Understanding matrix norms and spectral radius

matricesmatrix-normsspectral-radius

I thought I understood matrix norms and spectral radius after reading proof of $||A||\geq \rho(A)$. However, in the lecture notes I'm given the following line, and asked what is wrong with the reasoning?

$$
|\lambda|||x||=||\lambda x||=||A x|| \leq ||A||\ ||x||
$$

Now, I suspect the part $||\lambda x|| = ||Ax||$ to be wrong, however, the notes aren't giving a definite answer. Could someone point me in the right direction or try explain what I'm looking for?

First we are given the statement that $||A||\geq\rho(A)$ followed by a proof where $x$ is a eigenvector of A associated to an eigenvalue $\lambda$. We are given a proof utilizing norms from
$$
A(xy^T)=(Ax)y^T=\lambda (xy^T)
$$

where $y \in C^n$ to arrive at the definition $|\lambda| \leq ||A||$ and since its true for every eigenvalue of A, we also get $\rho(A)\leq ||A||$.

Then the following appears after the proof (note the etc. at the end):

Question to the reader: What is wrong with the following line of reasoning
$|\lambda|\ ||x||=||\lambda x||=||A x|| \leq ||A||\ ||x||$ etc.

Best Answer

To make it more clear: Let's denote the eigenvektor by $y$, then we have:

$$|\lambda|||y||=||A y||$$ hence we get $$|\lambda| = \frac{||Ay||}{||y||} \le \sup_{||x|| \not= 0}\frac{||Ax||}{||x||} = ||A||$$

Multiplying with ||x|| gives the result.

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