$T: V \to V $ is a linear operator on $V$.
A scalar $\lambda \in F $ is called an eigenvalue of $T $ if there exists a non-zero vector $x \in V$ s.t. $T(x) = \lambda x$.
The non-zero vector $x$ is called an eigenvector of $T$ associated to the eigenvalue $\lambda $.
$A_{n \times n} \in M_{n \times n }(F) , \lambda \in F$ is called an eigenvalue, if $\exists v \in F^n, v \neq {\Vcr{ 0 }{ 0 }{ \vdots }{ 0 }}, s.t. Av = \lambda v . v$ is called the eigenvector of $A $ with respect to $\lambda $.
Let $T \R^2 \to \R^2 $,
$T(x,y) = (x + 3y, 4x + 2y)$
$v_1 = (1, -1), v_2 = (3,4)$
Are $v_1$ and $v_2$ eigenvectors?
$T(v_1) = (-2, 2) = -2(-1, 1) = -2v_1 $
$\therefore -2$ is an eigenvalue, $v_1$ is the corresponding eigenvector.
$T(v_2) = (15, 20) = 5(3, 4) = 5v_2$
$5$ is an eigenvalue and $v_2 $ is a corresponding eigenvector.
Let $T:V \to V$,
Fix an eigenvalue $\lambda$ and an eigenvector $v$ corresponding to $\lambda $, then $av$ is also an eigenvector of T corresponding to $\lambda \forall a \neq 0, a \in F$ .
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Eigenvectors are not unique.
$A_{n \times n} \in M_{n \times n} (F) $ The characteristic polynomial of $A$ is the degree n-polynomial $f(t) $ where $f(t) = \det(A - I_n) , I_n = $identity matrix.
$A_{n \times n} \in M_{n \times n}(F)$, Then $\lambda \in F $ is an eigenvalue of $A$ $\iff$ $\lambda$ is a root of the character polynomial of $A_{i \times j} f(\lambda) = 0$ i.e. $\det{A - \lambda I_n} = 0 $
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<span class="proof__expand"><a>[expand]</a></span>
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<span class="proof__expand"><a>[expand]</a></span>
Similar matrices have have same characteristic polynomial.
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<span class="proof__expand"><a>[expand]</a></span>
If $A_{n \times n}, B_{n \times n} \in M_{n \times n}(F)$.
If $A$ and $B$ are similar, i.e.. $\exists$ an $Q $ s.t. $B = \inv{ Q } A Q$ then
$$\det(A - tI_n) = \det (B - t I_n)$$
Let $T $ be a linear operator on $V $ and $\dim V = n$. The characteristic polynomial of $T $ is defined as the characteristic polynomial of $[ T ]_{ \beta } $ with same ordered basis of $V $.
If $B' $ is another ordered basis of $V $, then $\exists a_{n \times n}$ invertible s.t. $[ T ]_{ \beta' } = \inv{ Q } [ T ]_{ \beta } Q$.
Then by the previous lemma,
$$\det([ T ]_{ \beta' } - t I_n) = \det ([ T ]_{ \beta } - t I_n) $$.
WILL APPEAR ON EXAM: FINDING eigenvector and eigenvalue, DIAGONALIZABILITY
$T: \P_2(x)/ \R \to \P_2 (x) $
$T(f(x)) = f(x) + (x+1) f'(x) $
Find all eigenvalues and eigenvectors.
$\beta = \set{ 1, x, x^2 }$
$$ T(1) = 1 + (x+1) \cdot 0 = 1 = 1 \cdot x + 0x + 0x^2 T(x) = x + (x+1) \cdot 1 = 1+2x = 1 \cdot 1 + 2x + 0x^2 T(x^2) = x^2 + (x+1) \cdot 2x = 3x^2+2x = 0 \cdot 1 + 2x + 3x^2 $$
$$A = [ T ]_{ \beta } = {\Mee{ 1 }{ 1 }{ 0 }{ 0 }{ 2 }{ 2 }{ 0 }{ 0 }{ 3 }} $$
Characteristic polynomial of T is:
$$\begin{align*} &=\det (A - t I_3) \br &= \det {\Mee{ 1-t }{ 1 }{ 0 }{ 0 }{ 2-t }{ 2 }{ 0 }{ 0 }{ 3-t }} \br &= (1-t)(2-t)(3-t) \end{align*}$$
solve $\det (A - t I_3) = 0 $
$t = 1, 2,$ or $3 $
The eigenvalue of $T$ are 1,2 and 3.
$t= 1$, Let $v = {\Vce{ x_1 }{ x_2 }{ x_3 }}$ be a eigenvector of $A $ with respect to $t = 1$.
$$\begin{align*} Av &= 1v \br {\Mee{ 1 }{ 1 }{ 0 }{ 0 }{ 2 }{ 2 }{ 0 }{ 0 }{ 3 }} {\Vce{ x_1 }{ x_2 }{ x_3 }} &= {\Vce{ x_1 }{ x_2 }{ x_3 }} \br {\Vce{ x_1 + x_2 }{ 2 x_1 + 2 x_3 }{ 3x_3 }} &= {\Vce{ x_1 }{ x_2 }{ x_3 }} \end{align*}$$
$\implies \begin{cases} x_1 = x_1 \br x_2 = 0 \br x_3 = 0 \end{cases} $
$v = {\Vce{ x_1 }{ x_2 }{ x_3 }} = {\Vce{ x_1 }{ 0 }{ 0 }} = x_1{\Vce{ 1 }{ 0 }{ 0 }} $
${\Vce{ 1 }{ 0 }{ 0 }}$ is an eigenvector of A with respect to $t = 1 $.
$1 \cdot + 0 \cdot x + 0 \cdot x^2 = 1 $
$\therefore f(x) = 1$ is an eigenvector of $T$ with respect to $t = 1 $.
$t = 2, v = {\Vce{ 1 }{ 1 }{ 0 }} \implies 1 \cdot 1 + 1 \cdot x + 0 \cdot x^2 = 1 + x$
$t=3, v = {\Vce{ 1 }{ 2 }{ 1 }} \implies 1 \cdot + 2 \cdot x + 1 \cdot x^2 = 1 + 2x + x^2 $
Let $T $ be a linear operator on $V $, and $\lambda$ is a eigenvalue of $T $.
Then a vector $v \in V $ is an eigenvector of T with respect to $\lambda \iff v \neq 0$ and $v \in N(T - \lambda I_v) $
$$\begin{align*} &\iff v \neq 0, T(v) = \lambda v \br & \iff Tv = \lambda v = 0\br & \iff (T - \lambda I_v) (v) = 0\br & \iff v \in N(T - \lambda I_v) \end{align*}$$