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Prove that $\frac{\|a\|^2}{a^TAa}\le \|A^{-1}\|$ where $a\in \mathbb{R}^n$, $A\in \mathbb{R}^{n\times n}$ is a symmetric positive definite matrix and $\|x\|$ is the Euclidean norm $\sqrt(x_1^2+\cdots+x_n^2),x\in\mathbb{R}^n$.

For a matrix $A$, $\|A\|=\max \{\|Ax\|:\|x\|=1\}$, for a symmetric matrix we have $\max\{|x^TAx|:\|x\|=1\}$ and also the maximum absolute value of the eigenvalues of $A$.

I see that $\frac{a^TAa}{\|a\|^2}=\frac{a^T}{\|a\|}A\frac{a}{\|a\|}\le \|A\|$. So $\frac{\|a\|^2}{a^TAa}\ge\|A\|^{-1}$.

We also have $\|A\|\|A^{-1}\|\ge1$.

Picasso
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2 Answers2

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We have from this min-max theorem that $\lambda_{min}\|x\|^2\leq x^THx \le \lambda_{max}\|x\|^2$ where $\lambda_{min},\lambda_{max}$ are the smallest and largest eigenvalues of the Hermitian matrix $H$.

Thus we have $\frac{a^TAa}{\|a\|^2} \ge \lambda_{min}$ so $\frac{\|a\|^2}{a^TAa} \le \frac{1}{\lambda_{min}}=\|A^{-1}\|$ as $\frac{1}{\lambda_{min}}$ is the largest eigenvalue of $\|A^{-1}\|$.

Picasso
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Hint. Use the spectral theorem for symmetric matrices (see here or here). This theorem tells us that if $A$ is a symmetric matrix there are a diagonal $n\times n$ matrix $\Lambda=\mathrm{diag}(\lambda_1,\lambda_2,\ldots,\lambda_n)$ and an orthogonal $n\times n$ matrix $Q$ (i.e. $ Q^TQ=I$), such that $Q^T=Q^{-1}$ and: $$ A=Q^T\Lambda Q. $$ Here, $\lambda_1 > 0,\quad \lambda_2> 0,\quad,\ldots,\;\lambda_n> 0$ are de eigenevalues of $A$ and the colums of $Q$ are the eigenevectors of $A$.

Update: see too Eigendecomposition of a matrix.

Elias Costa
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