Eigenvalues and eigenvectors

$\lambda$ is an eigenvalue of $A$ if there exists a non-zero vector $\vec{v}$ such that
$A\vec{v}=\lambda \vec{v}$.
Vector $\vec{v}$ is called an eigenvector of $A$ corresponding to $\lambda$.

$A\vec{v}=\lambda \vec{v}\\
(A-\lambda I)\vec{v}=\vec{0}$

Note that $A-\lambda I$ must be singular, implying that $\det(A-\lambda I)=0$.
[If $A-\lambda I$ is invertible, $(A-\lambda I)^{-1}(A-\lambda I)\vec{v}=(A-\lambda I)^{-1}\vec{0} \Rightarrow \vec{v}=\vec{0}$. Contradiction. We want a non-zero vector.]
After dealing with some algebra, we know the eigenvalues.

Sidenote: the set of all vectors $\vec{v}$ satisfying $A\vec{v}=\lambda \vec{v}$ is called the eigenspace of $A$ corresponding to $\lambda$.

Application: solving differential equations

Example:
$\frac{dx}{dt}=4x-y\\ \frac{dy}{dt}=2x+y$
$\frac{dX}{dt}=AX$ where $X=\begin{pmatrix}x\\y \end{pmatrix},A=\begin{pmatrix}4&-1\\2&1 \end{pmatrix}$

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$\frac{d}{dt}PY=APY\\
P\frac{dY}{dt}=APY\\
\frac{dY}{dt}=P^{-1}APY$
Characteristic polynomial $\delta(t)$ of $A$:
$\delta(t)=|tI-A|=\begin{vmatrix} t-4 & 1\\ -2 & t-1 \end{vmatrix}=t^2-5t+6=(t-3)(t-2)$
Eigenvalues of $A$: $2,3$
Substitute $t=3$, we have the homogeneous system $-x+y=0$ and $-2x+2y=0 \Rightarrow \vec{v_1}=(1,1)$.
Substitute $t=2$, $-2x+y=0 \Rightarrow \vec{v_2}=(1,2)$.
$P=(\vec{v_1}\vec{v_2})=\begin{pmatrix}1&1\\1&2\end{pmatrix}$
$B=P^{-1}AP=\begin{pmatrix}3&0\\0&2\end{pmatrix}$
Diagonalise the system by changing variables using $P$.
$\begin{pmatrix}x\\y \end{pmatrix}=P\begin{pmatrix}r\\s \end{pmatrix}\\
x = r + s\\
y = r + 2s\\
\frac{d}{dt}\begin{pmatrix}r\\s \end{pmatrix}=\begin{pmatrix}3&0\\0&2\end{pmatrix}\begin{pmatrix}r\\s \end{pmatrix}\\
\begin{pmatrix}\frac{dr}{dt}\\\frac{ds}{dt} \end{pmatrix}=\begin{pmatrix}3r\\2s \end{pmatrix}\\
\therefore x=ae^{3t}+be^{2t}\\
y=ae^{3t}+2be^{2t}$

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