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An interesting statement without proof in my textbook is that, $\forall A\in M_n(\mathbb{C})$, there exist unitary matrices $U,V\in U_n(\mathbb{C})$ such that $U^{-1}AV$ is diagonal.

After a few experiments I was shocked that this is true. Even if $A$ is not diagonalizable, you can still find some $U,V$ to make it diagonal.

But is there anyway I can prove it?

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You can even take $U=V$ if I remember correctly – Belgi Sep 19 '12 at 19:51
@Belgi Not always, for example if $A$ is not diagonalizable. – Alex Becker Sep 19 '12 at 19:52
@AlexBecker I thought I read the matrix is hermitian, my mistake – Belgi Sep 19 '12 at 19:53
In the literature you can look for "singular value decomposition". – Long Sep 19 '12 at 20:49
Why restrict yourself to square matrices? Let $U$ and $V$ have different sizes. – Michael Hardy Sep 19 '12 at 22:00
up vote 4 down vote accepted

It suffices to show that for some unitary matrix $W$, $AW$ is normal, i.e. $$(AW)^*(AW)=(AW)(AW)^*$$ where $*$ denotes conjugate-transpose. By the spectral theorem (see the linked wikipedia article), we can then write $D=U^{-1}A(WU)$ for some unitary $U$ and diagonal $D$.

The condition $(AW)^*(AW)=(AW)(AW)^*$ simplifies to $W^*A^*AW=AA^*$ since $WW^*=I$. Since $A^*A$ and $AA^*$ are Hermitian, by the spectral theorem they can be diagonalized by unitary matrices, so we have $X^*A^*AX=D$ and $Y^*AA^*Y=D'$ for some unitary $X,Y$ and diagonal $D,D'$. But the entries of $D$ and $D'$ are the eigenvalues of $A^*A$ and $AA^*$, and it is well-known that the eigenvalues of $AB$ and $BA$ coincide for any square matrices $A,B$. Thus $D=D'$ so $X^*A^*AX=Y^*AA^*Y$. Letting $W=XY^*$ which is unitary we get $$W^*A^*AW=AA^*=Y(X^*A^*AX)Y^*=AA^*$$ as desired.

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You can prove this by first proving that a square matrix can be made upper triangular using only row operations. Then you say that a left multiplication by a unitary matrix is equivalent to a sequence of row operations. Once this is done, you take a transpose of the resulting upper triangular matrix and apply the same method to the transposed matrix and then use the property that $(AB)^{t} = B^{t}A^{t}$ along with the property that a right multiplication is equivalent to a sequence of column operations. Alternatively, you can prove that the column transformation can be used to transform the matrix into a lower triangular one.

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