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Unitary matrix - linear alg

  • Thread starter tas3113
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  • #1
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I am going over old practice exams and came across this question:

Find the unitary matrix which diagonalizes the matrix

.......i 1 0
A = (-1 i 0 )
...... 0 0 -i


First off, can someone explain to me about unitary matrices and get me started on this question? I do not know where to begin solving it. Thank you!
 

Answers and Replies

  • #2
Cyosis
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A unitary matrix is a matrix for which [itex]U^{\dagger}U^{-1}=U^{-1}U^\dagger=I[/itex]. So [itex]U^\dagger=U^{-1}[/itex]. A matrix B is diagonalizable if there exists a matrix C such that [itex]C^{-1}BC[/itex] is diagonal.
 
  • #3
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so i found the conjugate transpose of the matrix and got

(-i -1 0)
(1 -i 0)
(0 0 i)

is that a start?
 
  • #4
Cyosis
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No, the idea is that you find [itex]U^\dagger A U[/itex]. A need not be unitary. The first step is to find the eigenvalues of A and their respective eigenvectors. You can however use [itex]A^\dagger[/itex] to check if A is unitarily diagonalizable. A matrix A is unitarily diagonalizable if and only if [itex]A^\dagger A=A A^\dagger[/itex]. Perhaps it's nice to check this for A, although of course A is unitarily diagonalizable.

After you have the eigenvectors check if they form an orthonormal set. If they don't you have to make an orthonormal set by the Gram-Schmidt process. Now make a matrix with its columns being the eigenvectors.
 
Last edited:
  • #5
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so i came up with the equation:
-h(h^2+1)+h^2 i

so the eigenvalues are 0, -i, and something else but i dont know how to find that. 0 = ih^2

the eigenvectors are (1,- i,0) , (0,0,1), and something..
 
  • #6
Cyosis
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That equation doesn't look right to me. I got [itex](-i-\lambda)[(i-\lambda)(i-\lambda)+1]=0[/itex]
 
  • #7
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so its
[tex]
\left|\begin{array}{ccc}i-\lambda & 1 & 0 \\ -1 & i-\lambda & 0 \\ 0 & 0 & -i-\lambda\end{array}\right|
[/tex]

so then
[itex]
[(i-\lambda)(i-\lambda)(-i-\lambda)]-[(-1)(1)(-i-\lambda)]=0
[/itex]
oo.. i forgot the lambda for the last (-i) before.. ok so
[itex]
(i-\lambda)(i-\lambda)(-i-\lambda)-(i+\lambda)=0
[/itex]
and so
[itex]
(-i-\lambda)[(i-\lambda)(i-\lambda)+1]=0
[/itex]
is right

so eigenvalues are 0, -i, 2i
with eigenvectors: (1,- i,0), (0,0,1), (1, i,0)

ok... so as for orthonormal set:
(1,- i,0), (0,0,1) are perpendicular
(0,0,1), (1, i,0) are perpendicular

then? lost from there
 
  • #8
Cyosis
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Well this set of eigenvectors is orthogonal and we want an orthonormal set. So you need to normalize them. After that you can simply form a matrix with the eigenvectors as columns. Then test if it's unitary and diagonalize the matrix A.

A tip when it comes to calculating determinants. You don't always have to start from the top left and work to the top right when you write a 3x3 determinant as the sum of 2x2 determinants. Like in this case the fastest way is:

[tex]
\left|\begin{array}{ccc}i-\lambda & 1 & 0 \\ -1 & i-\lambda & 0 \\ 0 & 0 & -i-\lambda\end{array}\right|=(-i-\lambda)\left|\begin{array}{cc}i-\lambda & 1 \\ -1 & i-\lambda & \end{array}\right|
[/tex]

Start at the bottom right, because that row/column has the most zeros.
 
  • #9
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aren't they already normalized?
u1 = (1,- i,0)
u2 = (0,0,1) - (0+0+0)/... doesnt matter
u3 = (1, i,0) - (0+0+0)/... doesnt matter

so then i would divide by ||u||
u1: squareroot(1^2+(-i)^2+0) = 0 --well then itd be undef if u divide by 0
u2: 1
u3: 0
 
  • #10
Cyosis
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A normalized vector has length 1. So [itex]||u_1||=\sqrt{(u_1,u_1)}=\sqrt{2}[/itex]. Normalizing u1 gives [itex]1/\sqrt{2}(1,-i,0)[/itex]. It works the same for the others.
 
  • #11
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ok. i ended up with the vectors
[itex]
(1/\sqrt{2}, -i/\sqrt{2}, 0), (0, 0, 1), (1/\sqrt{2}, i/\sqrt{2}, 0)
[/itex]

which gives me the matrix
[tex]
\left|\begin{array}{ccc}1/\sqrt{2} & 0 & 1/\sqrt{2} \\ -i/\sqrt{2} & 0 & i/\sqrt{2} \\ 0 & 1 & 0\end{array}\right|
[/tex]

and the conjugate transpose is
[tex]
\left|\begin{array}{ccc}1/\sqrt{2} & i/\sqrt{2} & 0 \\ 0 & 0 & 1 \\ 1/\sqrt{2} & -i/\sqrt{2} & 0\end{array}\right|
[/tex]

then
[tex]
\left|\begin{array}{ccc}1/\sqrt{2} & 0 & 1/\sqrt{2} \\ -i/\sqrt{2} & 0 & i/\sqrt{2} \\ 0 & 1 & 0\end{array}\right|
[/tex] * [tex]
\left|\begin{array}{ccc}1/\sqrt{2} & i/\sqrt{2} & 0 \\ 0 & 0 & 1 \\ 1/\sqrt{2} & -i/\sqrt{2} & 0\end{array}\right|
[/tex] = [tex]
\left|\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right|
[/tex]
which is already diagonalized

so then the unitary matrix/answer is:
[tex]

\left|\begin{array}{ccc}1/\sqrt{2} & i/\sqrt{2} & 0 \\ 0 & 0 & 1 \\ 1/\sqrt{2} & -i/\sqrt{2} & 0\end{array}\right|

[/tex]
?
 
  • #12
Cyosis
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Yep that is correct. You've shown that [itex]U^\dagger U=I[/itex]. Therefore U is unitary. You can test if U indeed diagonalizes A by calculating [itex]U^\dagger A U[/itex].
 
  • #13
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ok. thank you!
 

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