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Thursday, 15 January 2015

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}

\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}

More applications

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