Showing that the 2 norm of a matrix is at most 1

Question

Suppose that $A$ is an invertible matrix whose eigenvalues are at most 1. Can it be shown that $\|A\|_2 \le 1$?

I know that $A = P \Lambda P^{-1}$ which means that $\|A\|_2 \le \|P\|_2 \|\Lambda\|_2 \|P^{-1}\|_2$.

I also know that $\|\Lambda\|_2 $ is the maximum of the absolute value of the diagonal entries (which is 1).

If $A$ is symmetric, I know that $P$ is an orthonormal matrix and so the inequality holds. What about for the general case?

Answer 1

No. $\left\|\pmatrix{1&1\\ 0&1}\right\|_2=\frac{1+\sqrt{5}}{2}>1$. More generally, $\left\|\pmatrix{1&t\\ 0&1}\right\|_2\ge\left\|\pmatrix{t\\ 1}\right\|_2\to+\infty$ when $t\to+\infty$.