I have a matrix $A=\begin{pmatrix} a & b\\ 0 & a\end{pmatrix}\in M_2(\mathbb{C})$. Given that $|a|<1$ and $|b|\leq 1-|a|^2$, I am supposed to show that $\|A\|\leq 1$ (operator norm). I can't get a clear answer out of computing the operator norm as $\|A\|=\displaystyle \sup_{|x|=1}|Ax|$, and the same holds for trying to find eigenvalues of $A^*A$ (or the numerical range of $A^*A$. Since the matrix is not normal, I can't use the spectral theorem. At best, all my approaches do no better than just taking $\|A\|\leq \|aI\|+\left\|\begin{pmatrix} 0 & b\\ 0&0\end{pmatrix}\right\|\leq 1-|a|^2+a$. Is there anything else I can try here?
I should say that this is part of a larger problem in Paulsen, which asks you to show that $\|A\|\leq 1$ iff $|b|\leq 1-|a|^2$. I have already proved that, for $f(z)=\displaystyle\frac{z-a}{1-\overline{a}z}$, and polynomials $p_n(z)=(z-a)\displaystyle\sum_{k=0}^n (\overline{a}z)^k$, which converge to $f$ in $A(\mathbb{D})$ that $p_n(A)=\begin{pmatrix} p_n(a)&bp'_n(a)\\ 0&p_n(a)\end{pmatrix}=\begin{pmatrix} 0&b\sum_{k=0}^n|a|^k\\0&0\end{pmatrix}$ converges in norm to $\begin{pmatrix} f(a)&bf'(a)\\0&f(a)\end{pmatrix}= \begin{pmatrix} 0&\frac{b}{1-|a|^2}\\0&0\end{pmatrix}$. However, this (plus Von Neumann's Inequality) is only sufficient to prove $\|A\|\leq 1\Rightarrow |b|\leq 1-|a|^2$. This is a homework problem, so I'd prefer to have just some guidance.
I don't understand why you cannot get an answer from the eigenvalues of $A^*A$? Evaluating directly the maximal eigenvalue of this matrix gives $$ \|A\|_2^2=|a|^2+\frac{|b|^2}{2}+\frac{|b|}{2}\sqrt{4|a|^2+|b|^2}. $$ Now put the assumption $|b|\leq 1-|a|^2$ (with the one that $|a|\leq 1$ to make it possible) in there to verify that $\|A\|_2\leq 1$. The explicit verification is excluded to avoid homework spoiler but it uses simple algebraic manipulations.