Norm Property of a symmetric positive definite matrix

Question

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

Answer 1

Since $A$ is positive-definite and symmetric, you have that $A=B^TB$ for a certain $B$. Then, using Cauchy-Schwarz, $$ |a_{ij}|=\left|\sum_k b_{ki}b_{kj}\right|\leq\left(\sum_k |b_{ki}|^2\right)^{1/2}\left(\sum_k |b_{kj}|^2\right)^{1/2}=\sqrt{a_{ii}a_{jj}} $$

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

Answer 2

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