If $2\times 2$ matrix $A$ is symmetric, why can we find a pair of orthonormal eigenvectors $q_1=(x_1,-y_1)^T$, and $q_2=(y_1,x_1)^T$?

Question

I am okay with the fact that we can find a pair of orthonormal eigenvectors, but how do we know that we can find some of the form $q_1=(x_1,-y_1)^T$, and $q_2=(y_1,x_1)^T$?

NOTE: I'm not sure if we need $A$ to be real matrix here, but if so feel free to assume it.

Edit:In response to being possible duplicate... that question is about diagonalization and why eigenvectors of a real symmetric matrix are orthogonal. my question here is about why there are orthogonal eigenvectors of a certain form.

That is, we know that there are orthogonal eigenvectors. But how come the orthogonal vectors $q_1=(x_1,-y_1)^T$, and $q_2=(y_1,x_1)^T$ are eigenvectors (as opposed to just being orthogonal vectors)?

Answer 1

Pick any eigenvector $\mathbf v=(y_1,x_1)$ of $A$. What do the vectors $(x,y)$ that are orthogonal to it look like? Well, we must have $(y_1,x_1)\cdot(x,y)=y_1x+x_1y=0$. This is the equation of a line through the origin with direction vector $(x_1,-y_1)$, i.e., every vector orthogonal to $\mathbf v$ is a scalar multiple of $(x_1,-y_1)$. You agree that $A$ has orthogonal eigenvectors, which means that $(x_1,-y_1)$ in particular is itself an eigenvector of $A$. If in addition $\mathbf v$ is a unit vector (which we can always arrange by normalizing any eigenvector of $A$), then it should be clear that $(x_1,-y_1)$ is also a unit vector.

Answer 2

Let $( q_1, q_2 )$ be any orthonormal basis of eigenvectors of our real matrix $A$ (the existence of such a basis follows from the Spectral Theorem). Let us denote the entries of $q_1$ and $q_2$ by $q_1 = \begin{pmatrix} a \\ b \end{pmatrix}$ and $q_2 = \begin{pmatrix} y_1 \\ x_1 \end{pmatrix}$ where $y_1, x_1$ are known fixed entries, like in your notation.

Now we would like to show that we can choose $q_1$ such that $a = x_1$ and $b = -y_1$.

• First of all we note that the set $$ U = \{ x \in \mathbb{R}^2 \: : \: \langle x, q_2 \rangle = 0 \} $$ is a one-dimensional subspace of $\mathbb{R}^2$ which is generated by $q_1$ (this follows from the fact that $(q_1,q_2)$ constitutes an orthonormal basis, so the scalar product of $q_1$ and $q_2$ is $0$, i.e. $q_1 \in U$).

• The basis $(q_1, q_2)$ being orthonormal also means that the norms of $q_1$ and $q_2$ are $1$, i.e. $$ 1 = \langle q_1, q_1 \rangle = \langle \begin{pmatrix} a \\ b \end{pmatrix} , \begin{pmatrix} a \\ b \end{pmatrix} \rangle = a^2 + b^2 $$ and $$ 1 = \langle q_2, q_2 \rangle = \langle \begin{pmatrix} a \\ b \end{pmatrix} , \begin{pmatrix} a \\ b \end{pmatrix} \rangle = y_1^2 + x_1^2. $$

• We claim now that there are only two vectors in $U$ that have norm $1$, namely $q_1$ and $-q_1$:
  
  Indeed, let $u$ be any vector in $U$. Because $U$ is generated by $q_1$ there exists a scalar $\lambda \in \mathbb{R}$ such that $u = \lambda q_1 = \begin{pmatrix} \lambda a \\ \lambda b \end{pmatrix}$. Therefore, we get $\langle u , u \rangle = \lambda^2 (a^2 + b^2) = \lambda^2 $ and the scalar product is $1$ if and only if $\lambda \in \{1,-1\}$, i.e. $u = q_1$ or $ u = -q_1$.

• We observe that the vector $\begin{pmatrix}x_1 \\ -y_1 \end{pmatrix} $ has norm $1$, as $$ \langle \begin{pmatrix}x_1 \\ -y_1 \end{pmatrix}, \begin{pmatrix}x_1 \\ -y_1 \end{pmatrix} \rangle = x_1^2 + y_1^2 = 1 $$ as we can see from the second bullet.
  

• Furthermore, the same vector lies in $U$ because $$ \langle \begin{pmatrix} x_1 \\ -y_1 \end{pmatrix} , q_2 \rangle = \langle \begin{pmatrix} x_1 \\ -y_1 \end{pmatrix} , \begin{pmatrix} y_1 \\ x_1 \end{pmatrix} \rangle = x_1 y_1 - y_1 x_1 = 0. $$

• But we already know from the third bullet that $q_1$ and $-q_1$ are the only two vectors in $U$ with norm $1$. That means that either $q_1$ or $-q_1$ equals $\begin{pmatrix}x_1 \\ -y_1 \end{pmatrix}$.

• Case 1: If $q_1$ is equal $\begin{pmatrix}x_1 \\ -y_1 \end{pmatrix}$, then we are done.

• Case 2: If $q_1$ is not $\begin{pmatrix}x_1 \\ -y_1 \end{pmatrix}$, then $-q_1$ must be $\begin{pmatrix}x_1 \\ -y_1 \end{pmatrix}$. But then this is also okay, because we can choose $-q_1$ instead of $q_1$, as $(-q_1,q_2)$ is also an orthonormal basis consisting of eigenvectors of $A$.