Diagonalization and Unitary Matrices

[Dewgale]

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Homework Statement

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.

Homework Equations

N/A

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!

 

[Mark44]

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Homework Statement

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.

Homework Equations

N/A

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 $<1,\frac{-i}{2},\frac{1}{2}>$.

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.

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!

 

[Dewgale]

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.

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.

 

[Mark44]

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 ($V_1 = <-1, -i, 1>, V_2 = <1, \frac{-i}{2}. \frac{1}{2}>$),

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.

but the third one, $V_3 = < 0, -i, 1>$ has a magnitude of zero

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.

, so I don't know how to normalize it... I'm not sure where to go from here.

 

[ehild]

I have found the Eigenvalues (0, 2, -1) !

Are you sure 2 and -1 are eigenvalues?

 

[Mark44]

Are you sure 2 and -1 are eigenvalues?

That's a very good question.

 

[Dewgale]

Are you sure 2 and -1 are eigenvalues?

Yep, about that I'm certain.

$$(Q - \lambda I) =
\left[ \begin{array}{ccc} 1-\lambda & i & 1\\ -i & -\lambda & 0 \\ 1 & 0 & -\lambda \end{array} \right]$$
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, $\lambda = 0, 2, -1$.

 

[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!

 

[Samy_A]

Yep, about that I'm certain.

$$(Q - \lambda I) =
\left[ \begin{array}{ccc} 1-\lambda & i & 1\\ -i & -\lambda & 0 \\ 1 & 0 & -\lambda \end{array} \right]$$
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, $\lambda = 0, 2, -1$.

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

 

[Dewgale]

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

 

[Ray Vickson]

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.

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