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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: What is the definition of "Unitary operator" that you have? There are several alternative definitions, so exactly which is your basic one would be relevant.2011-02-10
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    Isn't "unitary space" bordering on *ancient* terminology? The modern term is "inner product space"2011-02-10
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    @kahen: Ah, I was about to ask what it was. It may not be ancient, but rather the terminology used elsewhere. Notice that the OP mentions that this is a translation.2011-02-10
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    Hila, apparently $V$ is finite dimensional, correct? Have you seen that every operator on $V$ can be unitarily triangularized?2011-02-10
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    @Arturo http://en.wikipedia.org/wiki/Unitary_operator this one.2011-02-10
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    @kahen @Arturo As far as I've seen, the terminology differs between books (and even between professors in the same institute!). Either way, I think my professor is quite old (his publications list dates back to before the time I was born), so it may be and old terminology.2011-02-10
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    @kahen @Arturo Forgot to add - we use the term "unitary space" for "inner product space over the complex numbers", and "euclidean space" for "inner product space over the real numbers", and just "inner product space" when we don't care about the field.2011-02-10
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    @Jonas Yes, we work only with finite dimensional spaces for now. I know that every operator can be triangularized (assuming the field is algebraically closed), but why unitarily?2011-02-10
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    @Hila: Ehr: the page lists **three** equivalent definitions: $UU^* = U^*U = I$; the range of $U$ is dense; and "$U$ respects the inner product". I am going to guess you mean the first, but you really could have saved some trouble by simply typing it...2011-02-11
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    @Arturo: Dense range and respect of inner product combine to give just one characterization (and in this case dense range is redundant).2011-02-11
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    @Jonas: Yes, I realize that. I was pointing out that when Hila said "this" definition, she pointed to a page that had **three** alternative definitions, not one. With problems like this, exactly which version of the definition you are aiming for is important.2011-02-11
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    @Arturo: I am nitpicking, and I understand your point. (It is likely that Hila has seen the equivalence of these characterizations, but we are not mind readers.) However, I don't see what third definition you are referring to.2011-02-11
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    @Jonas: Oh, I see what you mean; the list of two items are not two equivalent definitions, it's two conditions that make up one condition. Sorry for the confusion...2011-02-11
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    @Arturo Sorry about that. You are right, of course, just wasn't paying the right amount of attention. I am not used to reading math in English, so my eyes traveled to the shiny equation at the top and, well, stayed there...2011-02-11
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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

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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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    Given that $T$ can be triangularizable, it is easy to see that after diagonalizing and computing $T^\dagger T$ you get identity. But it seems that the condition $\parallel Tv\parallel\leq \parallel v\parallel$ is not being used here.2011-02-11
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    @Marcos Villagra: In order to show that $T$ is actually diagonalized, the condition $\|Tv\|\leq\|v\|$ must be used. (Otherwise the result would be false, e.g. consider a linear transformation on a 2-dimensional space that has matrix $\begin{bmatrix}1 & 1\\0&1\end{bmatrix}$ with respect to some orthonormal basis.) The condition is used above at "$1\geq\|Te_k\|^2$".2011-02-11
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    ahh I see, crytal clear now :-)2011-02-11
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    my mistake was that I thought that the eigenvectors of the triangularized matrix automatically diagonalized $T$2011-02-11
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    @Marcos: I see. There is no knowledge a priori that the $e_k$'s are eigenvectors. In order to have an orthonormal basis of eigenvectors an operator must be normal, and that is not given. But in this case, they do turn out to be eigenvectors.2011-02-11
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    Thanks a lot, you really helped me!2011-02-11
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    @Hila: You're welcome.2011-02-11
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    @Jonas 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