# Bounding the condition number by the eigenvalues

Let $$M$$ be an invertible $$n\times n$$ matrix. It is about bounding the condition number, $$cond (M) = \|M\| \|M^{-1}\|,$$ by it's eigenvalues given some matrix norm $$\|\ \|$$ and assuming $$\|Mx\| = \|M\| \|x\|, x\in \mathbb{R}^n.$$ I want to show, $$cond(M)\geq |\frac{\lambda_{max}}{\lambda_{min}}|,$$ for $$\lambda_{max},\lambda_{min}$$ the largest and smallest eigenvalue of $$M.$$ In case of the spectral norm I can get an equality by using the spectral radius. I do not see why there would rather hold in general an inequality. Can one provide some suggestion or a solution proposal ?

If $$\lambda$$ is an eigenvalue of $$M$$ and $$v\ne0$$ is a corresponding eigenvector, then you have that $$\|M\| = \sup_{x\ne 0}\frac{\|Mx\|}{\|x\|} \ge \frac{\|M v\|}{\|v\|} = \frac{|\lambda| \|v\|}{\|v\|} = |\lambda|$$

Since the eigenvalues of $$M^{-1}$$ are the inverses of the eigenvalues of $$M$$, you know that, for every eigenvalue $$\lambda$$ of $$M$$, $$\|M^{-1}\| \ge \frac{1}{|\lambda|}$$.

Finally,

$$cond(M) = \|M\| \|M^{-1}\| \ge |\lambda_{\max}| \cdot \frac{1}{|\lambda_{\min}|}.$$

Your assumption on the Matrix $$M$$ makes this identity in fact kind of trivial. Defining $$\tilde{M} = \frac{1}{\lVert M \rVert} M$$ we see that $$\tilde{M}$$ is isometric and bijective, i.e. orthogonal. Thus its eigenvalues are located on the unit circle. Since the eigenvalues of $$M$$ correspond to the eigenvalues of $$\tilde{M}$$ multiplied by $$\lVert M \rVert$$, one observes that $$\left\lvert \frac{\lambda_{max}}{\lambda_{min} } \right\rvert$$=1. But it also holds $$1 = \lVert \tilde{M}^{-1} \rVert = \lVert M \rVert \lVert M^{-1} \rVert.$$ Hence, both sides are equal to one.

• Dividing a matrix by a its norm does not make the matrix generally orthogonal nor does it move its eigenvalues on the unit circle. Just consider a diagonal matrix with distinct diagonal entries. Commented Jan 29 at 3:33
• Of course it does not. But the OP's assumption on the Matrix $M$ was $\lVert Mx \rVert = \lVert M \rVert \lVert x \rVert$ in which case this does indeed hold. Commented Jan 30 at 17:51
• I see, I missed that. Though it seems to me that it’s a typo in the question. Commented Jan 31 at 6:27