$\|A^{-1}\|=\frac{1}{\sqrt{\lambda_{min}(A^2)}}$ - Symmetrical, Invertible, Positive Defenit Matrix and Euclidean Norm

Question

Let $A \in \mathbb{R}^{n \times n}$ a symmetrical and invertible Matrix, and let $\| \cdot \|$ be by the euclidian norm (2-norm) induced matrix norm. Furthermore, $\lambda_{min}(A^2)$ is the smallest Eigenvalue of $A^2$. Prove:

$$\|A^{-1}\|=\frac{1}{\sqrt{\lambda_{min}(A^2)}}$$

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Well, I tried showing

$$\|A^{-1}\| \leq \frac{1}{\sqrt{\lambda_{min}(A^2)}}$$ and $$\|A^{-1}\| \geq \frac{1}{\sqrt{\lambda_{min}(A^2)}}$$

but to no avail.

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Since $A$ is symmetrical and invertible, I know that the following equality with $X$ as a transformation holds. And furthermore, $A$ is positive definit, therefore, all Eigenvales must be real and positive.

$$\|A^{-1}\| = \|X \cdot D^{-1} \cdot X^{-1}\|$$

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How should I tackle this proof? Thank you.

Answer 1

Hint: Showing that $\|A^{-1}\| \geq \frac{1}{\sqrt{\lambda_{min}(A^2)}}$ is the easier part. In particular, it suffices to show that we can find a unit vector $x$ for which $\|A^{-1}x\| = \frac{1}{\sqrt{\lambda_{min}(A^2)}}$.

The other direction of the inequality is a bit trickier. In particular, you'll need to use the fact that I refer to in my comment. The spectral theorem tells us that for a symmetric matrix $A$, there exists an orthogonal matrix $X$ (that is, $X$ satisfies $X^TX = I$) for which $A = XDX^{-1}$