How do I prove that for any $2\times2$ matrix $A$, $A^2$ can be written in the linear form $aA + bI$ where $a$ and $b$ are scalars?

Question

How do I prove that for any $2\times2$ matrix $A$, $A^2$ can be written in the linear form $aA + bI$ where $a$ and $b$ are scalars?

I've tried letting the elements of A be $w$, $x$, $y$, and $z$ then finding $A^2$ but I don't know where to go from there. Please help.

Answer 1

Here is a well-known theorem (Cayley-Hamilton):

Suppose that $A$ is an $n \times n$ matrix. Suppose that $f(x) = \det {(xI - A)} = x^n + a_1 x^{n-1} + \dots + a_n$. Then $$ A^n + a_1 A^{n-1} + \dots + a_n I = 0. $$

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Suppose that $$ A = \begin{bmatrix} w & x \\ y & z \\ \end{bmatrix}. $$ I leave the proof of the following identity as your exercise: $$ A^2 - (w + z) A + (wz - xy) I = 0. $$ Hence $$ A^2 = (w + z) A - (wz - xy) I. $$

Answer 2

For a $2 × 2$ matrix, $$ A={\begin{pmatrix}a&b\\ c&d\\ \end{pmatrix}}, $$ the characteristic polynomial is given by $p(λ) = λ^2 − (a + d)λ + (ad − bc)$, so the Cayley–Hamilton theorem states that $P(A)=0$, i.e., that $$ A^{2}-(a+d)A+(ad-bc)I_{2}={\begin{pmatrix}0&0\\0&0\\\end{pmatrix}}. $$ The last formula can be verified explicitly, independently of Cayley-Hamilton.