How to Determine a Unitary Matrix that Diagonalizes a Given Matrix?

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In summary: You want the columns of U to be eigenvectors of A, so if you divide each eigenvector by its length, the columns of U are guaranteed to be orthonormal, which is what you need to get a unitary matrix.In summary, to find the unitary matrix U that diagonalizes matrix A, first find the eigenvalues of A. Then, calculate the normalized eigenvectors for each eigenvalue. The columns of U should be the normalized eigenvectors of A. The order of the columns in U will determine the order of the eigenvalues on the diagonal of U^\dagger AU, which should result in a diagonal matrix with the eigenvalues along the diagonal. The order of the columns and eigenv
  • #1
Brewer
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Homework Statement


Given the matrix A =
(1 1 -1)
(1 -1 1)
(-1 1 1)
write down the unitary matrix U which diagonalises A. Verify that [tex]U^\dagger AU[/tex] is diagonal with the eigenvalues of A along the diagonal.

Homework Equations


The Attempt at a Solution


The eigenvalues were calculated earlier in the question, and found to be -2, 1 and 2. I know these are correct.

For U I would have said that the question (phrased as it is with "write down" so should not require any thinking) wants me to say that U has the same diagonal as A, but zeros in the other 6 elements of the matrix. And because this is a real matrix the adjoint of the matrix will be itself, because it is its own transpose. However when I come to calculate the final part of the question, I don't get a diagonal matrix, or the eigenvalues appearing anywhere, so I'm a little confused. Any pointers in the right direction wuold be appreciated.
 
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  • #2
You should also find the normalized eigenvectors of A. Call them v1, v2 and v3. If the standard basis is e1, e2 and e3, then U should be the matrix such that U(e1)=v1, U(e2)=v2 and U(e3)=v3. Ie, the matrix whose columns are v1, v2 and v3.
 
  • #3
If the eigenvalues are -2, 1, and 2, what is are unit eigenvectors corresponding to each eigenvalue? The columns of U are those eigenvectors.

It works for me. Be sure you do the multiplication in the right order- [tex]U^\dagger AU[/tex] will give you a diagonal matrix. [tex]UAU^\dagger[/tex] will not.
 
  • #4
Going back to finding the eigenvectors of the equation:

I can say that [tex]Ax = \lambda x[/tex] which can be re-written in the form of a secular equation and [tex]\lambda[/tex] can be found by solving the cubic secular equation that results to give values of -2, 1 and 2.

Now in the case of [tex]\lambda[/tex] = -2, I have stated that [tex](A-\lambda)x = 0[/tex] and solved this getting that [tex]x_1 = \frac{-x_2}{2} = x_3[/tex] I would have said that from here the eigenvector is (1,-1/2,1), but I used a matrix calculator on the web to check my answer and it tells me that the answer is (1,-2,1).

Why is there this discrepancy with my answers?

And back to the original point of the question, does the order matter when entering these eigenvectors into the unitary matrix? Other than making sure that it is its transpose I guess.
 
  • #5
Brewer said:
Going back to finding the eigenvectors of the equation:

I can say that [tex]Ax = \lambda x[/tex] which can be re-written in the form of a secular equation and [tex]\lambda[/tex] can be found by solving the cubic secular equation that results to give values of -2, 1 and 2.

Now in the case of [tex]\lambda[/tex] = -2, I have stated that [tex](A-\lambda)x = 0[/tex] and solved this getting that [tex]x_1 = \frac{-x_2}{2} = x_3[/tex] I would have said that from here the eigenvector is (1,-1/2,1), but I used a matrix calculator on the web to check my answer and it tells me that the answer is (1,-2,1).

Why is there this discrepancy with my answers?
Since you don't say how you solved these equations, we can't say where you went wrong- but you did solve them wrong.
With [itex]\lambda[/itex]= -2, the equation Ax= -2x gives:
x+ y- z= -2x, x- y+ z= -2y, and -x+ y+ z= -2z. The first is equivalent to 3x+ y- z= 0, the second x+ y+ z= 0, and the third -x+ y+ 3z= 0. If you subtract the third equation from the first you eliminate y and get 4x- 4z= 0 or z= x. Putting that into the second equation, x+y+ x= 2x+ y= 0 so y= -2x. Taking x= 1, an eigenvector is < 1, -2, 1> as your "matrix calculator" said. To get a unitary matrix, you will need to divide that by its length, and use that vector as a column.

And back to the original point of the question, does the order matter when entering these eigenvectors into the unitary matrix? Other than making sure that it is its transpose I guess.
Changing the order of the columns changes the order of the eigenvalues on the diagonal. The order of the eigenvalues will be the same as the order of their corresponding eigenvectors.
 
  • #6
Yes, I subsequently found out how to find the eigenvectors, giving me the same answer as the "calculator".

I think this helps that you've just said. I'll give it a go like this and see what happens.

By dividing by the length of the vector, does that make it a "normalised" eigenvector. My notes tell me to do that involves the sum over the square of the modulus of the matrix elements, but I don't follow what it means. Although thinking about what you've said that makes sense to me now.

Thank you for your help.
 
  • #7
Yes, divide the eigenvector by its length. The result is an eigenvector that has length one. That's all normalized means.
 

What is a unitary matrix?

A unitary matrix is a square matrix where the conjugate transpose of the matrix is equal to its inverse. In other words, if U is a unitary matrix, then U^H * U = U * U^H = I, where I is the identity matrix.

What are the properties of unitary matrices?

Unitary matrices have the following properties:

  • They preserve vector length, meaning that the length of a vector multiplied by a unitary matrix will remain the same.
  • They preserve angles, meaning that the angle between two vectors will remain the same after being multiplied by a unitary matrix.
  • They have orthonormal columns, meaning that the columns of a unitary matrix are orthogonal to each other, and each column has a length of 1.
  • They have orthonormal rows, meaning that the rows of a unitary matrix are also orthogonal to each other and have a length of 1.

What is the significance of unitary matrices in linear algebra?

Unitary matrices are important in linear algebra because they represent linear transformations that preserve both vector length and angles. They are also used in many applications, such as in quantum mechanics, signal processing, and data compression.

How are unitary matrices related to Hermitian matrices?

A Hermitian matrix is a square matrix that is equal to its own conjugate transpose. In other words, if A is a Hermitian matrix, then A^H = A. Unitary matrices can be used to diagonalize Hermitian matrices, meaning that they can be transformed into a diagonal matrix with real eigenvalues.

Can any matrix be a unitary matrix?

No, not all matrices can be unitary. A matrix can only be unitary if it meets the requirements of having orthonormal columns and rows, and if its inverse is equal to its conjugate transpose.

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