Trace of matrix and Cayley Hamilton

Let $A \in \mathbb{R}^{2 \times 2}$. Prove that:
$$A^2-\text{tr}\left(A\right)A+\det\left(A\right) \mathbb{I}=0$$

Solution



Suppose that $ A= \begin{pmatrix}
a &b \\
 c&d
\end{pmatrix}$. Then by definition the trace of the matrix $A$ is the sum of the elements of the diagonal, that is:

$${\rm tr}(A)= a+d$$

The determinant of the matrix is simply $\det A = ad - bc$.

The characteristic polynomial of $A$ is:

$$ \chi_A(x)=x^2-{\rm tr}(A)x+\det(A)$$

From Cayley - Hamilton we get that $A^2 - {\rm tr}(A) A +\det A \mathbb{I}=0$ which is the wanted result.