 98$

d) $(tr(N))^2 + tr(M) = 49 + 2 = 51$

Thus, none of variants matches.

For any matrix $M$, it might be calculated that $tr(NMN^T) = tr(N^T N M) = 25(M_{11} + M_{22}) + M_{33}$ .