Problem 1 - Characteristic Polynomials

For a $2 \times 2$ matrix $A = \left( \begin{tabular}{cc} a & b \\ c & d \\ \end{tabular} \right)$, answer the following questions.

1.

Compute the characteristic polynomial $f_A (x) = \vert xI - A\vert$, where $\vert\vert$ is the determinant of a matrix, and $I$ is a $2 \times 2$ unit matrix.

2.

Compute $f_A (A)$.

3.

A polynomial $f(x)$ is called a minimal polynomial of $A$ if $f(A)=0$, its coefficients are scalar, the coefficient of highest degree is 1, and the degree is minimum with respect to these properties. The minimal polynomial of $A$ is denoted by $\phi_A(x)$. Show that $\phi_A(x)$ divides $f_A (x)$.

4.

For an eigenvalue $\lambda$ of $A$, show $\phi_A(\lambda) = 0$.

1.

$$f_A(X) = \left\vert xI - A \right\vert = \left\vert \begin{tabula... ...\ $c$ & $x-d$ \end{tabular}\right\vert = (x-a)(x-d)-bc = x^2 - (a+d)x + ad -bc$$

2.

$$f_A(A) = \left\vert AI - A \right\vert = 0$$

3.

$f_A(x) = \phi_A(x)Q(x) + R(x), \; \hbox{with} \; \deg R(x) < \deg \phi_A(x)$.
$f_A(A)=0$ and $\deg f_A = 2$.
$\phi_A$ is a minimal degree polynomial such that it has a zero value when applied to $A$.
It means that $deg \phi_A \leq \deg f_A$.
Let us first assume that $deg \phi_A = 1$.
Then it means that $\deg R = 0$.
Then, since $f_A(A)=0$, it implies that $R(A)=0$, and finally $R=0$.
Thus, $\phi_A$ divides $f_A$.
Let us now assume that $\deg f_A = 2$.
Then it means that $\deg R = 1$.
For instance, $R(x) = \alpha x + \beta$.
Since, $f_A(A)=0$, then $R(A)=0$.
Thus, $R$ is a polynomial that has a zero value in $A$ and whose degree
is inferior to the one of $\phi_A$, which is not possible by definition.
Therefore, $\deg R = -\infty$ and $\phi_A$ divides $f_A$.

4.

Let $\phi_A$ be $\phi_A(X) = X^2 + \alpha X + \beta$.
We have $\phi_A(A) = A^2 + \alpha A + \beta = 0$.
$\lambda$ is an eigenvalue of $A$. Thus $\exists X \in \mathcal{M}_{2,1}(K)$ so that:
$X \neq 0$ and $AX=\lambda X$, $A^2X=\lambda^2 X$, etc.
Therefore, $\phi_A(A)[X] = \lambda^2 X + \alpha \lambda X + \beta X = \left( \phi_A(\lambda) \right) X$.
Since $\phi_A(A) = 0$ and $X \neq 0$, we have $\phi_A(\lambda) = 0$.