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Let A be symmetric and positive definite. Show that $||a_{ij}|| < \sqrt{a_{ii}a_{jj}}$

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  • $\begingroup$ What is the norm of a complex number? Do you mean its modulus? $\endgroup$ – user1551 Dec 5 '16 at 4:24
  • $\begingroup$ Yes, that's how it's dealt with in linear algebra. Isn't it? @user1551 $\endgroup$ – Rama Dec 5 '16 at 4:35
  • $\begingroup$ @Rama For just a single number, $|z|$ is used for the modulus of $z$. $\endgroup$ – Sungjin Kim Dec 6 '16 at 19:01
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Since $A$ is positive-definite and symmetric, you have that $A=B^TB$ for a certain $B$. Then, using Cauchy-Schwarz, $$ |a_{ij}|=\left|\sum_k b_{ki}b_{kj}\right|\leq\left(\sum_k |b_{ki}|^2\right)^{1/2}\left(\sum_k |b_{kj}|^2\right)^{1/2}=\sqrt{a_{ii}a_{jj}} $$


If you want to "start from scratch":

  1. Considering the matrix as a complex matrix, show that it always has an eigenvalue.

  2. Using that $A$ is symmetric, show that the orthogonal subspace of an eigenvector is invariant for $A$

  3. Show inductively that $A$ admits a basis of eigenvectors.

  4. Using said basis of eigenvectors, show that $A$ admits a positive-definite, symmetric square root $B$.

  5. Write $A=B^2=B^TB$.

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of course. Every two by two principal subform must also be positive definite

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