TITLE: Linear transformations and norm
QUESTION [7 upvotes]: I am studying Linear Algebra II, and I came across several questions in which, for a certain linear transformation ($T\colon\mathbf{V}\to\mathbf{V}$) I was told that:
$$||T(a)|| \leq ||a||.$$
I am not completely certain how to use this information. For instance, consider the following question (please forgive my translation, it's the first time I write math in English):

For a linear transformation $T\colon\mathbf{V}\to\mathbf{V}$in a unitary space [i.e., complex inner product space], such that

$|c|=1$ for every eigenvalue $c$ of $T$;
$||T(a)|| \leq ||a||$ for every vector $a$ in $\mathbf{V}$;
  prove that T is a unitary operator.


How does the fact that $||T(a)|| \leq ||a||$ help me?
Thanks.

REPLY [4 votes]: Every linear transformation on a finite dimensional complex inner product space is unitarily triangularizable.  See for example Hogben's Handbook of linear algebra.  This means you can find an orthonormal basis $e_1,\ldots e_n$ such that $Te_k=\sum_{i\leq k}a_{ik}e_i$.  The eigenvalues of a triangular matrix are the diagonal entries, so $|a_{kk}|=1$ for each $k$.  Thus $1\geq\|Te_k\|^2=\sum_{i\leq k}|a_{ik}|^2\geq|a_{kk}|^2=1$, forcing $a_{ik}=0$ for $i\lt k$.  So the basis actually diagonalizes $T$, and it is now straightforward to show that $T$ is unitary regardless of which definition you use.