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

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:

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

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.

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.

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

How should I tackle this proof? Thank you.

• You can say a little bit more than the fact that $A$ is diagonalizable. In particular, you should use the fact that we can select an orthogonal $X$ such that $A = XDX^{-1}$. – Ben Grossmann Oct 29 '18 at 16:36

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 $$A = XDX^{-1}$$