Proof If $A_{2x2}$ and $\lambda$ real number then $|\lambda I-A|=\lambda^2-(\operatorname{tr}A) \lambda+|A|$

Question

I have this problem :

Proof If $A_{2x2}$ and $\lambda$ real number then $|\lambda I-A|=\lambda^2-(\operatorname{tr}A) \lambda+|A|$

This is what I did :

I took an arbitrary $A$

$$ A= \left( {\begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \\ \end{array} } \right) $$

Now I try to proof $|\lambda I-A|=\lambda^2-(\operatorname{tr}A) \lambda+|A|$.

Lets start with $|\lambda I-A|$.

$$|\lambda I-A| =\\ det \left( {\begin{array}{cc} \lambda-a_{11} & -a_{12} \\ -a_{21} & \lambda-a_{22} \\ \end{array} } \right)=\\ (\lambda-a_{11})(\lambda-a_{22})-(a_{12})(a_{21})=\\ -\lambda(a_{11}a_{22})-(a_{12}a_{21}) $$

Now this : $\lambda^2-(\operatorname{tr}A) \lambda+|A|$.

$$\lambda^2-(\operatorname{tr}A) \lambda+|A|=\\ \lambda^2-\lambda a_{11}-\lambda a_{22}+a_{11}a_{22}-a_{12}a_{21}=\\ \lambda^2-\lambda (a_{11}+a{22})+a_{11}a_{22}-a_{12}a_{21} $$

But that doesn't proof that $|\lambda I-A|=\lambda^2-(\operatorname{tr}A) \lambda+|A|$

Any ideas?

Any help will be appreciated.

Answer 1

You made a mistake in the first calculation. $\lambda I - A$ means $$\left[\begin{array}{cc}\lambda - a_{11} & -a_{12}\\-a_{21} & \lambda - a_{22}\end{array}\right];$$ the $\lambda$ doesn't appear in every term, only the diagonal.