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How to prove a unit norm matrix is the average of two unitary matrix

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which norm is unit? –  Igor Rivin Jan 2 '11 at 17:49
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Do you know this is true? How many dimensions are you working in? –  Peter Shor Jan 2 '11 at 17:55
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I think the only norm where this question makes sense is the spectral norm. I see the votes to close piling up, but since nobody has jumped in and said that this is a well-known elementary fact, I think if you add some details to your question, and flag it to be reopened, it probably will be. There's an easy proof for 2 dimensions and the spectral norm. My first reaction is that I wouldn't think it would be true for arbitrarily many dimensions, and the spectral norm, but I could easily be wrong. –  Peter Shor Jan 2 '11 at 18:24
    
Further to Peter Shor's comment: mathoverflow.net/howtoask –  Yemon Choi Jan 2 '11 at 21:41
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Moderators: Please rollback to the original statement of the question! –  Andres Caicedo Jan 6 '11 at 17:15

1 Answer 1

I assume that the norm is the spectral norm.

By the polar decomposition, any unit norm matrix can be written as $UD$ where $U$ is unitary and $D$ is matrix which, in some basis $E$, is diagonal with non-negative entries not greater than $1$. The diagonal entries can thus be written as $\cos(\theta_1),...,\cos(\theta_n)$ for real $\theta_i$'s. $UD$ is then the average of $U V$ and $U V^*$ where $V$ is the matrix which, in the basis $E$, is diagonal with diagonal entries $e^{i\theta_1},...,e^{i\theta_n}$.

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thanks for your answering this question. However I still doubt whether the basis E is orthogonal(unitary) basis. –  yaoxiao Jan 2 '11 at 20:41
    
@yaoxiao: the matrix $D$ is hermitian (it is even positive semid-efinite), and every hermitian matrix can be diagonalized in an orthonormal basis. –  Mikael de la Salle Jan 2 '11 at 20:48
    
I see what you have mean,great! I think it is really interesting, when we compare it with the similar claim “every matrix can be decompsed into linear combinations of four orthogonal matrix –  yaoxiao Jan 3 '11 at 4:27

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