# product of hermitian and unitary matrix

Could anyone tell me how to show that, for any $g\in GL_n(\mathbb{C})$, $\exists$ $R$ a hermitian matrix with positive eigenvalues and $U$ an unitary matrix such that $g=RU$?

And (I am not sure) can we use this to prove that $GL_n(\mathbb{C})$ is connected?Is the set of hermitian matrices with positive eigen values are path connected?

• See also this question about the path-connectedness of $GL_n(\mathbb C)$. – joriki May 2 '12 at 17:35
• Sounds like the polar decomposition of a matrix... – J. M. is a poor mathematician May 2 '12 at 17:36
• Dear joriki, actually I need to prove the result, I know the path connectedness of inveritble complex matrices. – Marso May 2 '12 at 17:52
• @Makuasi: I know you know, since you asked that question; but others might not know; it would have made sense to link to that question in this question. – joriki May 2 '12 at 19:55

1) the set of diagonal matrices is path connected: if $A=\sum a_j E_{jj}$, $B=\sum b_j E_{jj}$ we take the map $t\mapsto \sum (ta_j+(1-t)b_j) E_{jj}$, $t\in[0,1]$.

2) The set of unitaries is path connected. If $U,V$ are two unitaries, we can always write them as $U=e^{iA}$, $V=e^{iB}$ with $A,B$ hermitian. Then we can consider the map $t\mapsto e^{itA}e^{i(1-t)B}$, $t\in[0,1]$ which gives a path from $U$ to $V$ within the unitary group.

3) The set of invertible hermitian matrices with positive eigenvalues is path connected. If $A,B$ are like that, then $A=UD_AU^*$, $B=VD_BV^*$. By parts 1) and 2), there exist continuous $f,g:[0,1]\to M_n(\mathbb{C})$ with $f(0)=D_A$, $f(1)=D_B$, $g(0)=U$, $g(1)=V$. Then $t\mapsto g(t)f(t)g(t)^*$ is continuous and takes $A$ to $B$. Note that the way that $f$ is defined guarantees that $f(t)$ will have positive eigenvalues for all $t\in[0,1]$.

4) GL$_n(\mathbb{C})$ is connected: Given $A,B$ invertible, we can write them as $A=RU$, $B=SV$ with $R,S$ hermitian and positive, and $U,V$ unitaries. By 3) and 2) we can find continuous functions $f,g:[0,1]\to M_n(\mathbb{C})$ with $f(0)=R$, $f(1)=S$, $g(0)=U$, $g(1)=V$. Then the map $t\mapsto f(t)g(t)$ is a continuous path from $A$ to $B$ (note that $f(t)$ and $g(t)$ are invertible for every $t\in[0,1]$ and then so is their product).

It only remains to justify the polar decomposition $A=RU$. An easy way to see this is by using the singular value decomposition. We write $A=WDV$, with $W,V$ unitaries and $D$ diagonal with non-negative entries (positive if $A$ is invertible). Then we can write $$A=(WDW^*)WV=RU,$$ where $R=UDU^*$ is hermitian with positive eigenvalues (because $D$ is), and $U=WV$ is a unitary.

• I just added an explanation for the polar decomposition. – Martin Argerami May 2 '12 at 19:16
• thanx and great pleasure dear sir, i just want to know about 2)..how unitaries can be written as exponential hermitian? – Marso May 2 '12 at 19:18
• oh! I dont know about singular value decomposition, could you give me some reference to study about that? – Marso May 2 '12 at 19:20
• @Makuasi: "singular value decomposition" is a well-known linear algebra theorem. It should be in any book about matrices. – Martin Argerami May 2 '12 at 19:27
• For the unitaries, the argument is like this: any unitary $U$ is normal, so it is diagonalizable by a unitary: there exists $V$ unitary, $D$ diagonal with $U=VDV^*$. From $U^*U=I$ one sees that $D^*D=I$, i.e. each entry $D_{jj}$ is a complex number of modulus $1$; these can be written as $D_{jj}=e^{i F_{jj}}$ for real numbers $F_{jj}$. So $D=e^{iF}$ if we complete $F$ with zeroes off the diagonal. Finally, $U=VDV^*=Ve^{iF}V^*=e^{iVFV^*}$. – Martin Argerami May 2 '12 at 19:31

This is how I'd do it.

Assume that your matrix has an expression $A=HU,$ where $H$ is a positive-definite Hermitian matrix and $U$ is unitary (inverting the expression you obtain such a polar decomposition in the reverse, i.e. writing the polar decomposition for $A^{-1}$ and then inverting again one obtains $A=U' H'$).

If $A=HU$, where both $H, U$ have a peculiar behaviour with regards to the involution $A\mapsto A$, then how can we find $H, U$?

Apply $^*$ to the identity yields $$A^*=U^{-1}H.$$ It is thus tempting to multiply $$AA^*=H^2.$$

If $A$ is invertible, which is the case, the l.h.s. is positive definite! Indeed, it suffices to write down $$\langle x, AA^*x \rangle = \langle A^*x, A^* x\rangle > 0$$ for $x\neq 0.$

You would have to find positive-definite Hermitian matrices $H$ whose square is $A A^*.$ But then one has the following Lemma.

Lemma For every positive-definite Hermitian matrix $B$, there exists only one positive-definite Hermitian square root $H$.

Proof: Indeed, the eigenspace $Ker(B-\lambda^2)$, where $\lambda$ is the positive square root of an eigenvalue $\mu$ of $B$, plays a role here. One has

$$Ker(B-\lambda^2)=Ker(H^2-\lambda^2)=Ker((H+\lambda)(H-\lambda)).$$ Since $H$ is positive-definite, one has that $H+\lambda$ is non-singular, i.e. invertible, so

$$Ker(B-\mu)=Ker(H-\sqrt{\mu}).$$ Thus $\mathbb{C}^n$ is a direct sum of eigenspaces $Ker(H-\lambda)$ where $H$ acts as multiplying by $\lambda$, which renders $H$ unique.

As for $U$, you will have to prove that $U$ thus chosen so $A=HU$ is unitary (i.e. check whether the definition of unitarity holds for $U$).

• I forgot to add that connectedness of $GL(n,\mathbb{C})$ hinges on connectedness of $U(n)$, and that this is part of the philosophy that given a Lie group $G$ and a compact maximal subgroup $K$, the inclusion $$K\hookrightarrow G$$ is an homotopical equivalence (Hochschild, see also I. Satake's book on symmetric domains). – Theon Alexander Feb 8 '17 at 22:16