Non invertible matrix

Prove that for all matrices $A \in \mathbb{R}^{2\times 2}$ there are at most two values of $\ell \in \mathbb{R}$ such that the matrix $A+\ell \mathbb{I}_2$ is not invertible.

Solution

Let $A= \begin{pmatrix}
a &b \\
 c& d
\end{pmatrix}, \;\; a, b, c, d \in \mathbb{R}$ be the matrix. Then:

$$A+\ell \mathbb{I}_2 = \begin{pmatrix}
a &b \\
 c& d
\end{pmatrix}+ \ell \begin{pmatrix}
1 &0 \\
 0&1
\end{pmatrix} = \begin{pmatrix}
a+\ell &b \\
 c& d+\ell
\end{pmatrix}$$

Since we want the matrix $A+\ell \mathbb{I}_2$ not to be invertible then we have that $\det (A+\ell \mathbb{I}_2)=0$. However:

$$\det  \left ( A+\ell \mathbb{I}_2 \right )=0 \Rightarrow \left ( a+\ell \right )\left ( d+\ell \right ) - cb =0 \Leftrightarrow  ad +a\ell +\ell d +\ell^2 -cb =0 \Leftrightarrow  \\ \Leftrightarrow \ell^2 + \ell \left ( a+d \right )+ad -cb =0$$

We can see the last equation as a trinomial of $\ell$. The trinomial can have no more than two roots. If $\Delta>0$ we have exactly two, if $\Delta=0$ we have only one (actually a double one) and finally if $\Delta<0$ then we have no roots.

In any case there exist at most two values of $\ell$ that the matrix $A+\ell \mathbb{I}_2$ is not invertible.