How to find the eingenvalues of matrix $M$ with $\text{rank}(M)\leqslant1\;?$

Question

If a $\,n\times n\,$ matrix $M$ satisfies

$\text{Rank}(M)\leqslant1\;,\;$ then $\;\det(I+tM)=1+t\text{Tr}(M)\;,$
where $\;\text{Tr}(M)\;$ denotes the trace of $\,M$.

In this book it is stated that all eigenvalues of $M$ are zero except a single eigenvalue $\sum_{i=1}^{n}a_{i}b_{i}.$ But I don't know why.
I would appreciate your collaboration.

Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. — Another User, Jan 03 '23 at 16:12
Yes, the eigenvalue $0$ then has multiplicity at least $n-1$, see the duplicate. — Dietrich Burde, Jan 03 '23 at 16:30
Does this answer your question? Eigenvalues of the rank one matrix $uv^T$ — Dietrich Burde, Jan 03 '23 at 16:32

score 0 · Accepted Answer · answered Jan 03 '23 at 19:34

0

Deal the problem case wise. Case 1:Rank M=0. Then M =0. So all eigenvalues are zero. Case 2: Rank M =1. Then M =$AA^T$ where A is a column vector. The eigenvalues are then 0 repeated n-1 times and $A^TA$. Thus all n eigenvalues are determined.

answered Jan 03 '23 at 19:34

Lawrence Mano

2,136

How to find the eingenvalues of matrix $M$ with $\text{rank}(M)\leqslant1\;?$

1 Answers1