Show that a matrix $A=\pmatrix{a&b\\c&d}$ satisfies $A^2-(a+d)A+(ad-bc)I=O$

Let $A= \begin{bmatrix} a & b \\ c & d \end{bmatrix} ,a,b,c,d\in\mathbb{R}$ . Prove that every matrix $A$ satisfies the condition $$A^2-(a+d)A+(ad-bc)I=O .$$ Find $$ \begin{bmatrix} a & b \\ c & -a \end{bmatrix}^n .$$

For $a=1,b=2,c=3,d=4$ equality $A^2-(a+d)A+(ad-bc)I=O$ holds.

How to prove this equality for every $A= \begin{bmatrix} a & b \\ c & d \end{bmatrix}$?

The short answer is: The equality can be proofen by simply inserting A, performing a matrix multiplication and adding up the matrices. For the second part of your question, one may find a recursion formula.

The long answer including the calculation follows here: Assuming $I=\begin{bmatrix}1& 0 \\ 0 &1\end{bmatrix}$ and $O = \begin{bmatrix}0& 0 \\ 0 &0\end{bmatrix}$, the equality can be shown by plugging in $A$.

\begin{align} A^2 &- (a+d)A&+&(ad-bc)I &=& O\\ \rightarrow \begin{bmatrix}a& b \\ c &d\end{bmatrix}\begin{bmatrix}a& b \\ c &d\end{bmatrix}& -\begin{bmatrix}(a+d)a& (a+d)b \\ (a+d)c &(a+d)d\end{bmatrix}&+&\begin{bmatrix}ad-bc& 0 \\ 0 &ad-bc\end{bmatrix}&=& \begin{bmatrix}0& 0 \\ 0 &0\end{bmatrix} \end{align} Now we still need to do the matrix multiplication \begin{align} \begin{bmatrix}a& b \\ c &d\end{bmatrix}\begin{bmatrix}a& b \\ c &d\end{bmatrix} = \begin{bmatrix}a^2+bc& ab+bd \\ ac+dc &bc+d^2\end{bmatrix} \end{align}

Now, we may add the matrices component-wise: \begin{align} \begin{bmatrix}a^2+bc& ab+bd \\ ac+dc &bc+d^2\end{bmatrix}& -\begin{bmatrix}(a+d)a& (a+d)b \\ (a+d)c &(a+d)d\end{bmatrix}&+&\begin{bmatrix}ad-bc& 0 \\ 0 &ad-bc\end{bmatrix}&=& \begin{bmatrix}0& 0 \\ 0 &0\end{bmatrix} \end{align} \begin{align} \rightarrow \begin{bmatrix}a^2+bc -(a+d)a+ad-bc& ab+bd -(a+d)b \\ ac+dc -(a+d)c&bc+d^2-(a+d)d+ad-bc\end{bmatrix}= \begin{bmatrix}0& 0 \\ 0 &0\end{bmatrix} \end{align} After rearranging the terms, we see that the quation holds. \begin{align} \rightarrow \begin{bmatrix}a^2-a^2 -ad+ad+bc-bc& ab-ab -ab-db \\ ac+dc -ac-cd&bc-bc+d^2-d^2+ad-ad\end{bmatrix}= \begin{bmatrix}0& 0 \\ 0 &0\end{bmatrix} \end{align}

The second part is about how this helps evaluating $ \begin{bmatrix}a& b \\ c &-a\end{bmatrix}^n=(\tilde A)^n$. Here we defined $\tilde A= \begin{bmatrix}a& b \\ c &-a\end{bmatrix}$.

Using our equation we just proofed with d=-a, we get: \begin{align} (\tilde A)^2-(a+(-a))\tilde A+(-a^2-bc)I=&O\\ \rightarrow (\tilde A)^2+(-a^2-bc)I=&O \end{align} Multiplying this equation with $(\tilde A)^{n-2}$ gives the recursion formula \begin{align} (\tilde A)^n=(a^2+bc)(\tilde A)^{n-2} \end{align}

This is useful since you may plug the right side back in yielding \begin{align} (\tilde A)^n=\begin{cases}(a^2+bc)^{n/2}I&\text{if } n \text{ even} \\ (a^2+bc)^{(n-1)/2}\tilde A &\text{if } n \text{ odd} \end{cases} \end{align}

You can really just prove this by brute force, and its not that complicated.

$A= \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ so

$A^2= \begin{bmatrix} a^2-bc & ab-bd \\ ac-cd & bc-d^2 \end{bmatrix}$

$(a+d)A= \begin{bmatrix} (a+d)a & (a+d)b \\ (a+d)c & (a+d)d \end{bmatrix}$

$(ad-bc)I= \begin{bmatrix} (ad-bc) & 0 \\ 0 & (ad-bc) \end{bmatrix}$

just add them together and you get $$A^2-(a+d)A+(ad-bc)I=O$$

One can of course prove this directly by substituting and computing the entries of the $2 \times 2$ matrix on the left-hand side, but here's an outline for a solution that reduces the amount of actual computation one needs to do.

Hint

1. Prove the claim for diagonal matrices.
2. Prove the claim for matrices similar to diagonal matrices. (Recall that the trace and determinant of a matrix, which up to sign are the linear and constant terms of the polynomial in $A$, are invariant under similarity.)
3. Use the fact that the set of diagonalizable matrices (those similar to diagonal matrices) is dense in the set of all $2 \times 2$ matrices.

Remark This is anyway a special case of a more general fact: That any matrix $A$ satisfies $p_A(A)$, where $p_A$ is the characteristic polynomial $p_A(t) := \det(t I - A)$ of $A$.

Note that $$ A^2= \begin{bmatrix} a^2+bc&ab+bd\\ ac+cd&cb+d^2 \end{bmatrix} $$ and $$ (tr A)A=(a+d) \begin{bmatrix} a&b\\ c&d \end{bmatrix}= \begin{bmatrix} a^2+ad&ab+bd\\ ac+cd&ad+d^2 \end{bmatrix} $$ so $$ (tr A)A-A^2= \begin{bmatrix} ad-bc&0\\ 0&ad-bc \end{bmatrix} = (det A) I $$

Also note that $$ t^2-(tr A)t +det A=0 $$ is the characteristic equation of the $2\times 2$ matrix $A$ and, by Cayley-Hamilton theorem a matrix satisfies his characteristic equation.