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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.

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    @Hila: Not a problem; as you can read above, I too did not accurately report the contents of the page. (-:2011-02-11

1 Answers 1

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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.

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    @Jo$n$as Arrgh! I still don't get it (http://math.stackexchange.com/questions/21668/orthogonal-projection). At least I learnt how to format my questions here ;)2011-02-12