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If $A\in \Bbb C^{n\times n}$ is any matrix, then $$\|A\|_2^2\leqslant \|A\|_1\|A\|_\infty.$$

I know that $\|A\|_2^2=\max_i \lambda_i$ where $\lambda_i$ are the eigenvalues of $A^*A$, and I also know that $\|A\|_1$ is the largest column sum, and $\|A\|_\infty$ the largest row sum.

Yet I am unable to produce a proof. Could you tell me the right approach please?

John Steinbeck
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    I think this http://math.stackexchange.com/questions/487266/matrix-norms-inequality-proof might help, especially Algebraic Pavel's comment. – Alijah Ahmed Jan 02 '16 at 14:57
  • @AlijahAhmed, You think it might help? Don't you just mean it's a duplicate? – Eric S. Jan 02 '16 at 15:10
  • @EricS Well, you're right, it actually is a duplicate. – Alijah Ahmed Jan 02 '16 at 15:12
  • With the interpretation you gave, this is obviously false, for example $A=\left( \begin{smallmatrix} 0 & 1 \ 1 & 0 \end{smallmatrix} \right)$. However, this was almost certainly intended as an inequality between the Schatten norms, and then it becomes a standard fact (google it). –  Jan 02 '16 at 18:29
  • @ChristianRemling That is not a counterexample. – John Steinbeck Jan 04 '16 at 12:49
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    @AlijahAhmed Thanks a lot. Could you tell me why $\rho(A^A)\leq|A^A|_1$ holds? I don't see it. – John Steinbeck Jan 04 '16 at 12:50
  • @TheRealBritneySpears The reason why $\rho(A^A)\leq|A^A|_1$, which is the same as saying $\rho(M)\leq|M|_1$, where $M=A^*A$, is because the spectral radius of matrix $M$ will be less than its norm, as shown in Lemma 10 in http://www.math.drexel.edu/~foucart/TeachingFiles/F12/M504Lect6.pdf, for example. – Alijah Ahmed Jan 04 '16 at 21:44

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