Find all 2 by 2 matrices that are orthogonal, real, and also symmetric.

Question

The question is taken from PSet 6.4, Q27 of Strang's Introduction to Linear Algebra, 5th edition.

Find all 2 by 2 matrices that are orthogonal and also symmetric. Which two numbers can be eigenvalues of those two matrices?

I have assumed in what follows that he also meant to say real matrices, as I can find examples not included in the answer that are complex and also satisfy the conditions.

The answer states that orthogonality and symmetry mean that $|\lambda| = 1$, (Which is only true if the matrix is orthogonal and real, I think?) and that $\lambda$ must be real (Since the matrix is symmetric and real), so $\lambda = \pm 1$.

He then goes on to say that the matrix must be either $\pm I$, or

$ S = Q\Lambda Q^T = \begin{bmatrix} \cos{\theta} & -\sin{\theta}\\ \sin{\theta} & \cos{\theta} \end{bmatrix} \begin{bmatrix} 1 & 0\\ 0 & -1 \end{bmatrix} \begin{bmatrix} \cos{\theta} & \sin{\theta}\\ -\sin{\theta} & \cos{\theta} \end{bmatrix} = \begin{bmatrix} \cos{2\theta} & \sin{2\theta}\\ \sin{2\theta} & -\cos{2\theta} \end{bmatrix} $.

This seems to come from nowhere to me. It's clear that a matrix with such properties must have the form $S = Q\Lambda Q^T$, but the rotation matrix as the eigenvector matrix doesn't seem to have a motivation to me other than the fact that it just works.

I'm also not at all sure what the justification is for saying this is the only form such a matrix can take. I can't see anything wrong with it, so I'm sure this describes at least some of the required matrices, but why is it the case that it describes them all?

Answer 1

Orthogonal matrices are meant to be real matrices; their complex counterparts are known as unitary matrices.

Now, the only orthogonal $2×2$ matrices possible are $$\begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix}$$ and $$\begin{bmatrix} \cos\theta & \sin\theta \\ \sin\theta & -\cos\theta \end{bmatrix}$$ This you can prove by assuming the matrix to be $$\begin{bmatrix} a & b \\ c & d\end{bmatrix}$$ Then applying the orthogonality condition ,i.e. $AA^T=I$

Hope it helps:)

Answer 2

Answers to your questions, in order.

The answer states that orthogonality and symmetry mean that $|\lambda| = 1$, (Which is only true if the matrix is orthogonal and real, I think?)

By Strang's convention, orthogonal matrices are necessarily real. Indeed, a "complex-orthogonal" matrix (a complex matrix satisfying $A^TA = I$) need not satisfy this property.

This seems to come from nowhere to me. It's clear that a matrix with such properties must have the form $S = Q\Lambda Q^T$, but the rotation matrix as the eigenvector matrix doesn't seem to have a motivation to me other than the fact that it just works.

I agree that this step merits more explanation. A matrix has orthonormal columns if and only if it is orthogonal, and every $2 \times 2$ orthogonal matrix is either a reflection or a rotation matrix. If we have a reflection matrix, then multiplying the first column by $-1$ produces a rotation matrix.

In other words: given any orthonormal basis of $\Bbb R^n$ (such as the eigenbasis of our symmetric matrix), we can multiply the vectors in the basis by $\pm 1$ appropriately so that placing them as columns in a matrix (in the same order) gives us a rotation matrix.

Answer 3

As others have already pointed out, your first question is a matter of Strang's convention.

Here's how I would argue in response to your second question. First, $S$ is symmetric and therefore diagonalizable, so we have

$$ S = Q\Lambda Q^{-1}$$

for some eigenvector matrix $Q$ and diagonal eigenvalue matrix $\Lambda$. Since we already know all the eigenvalues are $\pm1$, $\Lambda$ can be $I$ (in which case $S=I$), $-I$ (in which case $S=-I$), or $\begin{bmatrix}1&0\\0&-1\end{bmatrix}$ (we can choose the order of the eigenvalues so we don't need to also consider the $\begin{bmatrix}-1&0\\0&1\end{bmatrix}$ case). So we already know $S=\pm I$ or $$S = Q\begin{bmatrix}1&0\\0&-1\end{bmatrix} Q^{-1}$$

The only thing that remains to show is that we can choose $Q$ to be an orthogonal matrix. We can always normalize the columns of $Q$, and eigenvectors of a symmetric matrix corresponding to different eigenvalues must be orthogonal, so the columns must be orthogonal.

Answer 4

It is perhaps worth observing that this question generalizes. First, we can let $n>2$. Second, we can consider complex matrices (instead of real ones) by changing symmetric to be Hermetian and changing orthogonal to be unitary. If you have not seen these terms before, if we let $A^*=\overline(A^T)$ be the complex conjugate of the transpose of $A$, then we change $A=A^T$ to $A=A^*$ and $A^{-1}=A^T$ to $A^{-1}=A^*$

Being Hermetian forces all the eigenvalues to be real. Being unitary forces all the eigenvalues to have norm $1$. Together, they forces all the eigenvalues to be $\pm 1$.

Then, the spectral theorem says not only that your matrix is diagonalizable, but actually that it can be diagonalized by a a unitary matrix. There isn't quite as nice a description of things as rotation matrices, but essentially, if we pick $n$ (unitarily) orthogonal directions and $n$ choices of sign, that will give us a matrix. However, a lot of these choices will yield the same matrix. For example, if all our signs are positive, then whatever directions we take, we will get the identity matrix, and if all our signs are negative, then whatever directions we take, we will get $-I$.

It turns out that the matrix will be completely determined by picking one subspace of $\mathbb C^n$ to be the eigenspace of the eigenvalue $1$. Its orthogonal complement will be the eigenspace of the eigenvalue $-1$, and with this information, there will be a unique Hermetian matrix with those eigenspaces.

With a $2\times 2$ real matrix, you can only end up with $\pm I$ or a reflection about a line. more generally, you end up with a "reflection about a subspace".

Let $V$ be the eigenspace of $1$, and $A_V$ the corresponding matrix. Let $P_V=(I+A_V)/2$. Then $P_V$ will be the orthogonal projection onto $V$, and $A_V=2PV-I$. Of course, I don't know if you find orthogonal projections easier to think about, but at least the option is there for you.