Just a quick question:
For the matrix $$ \begin{pmatrix} 2 & 2 \\ 2 & 2\\ \end{pmatrix} $$ Show that the eigenvectors of the matrix are orthogonal.
I can tell that the eigenvectors are $(-0.5, 1)$ and $(2, 1)$. I'm just not sure how to show they are orthogonal.
That result holds in general.
If $u$ and $v$ are eigenvectors corresponding to the eigenvalues $\lambda_u$ and $\lambda_v$ of a symmetric matrix $A$ ($A=A^T$) then we can use the relation that is always true:
$$(Au)\cdot v=u\cdot (A^Tv)$$
Using that $A=A^T$ we have
$$\lambda_u(u\cdot v)=\lambda_v(u\cdot v)\to(\lambda_u-\lambda_y)(u\cdot v)=0$$
but $\lambda_u\ne \lambda_v$ so
$$u\cdot v=0$$
Firstly, as pointed out by @amd, you've made a mistake with the eigenvectors. The eigenvalues are: $$(2-\lambda)^2-4=-4\lambda+\lambda^2=\lambda(\lambda-4)=0$$ Which implies that: $\lambda_1=0$ and $\lambda_2=4$. Therefore, the eigenvectors are: $$\begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}=0\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \implies x_1=-x_2 \implies v_1=\begin{bmatrix} -1 \\ 1 \end{bmatrix}$$ $$\begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}=4\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \implies x_1=x_2 \implies v_2=\begin{bmatrix} 1 \\ 1 \end{bmatrix}$$
Let's now show that the eigenvectors of the matrix are orthogonal. Let's start with the formula for the angle $\theta$ between two vectors $\mathbf{u},\mathbf{v}$: $$\mathbf{u}\cdot \mathbf{v}=\|\mathbf{u}\|\|\mathbf{v}\|\cos{\theta}$$ Now, for orthogonal vectors: $\theta=\frac{\pi}{2}$. This is because two Euclidean vectors are called orthogonal if they are perpendicular. Therefore: $$\mathbf{u}\cdot \mathbf{v}=0$$ Thus, you must show that the dot product of your two eigenvectors $v_1$ and $v_2$ is equal to zero.
