Matrix Theory...showing that matrix is Unitary

cylers89

New member
Let x be an element in space C be a given unit vector (x*x=1) and write x=[x,yT]T, where x1 is element in space C and y is element in Cn-1. Choose theta (element in space R) such that ei(theta)x1 greater than or equal to 0 and define z=ei(theta)x =[z1, BT]T, where z1 is element in R is non negative and B is element in Cn-1. Show that the matrix V is unitary.

V= [z1 B* ]
[B -I+((1)/(1+z1)) BB*]

Can someone help me get off on the right foot?

I know that to show unitary, I can prove that VTV=I.
So I can do the VTV.

What I don't understand is how to impliment what z= into my VTV. Would it be better to substitute in z1 in the beginning, or simplify VTV first, then plug in? Does (x*x=1) imply that it works for (B*B) as well?

Opalg

MHB Oldtimer
Staff member
Let $x$ be an element in space C be a given unit vector ($x^*x=1$) and write $x=[x_1,y^{\text{T}}]^{\text{T}}$, where $x_1$ is element in space $C$ and $y$ is element in $C^{n-1}$. Choose $\theta$ (element in space $\mathbb R$) such that $e^{i\theta}x_1$ greater than or equal to 0 and define $z=e^{i\theta}x =[z_1, B^{\text{T}}]^{\text{T}}$, where $z_1$ is element in $\mathbb R$ is non negative and $B$ is element in $C^{n-1}$. Show that the matrix $V$ is unitary.

$$V = \begin{bmatrix}z_1&B^* \\ B & -I + (1/(1+z_1))BB^* \end{bmatrix}$$

Can someone help me get off on the right foot?

I know that to show unitary, I can prove that $V^*V=I$.
So I can do the $V^*V$.

What I don't understand is how to impliment what z= into my $V^*V$. Would it be better to substitute in $z_1$ in the beginning, or simplify $V^*V$ first, then plug in? Does $x*x=1$ imply that it works for $B^*B$ as well?
The condition $xx^*x = 1$ tells you that $|x_1|^2 + y^*y=1$. Since $z$ is obtained from $x$ by multiplying by a scalar of modulus 1, it follows that the analogous result holds for $z$, namely $z_1^2 + B^*B = 1$. Notice that in this case we do not need to put mod signs around $z_1$, because $z_1$ is real and nonnegative. Now compute $V^*V$, and then substitute $1-z_1^2$ for $B^*B.$ You should find that $V^*V$ is the identity matrix.

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cylers89

New member
Okay, that makes sense. I do remember covering that in class. So I also know that BB*=(1-z1)(1+z1).

So for the matrix, after I multiply V*V...row1column1 easily is substituted to 1, since z12+B*B = 1.

Now I am getting tripped up on the canceling of row2column2 to 1.

The first one is: BB*+(-I+(1/1+z1)BB*)2
Substituting and squaring the binomial gets: (1-z12)+(I2-I+Iz1-I+Iz1+1-z1-z1+z12)
Combining like terms gives me: 2+I2-2I+2Iz1-2z1
This is where I am getting fuzzy...I cannot seem to figure out how to cancel this down to 1. It seems to overlap just enough that I am going in loops.

Now once I finish that part up, V*V will in fact be shown to be equal to the Identity matrix.

What does the directions mean by "choose theta in R such that ei(theta)x1 is greater than or equal to 0. and define z=ei(theta)x "...I'm not understanding what it is asking I guess.

Opalg

MHB Oldtimer
Staff member
Okay, that makes sense. I do remember covering that in class. So I also know that BB*=(1-z1)(1+z1).

So for the matrix, after I multiply V*V...row1column1 easily is substituted to 1, since z12+B*B = 1.

Now I am getting tripped up on the canceling of row2column2 to 1.

The first one is: BB*+(-I+(1/1+z1)BB*)2
(I may have confused you in my previous comment by writing $BB^*$ where it should have been $B^*B$. I have edited it to correct that.)

Okay, so the (2,2)-element of your matrix is $$BB^* + \Bigl(-I + \tfrac1{1+z_1}BB^*\Bigr)^2 = BB^* + I - \tfrac2{1+z_1}BB^* + \tfrac1{(1+z_1)^2}B(B^*B)B^*.$$

Replace the $B^*B$ that I have bracketed in that last term by $1-z_1^2$, and you will see that the whole thing reduces to $I$, as required. In a similar way, the off-diagonal terms in the matrix are both 0.

What does the directions mean by "choose theta in R such that ei(theta)x1 is greater than or equal to 0. and define z=ei(theta)x "...I'm not understanding what it is asking I guess.
In this problem, $x_1$ is a complex number, and any complex number can be multiplied by a complex number of modulus 1 (that is, a number of the form $e^{i\theta}$) so as to become real and nonnegative. The point here is that the computation to show that $V$ is unitary relies on the fact the element in the top left corner of the matrix is real. So you need a construction that replaces $x_1$ by $z_1$ in order to make $V$ unitary.

• cylers89