# How can a diagonalising matrix be unitary?

• George Keeling
George Keeling
Gold Member
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?

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.

aaroman
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)

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

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.

#### Attachments

• PA28 2x2 Hermitian matrix.pdf
184.1 KB · Views: 56
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.

George Keeling said:
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....

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

George Keeling
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.

Last edited:
aaroman, George Keeling, PeroK and 1 other person
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.

Last edited:
George Keeling said:
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
$$S^{-1} = \begin{pmatrix} \sqrt{\frac 23}e^{-i\pi/4} & -\sqrt{\frac 13} \\ \sqrt{\frac 13} & \sqrt{\frac 23}e^{i\pi/4} \end{pmatrix} = \begin{pmatrix} \sqrt{\frac 13}(1-i) & -\sqrt{\frac 13} \\ \sqrt{\frac 13} & \sqrt{\frac 13}(1+i) \end{pmatrix} = \frac 1{\sqrt 3}\begin{pmatrix} 1-i & -1 \\ 1 & 1+i \end{pmatrix}.$$

aaroman and George Keeling
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

George Keeling and WWGD
vanhees71 said:
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?

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.

WWGD and topsquark

## What is a unitary matrix?

A unitary matrix is a complex square matrix U that satisfies the condition U*U^H = U^H U = I, where U^H is the conjugate transpose of U and I is the identity matrix. This means that the columns (and rows) of a unitary matrix are orthonormal vectors.

## What does it mean for a matrix to be diagonalizable?

A matrix is diagonalizable if it can be written in the form A = PDP^(-1), where A is the original matrix, P is an invertible matrix, and D is a diagonal matrix. This means that there exists a basis in which the matrix A acts as a scaling operation along the directions of the basis vectors.

## How can a diagonalizing matrix be unitary?

A diagonalizing matrix can be unitary if the matrix being diagonalized is normal. A matrix A is normal if it commutes with its conjugate transpose, i.e., AA^H = A^H A. In this case, the matrix P that diagonalizes A can be chosen to be unitary, meaning P*P^H = I.

## What is the significance of a unitary diagonalizing matrix?

A unitary diagonalizing matrix is significant because it preserves the inner product structure of the vector space. This means that the transformation associated with the matrix does not distort angles or lengths, making it particularly useful in quantum mechanics and other areas of physics and engineering.

## Can any matrix be diagonalized by a unitary matrix?

No, not every matrix can be diagonalized by a unitary matrix. Only normal matrices can be diagonalized by a unitary matrix. Examples of normal matrices include Hermitian matrices (A = A^H), unitary matrices (U*U^H = I), and skew-Hermitian matrices (A = -A^H).

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