0
$\begingroup$

The trace of a matrix is the sum of the entries on its main diagonal. Prove that if $A$ is a $2 \times 2$ matrix, then the characteristic polynomial of $A$ is $x^2 − {c_1}x + c_2$ where $c_1$ is the trace of $A$ and $c_2$ is the determinant of $A$.

Can anyone explain this to me? So far, I only know that $C_a (x) = \operatorname{det}(A-xI)$, that the product of eigenvalues (counting multiplicity) is the $\operatorname{det}{A}$, and the sum of eigenvalues (counting multiplicity) equals the trace of $A$. I am just lost as to how to apply these.

$\endgroup$
2
$\begingroup$

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

Than it's characteristic equation is given by $|A-xI|=0$

i.e. $\begin{vmatrix} a-x & b \\ c & d-x \end{vmatrix}=0$

On solving the determinant : $(a-x)(d-x)-cb=0 \implies x^2-\underbrace{(a+d)}_{ tr.(\text{A})}x+\underbrace{(ad-cb)}_{\det(A)}=0$

Here $a+d = tr.(\text{A}), ~ad-cb = \det{(A)}$

$\endgroup$
1
$\begingroup$

Let $A=\begin{pmatrix} a&b\\c&d \end{pmatrix}$. The the characteristic polynomial is \begin{eqnarray} \begin{vmatrix} a-x& b\\ c& d-x \end{vmatrix}&=&(a-x)(d-x)-cb\\&=&x^2-(a+d)x+ad-bc\\&=&x^2-\text{Tr}(A)x+\det(A).\end{eqnarray}

$\endgroup$
1
$\begingroup$

Eigenvalues are the roots of the characteristic polynomial, and the characteristic polynomial is monic. This is a $2 \times 2$ matrix, so the characteristic polynomial has degree $2$. By the Fundamental Theorem of Algebra, there are two eigenvalues, let them be $\alpha_1$ and $\alpha_2$, not necessarily distinct. Then by the Factor Theorem, since the characteristic polynomial is monic it can be written in the form $$ (x-\alpha_1)(x-\alpha_2) = x^2 - (\alpha_1+\alpha_2)x + \alpha_1\alpha_2, $$ and then you can use the results you know to relate these to $c_1$ and $c_2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.