Proof that the characteristic polynomial of a $2 \times 2$ matrix is $x^2 - \text{tr}(A) x + \det (A)$

Question

Let $$ A=\begin{bmatrix} a_{11} & a_{12}\\ a_{21} & a_{22}\\ \end{bmatrix}$$

Let $C_{A}(x) := \det(xI-A)$ be the characteristic polynomial of A. Show that $$C_{A}(x)=x^2-\text{tr}(A)x+\det(A).$$

I know that $\text{tr}(A)=a_{11}+a_{22}$ and $\det(A)=a_{11}a_{22}-a_{21}a_{12}$. Plugging this into the above equation I get $$C_{A}(x)=x^2-(a_{11}+a_{22})x+a_{11}a_{22}-a_{21}a_{12}.$$

I'm not sure how to get past this. As you can tell, I'm not too good at proofs. Any help will be appreciated. Thanks.

Answer 1

You're there. It's like you've reached the front door without recognizing it. Just take the equation you've derived for the characteristic polynomial

$C_A(x) = \det(x I - A) = x^2 - (a_{11} + a_{22})x + a_{11}a_{22} - a_{21}a_{12}, \tag{1}$

plug in your formulas for trace and determinant:

$tr(A) = a_{11} + a_{22}, \; \; \det(A) = a_{11}a_{22} - a_{21}a_{12}, \tag{2}$

and voila!, you obtain

$C_A(x) = X^2 - tr(A) + \det (A); \tag{3}$

behold! As if Aladdin had said the magic word, the front door swings open, and you're in (to the room of the right answer)!

Hope this helps. Cheers,

and as ever,

Fiat Lux!!!

Answer 2

So far so good.

On the other hand, computing $C_A(x)$ directly using the definition gives $$ C_A(x) = \det(x I - A) = \det\begin{pmatrix}x - a_{11} & -a_{12} \\ -a_{21} & x - a_{22}\end{pmatrix} = (x - a_{11})(x - a_{22}) - (-a_{12})(-a_{21}) . $$ All that remains is to show that this agrees with the expression you already produced.