Show that for every symmetric $2 \times 2$ matrix $A$ there exists an $r \in \mathbb R$ such that $A-rI$ is not invertible. Make an example of a $2\times 2$ matrix for which such number $r$ does not exist.
My thoughts:
I tried to build $A-rI$ and have it's determinant, but I cant prove how it can always be $0$ in a smart way
On
Since $A=A^T$ so $A$ is diagonalisable.
Hence $A=PDP^{-1}$ where $P$ is non-singular and $D$ is a diagonal matrix.
Now $A-rI=PDP^{-1}-rI=PDP^{-1}-rPP^{-1}=P(D-rI)P^{-1}$
$\det (A-rI)=\det (D-rI)$ $=\det$\begin{bmatrix} \lambda_1-r &0\\ 0 &\lambda_2-r\end{bmatrix}
where $\lambda_1,\lambda_2$ are eigen values of $A$.
So $\det (A-rI)=(\lambda_1-r)(\lambda_2-r)$ which is $0$ if $r=\lambda_1$ or $r=\lambda_2$
Hence such an $r$ exists.
If $A=\left(\begin{matrix}a&b\\b&c\end{matrix}\right)$, then $A-rI=\left(\begin{matrix}a-r&b\\b&c-r\end{matrix}\right)$, whose determinant is $$(a-\color{red}{r})(c-\color{red}{r})-b^2=\color{red}{r^2}+(-a-c)\color{red}{r}-b^2+ac$$
If you want a choice for $\color{red}{r}$, so that the quadratic polynomial (in $\color{red}{r}$) equals zero, you must use the quadratic formula:
$$\color{red}{r}=\frac{a+c\pm\sqrt{(a+c)^2+4(b^2-ac)}}{2}$$
This will have at least one value for $r$ which works, unless the discriminant $$(a+c)^2+4(b^2-ac)$$ is negative. However, that discriminant rearranges to $(a-c)^2+4b^2$, which is never negative (if $a,b,c$ are real). To make the discriminant negative, you must use complex numbers, e.g. $a=c=0, b=i$.