Properties of matrices having same eigen values.

Question

Let $ P=\begin{bmatrix} x & y & z\\ \end{bmatrix} $ and $ M=\begin{bmatrix} 1 & 2 & 3\\ 3 & 1 & 2\\ 2 & 3 & 1\\ \end{bmatrix} $ and $ N=\frac{1}{2} \begin{bmatrix} 2 & 5 & 5\\ 5 & 2 & 5\\ 5 & 5 & 2\\ \end{bmatrix} $

Here $PMP'=PNP' \nRightarrow N=M$, why?

Only two things I can see, first that matrix $M$ and $N$ commute that is $MN=NM$ and second, they have the same eigen values.

Is this reason enough? What is the name for this kind of matrices?

I have also tried to premultiply by $P'$ as:

$$PMP'=PNP' \\ (P'P)MP'=(P'P)NP' \\ MP'=NP' \\ \text{(roughly assuming that inverse of $PP'$ exists)}$$

But since $MP'\ne NP'$ so what I had assumed that $(PP')^{-1}$ exists is wrong in general. Is this always the case? Because $P$ is a general row-matrix.

Answer 1

Yes, if $Z_{3 \times 3}=P_{3\times 1}.P_{1\times 3}$, then $\det|Z|=0$.

Answer 2

Same eigenvalues is different from common eigenvalues

The first matrix $M$ has eigenvalues of $[\omega_1 = -\frac{\sqrt(3) i+3}{2}, \omega_2 = \frac{\sqrt(3) i-3}{2}, \omega_3 = 6]$ While the second matrix $N$ has eigenvalues of $[\omega_1 = -\frac{3}{2}, \omega_2 = -\frac{3}{2} , \omega_3 = 6 ]$

By examination the two matrix there are made of $3$ elements, oriented in each row $$M = \begin{pmatrix} a&b&c\\ c&a&b\\ b&c&a\\ \end{pmatrix}$$ $$N = \begin{pmatrix} p&q&r\\ r&p&q\\ q&r&p\\ \end{pmatrix}$$ Sir can you work out some of the properties or prove that the multiplication here is commutative $$ M \cdot N = N \cdot M $$ Your matrix are another example of commutative matrix other than orthogonal matrixes