Question

If lambda max is the largest eigenvalue of a real symmetric matrix A, show that no diagonal entry of A can be larger than lambda max

Answer 1

Don't have an exact solution, but here is a sketch. The trace is the sum along the diagonal, but the trace is also equal to the sum of the eigenvalues.

For a symmetric matrix, the trace will consist of 2 copies of all values plus 1 more from the center (if 3x3 or 5x5).

No entry of A can be more than half the trace, half the sum of the eigenvalues.
That's all I've got on first impression.
HTH

Answer 2

OKay, thank you!

Answer 3

Scratch that. It's wrong. Sorry.

Answer 4

Okay... Can you walk me through what the answer it. Like give me a gist?

Answer 5

I don't see an answer right away. I can try to think about it. I see that if we had all 0 on the diagonal except for one entry, then this is true. There must be some results relating changes on the diagonal to changes in the eigenvectors (pretty sure there is), but I don't remember.

Answer 6

@eliassaab Can you please help, Sir?

Answer 7

*

Answer 8

Since A is symmetric, there exists an orthogonal matrix
M such that
\[
A= M D M^T
\]
Where D is the diagonal matrix of the eigenvalues of A.
let \( \lambda_{max}\) be the maximum of the eigenvalues.
We will show that \( a_{n,n}\le \lambda_{max}\) for every n.
We will show it for n=1. The general case is similar.
Let
\[u= (1,0,0,\cdots ,0)^T\\
w=M^T u=(w_1,w_2,\cdots,w_n)\\
||w||^2 =<w,w>=\sum_{i=1}^n w_i^2=<M^T u,M^T u>=\\ <u, MM^Tu>=<u,u>=1\\
a_{1,1}= u^T A u= u^T M D M^T u=\\
\left(M^T u\right)^T D\left( M^T u\right)=\\
w^T D w=\sum_{i=1}^n \lambda_i w_i^2\le \lambda_{max}
\sum_{i=1}^n w_i^2=\lambda_{max} (1)=\lambda_{max}
\]

Answer 9

@telliott99

Answer 10

@mukushla

Answer 11

Thanks. Will study..

Answer 12

Thank You very much!