# Homework Help: Diagonalization and Unitary Matrices

1. Feb 29, 2016

### Dewgale

\1. The problem statement, all variables and given/known data
Let B = $\left[ \begin{array}{ccc} -1 & i & 1 \\ -i & 0 & 0 \\ 1 & 0 & 0 \end{array} \right]$. Find a Unitary transformation to diagonalize B.

2. Relevant equations
N/A

3. The attempt at a solution
I have found both the Eigenvalues (0, 2, -1) and the Eigenvectors, which are $<0,i,1>,\ \ <2,-i,1>,$ and $<-1,-i,1>$. Vectors 1 and 3 are both already normalized, but I normalized vector two to be $<1,\frac{-i}{2},\frac{1}{2}>$.

They are now orthonormal, but the matrix formed from them,

$$U = \left[ \begin{array}{ccc} 0 & 1 & -1\\ i & \frac{-i}{2} & -i \\ 1 & \frac{1}{2} & 1 \end{array} \right]$$

is not a unitary matrix. It will still succesfully diagonalize B, but I don't know what I've done wrong. Thank you!

2. Mar 1, 2016

### Staff: Mentor

The first and third eigenvectors aren't normalized, nor is the last one you show. For example, the magnitude of the first vector is $\sqrt{2}$. The magnitude of a normalized vector is 1.

3. Mar 1, 2016

### Dewgale

Oh shoot, yeah. My "norm" function in maple was spitting out ones, but I realized that I was using the wrong one.

It looks like two out of my three vectors are fine and normalized ($V_1 = <-1, -i, 1>, V_2 = <1, \frac{-i}{2}. \frac{1}{2}>$), but the third one, $V_3 = < 0, -i, 1>$ has a magnitude of zero, so I don't know how to normalize it.... I'm not sure where to go from here.

4. Mar 1, 2016

### Staff: Mentor

No, they are not normalized. For a complex vector v, $||v|| = \sqrt{v \cdot \bar{v}}$. $||v_1|| = \sqrt{3}$, so it's clearly not normalized.
No. The only vector with a magnitude of zero is the zero vector. If Maple is giving you the wrong values, do them by hand.

5. Mar 1, 2016

### ehild

Are you sure 2 and -1 are eigenvalues?

6. Mar 1, 2016

### Staff: Mentor

That's a very good question.

7. Mar 1, 2016

### Dewgale

$$(Q - \lambda I) = \left[ \begin{array}{ccc} 1-\lambda & i & 1\\ -i & -\lambda & 0 \\ 1 & 0 & -\lambda \end{array} \right]$$
Therefore $det(Q - \lambda I) = 0$ is
$$(1-\lambda)(-\lambda)(-\lambda) + \lambda + \lambda = 0$$
which is simplified to
$$\lambda(\lambda - 2)(\lambda + 1) = 0$$
Therefore, $\lambda = 0, 2, -1$.

8. Mar 1, 2016

### Dewgale

Mark44, thank you so much for your help! I'd forgotten that when finding the norm of a complex vector, you need the conjugate. I'm getting the right answer now.

Thank you!

9. Mar 1, 2016

### Samy_A

In your matrix, the term in column 1, row 1 is -1, not 1 as you assume here.

10. Mar 1, 2016

### Dewgale

Oh wow. That's a typo in my original post, the number in the question is 1.

11. Mar 1, 2016

### Ray Vickson

Maple gives the correct answer if the command is used correctly.