if anything can be said about rank of matrices if eigenvalues are known

Question

If the eigenvalues of $A,M$ are all distinct $1,\lambda_2,\lambda_3$ and $1,m\lambda_2,m\lambda_3$, $B$ is any $3\times 3$ matrix with positive entries(0 is allowed), If the rank of $K=[B\hspace{0.5cm} AB\hspace{0.5cm} A^2B]$ is $3$ can we say anything about the rank of

$L=[B\hspace{0.5cm} MB\hspace{0.5cm} M^2B]$

well, for the time being, we can assume $B$ is $3\times 1$, just a column vector. — Myshkin, Jun 19 '17 at 11:06

score 1 · Answer 1 · answered Jun 19 '17 at 11:43

1

No, you cannot say anything. Take e.g. $\lambda_2=2$ and $\lambda_3=3$, and the eigenvectors of $A$ to be the unit vectory. For $B=(1,1,1)'$ you have rank of $K$ equal 3.

If you now set $M$ to have the same eigenvalues ($m=1$), but with respective eigenvectors $(1,1,1)', (1,0,0)', (0,1,0)'$ you have rank of $L$ = 1.

Of course $L$ can have rank 3 if you set $M=A$, the construction of an example of rank $2$ is left to the reader.

So even with $m=1$ you cannot say anything.

(I used $B$ as a vector as mentionned in the comments. It's the same for matrices by using three copies of $B$).

answered Jun 19 '17 at 11:43

wonko

571

well, what if $m\in(0,1)$ open interval – Myshkin Jun 20 '17 at 09:40
I think also with $m \in (0,1)$ it's pretty easy to construct all kind of examples. Just make sure B is not an eigenvector in one case and it is one in the other – wonko Jun 20 '17 at 13:37
what do u mean by eigenvectors of A to be unit vectory? – Myshkin Jun 20 '17 at 13:47
I really dont want the case when $m=1$ or $m=0$ – Myshkin Jun 20 '17 at 13:47

if anything can be said about rank of matrices if eigenvalues are known

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