How can a diagonalising matrix be unitary?

[George Keeling]

Homework Statement

This refers to problems A.28/29 from Quantum Mechanics – by Griffiths & Schroeter.
I am confused by the request for unitary diagonalising matrix.

Relevant Equations

T'=ST inverse(S), hermitian (U) = inverse (U)

This refers to problems A.28/29 from Quantum Mechanics – by Griffiths & Schroeter.

I’ve now almost finished the Appendix of this book and been greatly helped with the problems by Wolfram Alpha.

In problems A.28/29 we are asked to "Construct the unitary matrix S that diagonalizes T" where T is some matrix. The diagonal matrix is given by
$$\rm{}T^\prime=STS^{-1}$$The columns of $\rm{}S^{-1}$ are the eigenvectors of $\rm{}T$. $\rm{}S$ diagonalises $\rm{}T$.

A unitary matrix is one where the hermitian is the same as the inverse: $\rm{}U^\dagger=U^{-1}$.

In neither question did $\rm{}S^{-1}=S^\dagger$. So why are they asking for a "unitary matrix S"? Am I supposed to somehow manipulate $\rm{}S$ so it not only diagonalises $\rm{}T$ but is also unitary?

 

[vela]

First off, $T$ has to be Hermitian, so check that first.

If $T$ has repeated eigenvalues, you need to find eigenvectors for an eigenvalue that are orthogonal.

 

[George Keeling]

T is Hermitian in both questions. In first (T was a 2x2 matrix) there were two different eigenvalues. In the second (T was 3x3) 2 eigenvalues were the same but I found orthogonal eigenvectors (and even normalised them)

 

[vela]

You'll need to show your work then. We can't see where you're going astray.

 

[George Keeling]

In the first of the questions we had
$$T=\left(\begin{matrix}1&1-i\\1+i&0\\\end{matrix}\right)$$Part (d) of the question is "Construct the unitary diagonalizing matrix $S$ , and check explicitly that it diagonalizes $T$."

I got a diagonalising matrix
$$S=\frac{1}{6}\left(\begin{matrix}2&1-i\\2&-2+2i\\\end{matrix}\right),\ \ S^{-1}=\left(\begin{matrix}2&1\\1+i&-1-i\\\end{matrix}\right)$$Which give
$$T^\prime=STS^{-1}=\left(\begin{matrix}2&0\\0&-1\\\end{matrix}\right)$$Clearly $S^\dagger\neq S^{-1}$ so $S$ is not unitary. Back to my question: why are they asking for a "unitary diagonalizing matrix $S$"?

My three page workings are in the attached pdf. I have rechecked them with wolfram alpha. Links are provided.

A shortcut is to use wolfram alpha here to get the diagonalising matrix. It's a little confusing because wolfram swaps the $S,S^{-1}$ so its $S$ is my $S^{-1}$ which is formed from the eigenvectors of $T$. And wolfram's eigenvectors differ from mine by a constant factor.

Nevertheless wolfram's diagonalising matrix is not unitary either.

 

[Steve4Physics]

EDIT: It may not be significant but
https://www.emathhelp.net/en/calculators/linear-algebra/eigenvalue-and-eigenvector-calculator/?i=[[1,1-i],[1+i,0]]
shows (unnormalised) eigenvectors for $T$ which are different to the ones shown your Post #5 attachment.

By the way, for simplicity, $\lambda^2-\lambda-2 = (\lambda-2)(\lambda+1)$.

 

[George Keeling]

It's not significant. My eigenvector for eigenvalue $2$ is $(1+i)$ time emathhelp's. The other one is $(−0.5+0.5i)$ time emathhelp's. You can always multiply an eigenvector by a scalar to get another eigenvector for the same eigenvalue.

 

[Steve4Physics]

It's not significant. My eigenvector for eigenvalue $2$ is $(1+i)$ time emathhelp's. The other one is $(−0.5+0.5i)$ time emathhelp's. You can always multiply an eigenvector by a scalar to get another eigenvector for the same eigenvalue.

Yes, I was careless when I checked. Apologies.

Check out this video which solves a very similar problem. There are some differences in the way the problem is solved compared to your method.

Probably the important bit is from 9:00 if you don't want to watch the full 11 minutes.

 

[George Keeling]

The video does a diagonalisation but I think I got that right and it was confirmed by wolfram alpha as per my post #5. The point is that neither I nor wolfram found a unitary diagonalising matrix....
PS thanks for your diligence!

 

[martinbn]

$$S=\frac{1}{\sqrt{3}}\left(\begin{matrix}1+i&1\\-1&1-i\\\end{matrix}\right)$$

 

[vela]

You have to normalize the eigenvectors.

Because any non-zero multiple of an eigenvector is also an eigenvector, the diagonalizing matrix $S$ isn't unique. You'll need to figure out the correct form for the eigenvectors to make to make $S$ unitary.

 

[George Keeling]

Gosh thanks!
I didn't have to normalise the eigenvectors I don't think. I took the eigenvectors which are in the $S^{-1}$ of my #5 and multiplied each by a constant to get
$$S^{-1}=\left(\begin{matrix}2a&b\\\left(1+i\right)a&\left(-1-i\right)b\\\end{matrix}\right)$$$S^{-1}$ must be unitary as well of course so multiply it by its hermitian and we must get
$$\left(\begin{matrix}2a&b\\\left(1+i\right)a&\left(-1-i\right)b\\\end{matrix}\right)\left(\begin{matrix}2a^\ast&\left(1-i\right)a^\ast\\b^\ast&\left(-1+i\right)b^\ast\\\end{matrix}\right)=\left(\begin{matrix}1&0\\0&1\\\end{matrix}\right)$$Thanks wolfram too for finding an infinity solutions:
$$-\frac{1}{\sqrt6}Re{\left(a\right)}\ \le\frac{1}{\sqrt6},\ \ Im{\left(a\right)}=\pm\frac{\sqrt{1-6{Re{\left(a\right)}}^2}}{\sqrt6}$$$$-\frac{1}{\sqrt3}Re{\left(b\right)}\ \le\frac{1}{\sqrt3},\ \ Im{\left(b\right)}=\pm\frac{\sqrt{1-3{Re{\left(b\right)}}^2}}{\sqrt3}$$So I pick an easy one:
$$a=\frac{1}{\sqrt6},\ \ b=\frac{1}{\sqrt3}$$and
$$S^{-1}=\left(\begin{matrix}\frac{2}{\sqrt6}&\frac{1}{\sqrt3}\\\frac{1+i}{\sqrt6}&-\frac{1+i}{\sqrt3}\\\end{matrix}\right),\ \ S=\left(\begin{matrix}\frac{2}{\sqrt6}&\frac{1-i}{\sqrt6}\\\frac{1}{\sqrt3}&-\frac{1-i}{\sqrt3}\\\end{matrix}\right)$$how amazing, it is unitary and it works.

Not as neat as @martinbn's, but still.

Thanks again everyone. I'll leave the 3x3 beast until Monday.
Corrected error in general solution.

 

[vela]

Gosh thanks!
I didn't have to normalise the eigenvectors I don't think.

You effectively did. Note that the columns of your $S^{-1}$ are normalized now.

If the columns of $U$ are the eigenvectors, then each element of $U^\dagger U$ is the product of eigenvectors. Requiring $U^\dagger U = I$ is the same as saying the eigenvectors are orthonormal.

If you want a nicer looking $S^{-1}$, you can do something like this:
$$\begin{pmatrix}
\frac{2}{\sqrt 6} & \frac{1}{\sqrt 3} \\
\frac{1+i}{\sqrt 6} & \frac{-(1+i)}{\sqrt 3}
\end{pmatrix} = \begin{pmatrix}
\sqrt{\frac 23} & \sqrt{\frac 13} \\
\sqrt{\frac 13}e^{i\pi/4} & -\sqrt{\frac 23}e^{i\pi/4}
\end{pmatrix}.$$ Multiply the second column by $e^{-i\pi/4}$ (which preserves normalization) to get
$$S^{-1} = \begin{pmatrix}
\sqrt{\frac 23} & \sqrt{\frac 13}e^{-i\pi/4} \\
\sqrt{\frac 13}e^{i\pi/4} & -\sqrt{\frac 23}
\end{pmatrix}.$$ Alternately, you could instead multiplied the first column by $e^{-i\pi/4}$ and the second column by -1 to get
$$S^{-1} =
\begin{pmatrix}
\sqrt{\frac 23}e^{-i\pi/4} & -\sqrt{\frac 13} \\
\sqrt{\frac 13} & \sqrt{\frac 23}e^{i\pi/4}
\end{pmatrix} =
\begin{pmatrix}
\sqrt{\frac 13}(1-i) & -\sqrt{\frac 13} \\
\sqrt{\frac 13} & \sqrt{\frac 13}(1+i)
\end{pmatrix} =
\frac 1{\sqrt 3}\begin{pmatrix}
1-i & -1 \\
1 & 1+i
\end{pmatrix}.$$

 

[vanhees71]

A matrix $\hat{M} \in \mathbb{C}^{d \times d}$ can be diagonalized with a unitary transformation if and only if it's a normal matrix. A Matrix is called normal, if $\hat{M} \hat{M}^{\dagger}=\hat{M}^{\dagger} \hat{M}$. A very nice paper on this is

https://doi.org/10.1119/1.13860

 

[WWGD]

A matrix $\hat{M} \in \mathbb{C}^{d \times d}$ can be diagonalized with a unitary transformation if and only if it's a normal matrix. A Matrix is called normal, if $\hat{M} \hat{M}^{\dagger}=\hat{M}^{\dagger} \hat{M}$. A very nice paper on this is

https://doi.org/10.1119/1.13860

I guess a difference with Real matrices is that Diagonal Complex ones both stretch and rotate , though by different amounts in each component, rather than just stretching , like the Reals?

 

[vanhees71]

I don't know, what you mean by this. The point is that diagonal (real or complex) matrices commute. If you can diagonalize a matrix $\hat{M}$ with a unitary transformation, i.e., you have $\hat{M}=\hat{U} \hat{D} \hat{U}^{\dagger}$ with $\hat{D}$ being diagonal and since then $\hat{M}^{\dagger}=\hat{U} \hat{D}^{\dagger} \hat{U}^{\dagger}$ it follows that
$$\hat{M} \hat{M}^{\dagger} = \hat{U} \hat{D} \hat{D}^{\dagger} \hat{U}^{\dagger} = \hat{U} \hat{D}^{\dagger} \hat{D} \hat{U}^{\dagger} = \hat{M}^{\dagger} \hat{M},$$
i.e., if a matrix is diagonalizable with a unitary transformation, it's necessarily normal. The other way, i.e., that every normal matrix is diagonalizable with a unitary transformation can be proven by induction. It's not too complicated.