Matrix Theory....showing that matrix is Unitary

[cylers89]

Let x be an element in space C be a given unit vector (x*x=1) and write x=[x,y^T]^T, where x₁ is element in space C and y is element in C^n-1. Choose theta (element in space R) such that e^i(theta)x₁ greater than or equal to 0 and define z=e^i(theta)x =[z₁, B^T]^T, where z₁ is element in R is non negative and B is element in C^n-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 V^TV=I.
So I can do the V^TV.

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

 

[Opalg]

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 implement 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.

 

[cylers89]

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 e^i(theta)x₁ is greater than or equal to 0. and define z=e^i(theta)x "...I'm not understanding what it is asking I guess.

 

[Opalg]

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 e^i(theta)x₁ is greater than or equal to 0. and define z=e^i(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.

 

[blue_raver22]

To show that the matrix V is unitary, we need to prove that VTV = I, where I is the identity matrix.

First, let's simplify VTV. We have:

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

= [z1^2 + BB* B*]
[B*z1 + (-I+((1)/(1+z1)) BB*)B]

= [z1^2 + B*B* B*]
[B*z1 - B*B* + ((1)/(1+z1)) B*B*]

= [z1^2 + B*B* B*]
[B*z1 + (-z1 + ((1)/(1+z1))) B*B*]

= [z1^2 + B*B* B*]
[B*z1 + ((1)/(1+z1)-z1) B*B*]

Next, we can use the given information that x is a unit vector (x*x = 1) to substitute in for z1. This gives us:

VTV = [(x1)^2 + B*B* B*]
[B*x1 + ((1)/(1+(x1)^2)-x1) B*B*]

Now, we can use the definition of z to substitute in for x1. This gives us:

VTV = [(z1)^2 + B*B* B*]
[B*z1 + ((1)/(1+z1)-z1) B*B*]

= [(z1)^2 + B*B* B*]
[B*z1 + ((1-z1^2)/(1+z1)) B*B*]

= [z1^2 + B*B* B*]
[B*z1 + ((1-z1^2)/(1+z1)) B*B*]

= [z1^2 + B*B* B*]
[((1+z1)^2/(1+z1)) B*B*]

= [z1^2 + B*B* B*]
[B*B*]

= [z1^2 + B*B* B*]


= [z1