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Let $A$ be an $n$ by $n$ complex matrix. The singular value decomposition of $A$ gives: $A=U\Sigma V^*$, where $U$ and $V$ are unitary matrices, and $\Sigma$ is a diagonal matrix that contains the singular values of $A$. I am defining the new matrix: $Q=V^*U$. I am given that: $\left \| A^2 \right \|=\left \| A \right \|^2$. I want to prove that $Q$ is diagonal. Obviously $Q$ is unitary (and hence it is normal) since it is the product of two unitary matrices. I need only to prove that $Q$ is upper triangular, and use the theorem that says that if you have a normal matrix that is upper-triangular then it is diagonal. Does anyone know how to prove that $Q$ is upper-triangular?

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Which norm is $\lVert \cdot \rvert$? – Davide Giraudo Apr 20 '12 at 20:20
@Davide Giraudo: The operator norm – M.Krov Apr 20 '12 at 20:21
With respect to the Euclidian norm I presume? – Davide Giraudo Apr 20 '12 at 20:22
@Davide Giraudo: You're right. – M.Krov Apr 20 '12 at 20:24
up vote 2 down vote accepted

I believe this is false. Suppose the matrix $A$ is defined as follows $$ A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1/2 & 0 \\ 0 & 0 & 1/2 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1/\sqrt{2} & 1/\sqrt{2} \\ 0 & 1/\sqrt{2} & -1/\sqrt{2} \end{pmatrix} $$ The above is clearly an SVD; the first matrix on the right is $U$, the second is $\Sigma$, the third is $V^T$. Now $||A||_2=1$; moreover, $||A^2||_2=1$ because $A^2$ is the direct sum of the identity on the subspace $W={\rm span}\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ and a matrix on $W^\perp$ which has norm at most $1/2$; so $||A^2||_2 = ||A||^2$. On the other hand, $V^T U$ is not diagonal.

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I may be missing something, but I think it is not true:

Let $\Sigma=U=I$, and let $V=\frac{1}{\sqrt{2}}\left( \begin{array}{cc} 1 &-1 \\ 1 & 1 \end{array} \right)$. Then $||A^2|| = ||A||^2 = 1$, but is clearly not diagonal?

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