4

(Note: this is not a duplicate as suggested. I am asking for an inequality.)

We know that for any complex square matrix $A$, we have

$$\sum_{k=1}^n a_{kk} = \sum_{k=1}^n \lambda_k.$$

I can see this relation in trace property that it is equal sum of diagonal elements.

But how to show that the sum of absolute values of the diagonal elements is less than or equal to the sum of absolute values of the eigenvalues? That is, do we have

$$\sum_{k=1}^n |a_{kk}| \le \sum_{k=1}^n |\lambda_k|\ ?$$

Ka-Wa Yip
  • 830
  • 7
  • 17
  • 1
    $\operatorname{tr}(A)$ is the constant term of $\det(A-xI)$. – Hagen von Eitzen Oct 04 '15 at 10:37
  • 4
    Possible duplicate of [Proof that the Trace of a Matrix is the sum of its Eigenvalues](http://math.stackexchange.com/questions/546155/proof-that-the-trace-of-a-matrix-is-the-sum-of-its-eigenvalues) – CKM Oct 04 '15 at 10:37
  • 4
    @chandresh It's not a duplicate, because the "less than or equal to" relationship is not addressed in the proposed duplicate. However, the title here is misleading. – Patrick Stevens Oct 04 '15 at 10:47
  • 1
    I have changed "absolute square" to "absolute value". If you do mean squared moduli, please correct your inequality (to $\sum_{k=1}^n |a_{kk}|^2 \le \sum_{k=1}^n |\lambda_k|^2$, for instance). – user1551 Oct 04 '15 at 11:42

1 Answers1

3

It's not true. Take

$$ \left( \begin{array} &1 & 1 \\ -1 & -1 \end{array} \right) $$

which has eigenvalues $\lambda_1=\lambda_2=0$. But $$|1|+|-1| \leq |0|+|0|$$

does not hold.

abnry
  • 14,161
  • 2
  • 33
  • 75