Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?


share|cite|improve this question
Isn't "unitary space" bordering on ancient terminology? The modern term is "inner product space" – kahen Feb 10 '11 at 21:16
Hila, apparently $V$ is finite dimensional, correct? Have you seen that every operator on $V$ can be unitarily triangularized? – Jonas Meyer Feb 10 '11 at 21:24
@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. – Hila Feb 10 '11 at 22:37
@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... – Arturo Magidin Feb 11 '11 at 1:01
@Arturo: Dense range and respect of inner product combine to give just one characterization (and in this case dense range is redundant). – Jonas Meyer Feb 11 '11 at 1:28
up vote 3 down vote accepted

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.

share|cite|improve this answer
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. – Marcos Villagra Feb 11 '11 at 1:53
@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 "$1\geq\|Te_k\|^2$". – Jonas Meyer Feb 11 '11 at 1:58
ahh I see, crytal clear now :-) – Marcos Villagra Feb 11 '11 at 2:05
my mistake was that I thought that the eigenvectors of the triangularized matrix automatically diagonalized $T$ – Marcos Villagra Feb 11 '11 at 2:13
@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. – Jonas Meyer Feb 11 '11 at 2:23

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.