Determining Unitarily Equivalent Matrices

[Shackleford]

I've had the flu all week.

Of course, the book defines unitary equivalent, but it doesn't talk about an efficient method of determining if two matrices are unitarily equivalent.

If A and B are similar and tr(B*B) = tr(A*A), then A and B are unitarily equivalent.

If A and B are normal matrices and have the same eigenvalues, then they are unitarily equivalent.

Is there an efficient way to determine if these matrices are unitarily equivalent?

<br /> \begin{bmatrix}<br /> 0 &amp; 1 &amp; 0\\ <br /> -1 &amp; 0 &amp;0 \\ <br /> 0 &amp;0 &amp;1 <br /> \end{bmatrix}

<br /> \begin{bmatrix}<br /> 1 &amp; 0 &amp; 0\\ <br /> 0 &amp; i &amp;0 \\ <br /> 0 &amp;0 &amp;-i <br /> \end{bmatrix}

 

[micromass]

You can easily find the eigenvalues, no?

 

[Shackleford]

You can easily find the eigenvalues, no?

How does that relate to unitary equivalence?

 

[Shackleford]

I found that A and B are unitarily equivalent if they have the same sets of eigenvalues, counting multiplicity.

A = P*BP (unitarily equivalent)

det(A) = det(P*BP) = det(P*)det(B)det(P) = det(P*)det(P)det(B) = det(B)
det(A) = det(B)

Their characteristic polynomials must be equal.

 

[HallsofIvy]

I found that A and B are unitarily equivalent if they have the same sets of eigenvalues, counting multiplicity.

No, that is not true. The matrices
A= \begin{bmatrix}1 &amp; 0 \\ 0 &amp; 1\end{bmatrix}
and
B= \begin{bmatrix}1 &amp; 1 \\ 0 &amp; 1\end{bmatrix}
have the same eigenvalues (1 with multiplicity two) but are not unitarily equivalent because they do not have the same eigenvectors. A has every vector as eigenvector while B has only multiples of <1, 0> as eigenvectors.

Two matrices are "unitarily equivalent" if and only if they have the same eigenvalues and the same corresponding eigenvectors.

A = P*BP (unitarily equivalent)

det(A) = det(P*BP) = det(P*)det(B)det(P) = det(P*)det(P)det(B) = det(B)
det(A) = det(B)

Their characteristic polynomials must be equal.

 

[Shackleford]

No, that is not true. The matrices
A= \begin{bmatrix}1 &amp; 0 \\ 0 &amp; 1\end{bmatrix}
and
B= \begin{bmatrix}1 &amp; 1 \\ 0 &amp; 1\end{bmatrix}
have the same eigenvalues (1 with multiplicity two) but are not unitarily equivalent because they do not have the same eigenvectors. A has every vector as eigenvector while B has only multiples of <1, 0> as eigenvectors.

Two matrices are "unitarily equivalent" if and only if they have the same eigenvalues and the same corresponding eigenvectors.

Okay, so I found the eigenvalues of each of the matrices: 1, -i, +i. Now I have the tedious job of finding the eigenvectors. -_-

 

[Dick]

Okay, so I found the eigenvalues of each of the matrices: 1, -i, +i. Now I have the tedious job of finding the eigenvectors. -_-

Halls is definitely wrong to say that they have to have the same eigenvectors. You just have to have the same number of linearly independent eigenvectors for every eigenvalue. You have three distinct eigenvalues. That means you don't have to compute the eigenvectors. Why?

 

[Shackleford]

Halls is definitely wrong to say that they have to have the same eigenvectors. You just have to have the same number of linearly independent eigenvectors for every eigenvalue. You have three distinct eigenvalues. That mean you don't have to compute the eigenvectors. Why?

Ah, you're right. The dimensions of the eigenspaces are equal - 3.