Andrew Thangaraj
Aug-Nov 2020
\(T:V\to V\) is an operator
A subspace \(U\subseteq V\) is said to be invariant under \(T\) if \(Tu\in U\) for all \(u\in U\).
Examples
\(A=\begin{bmatrix} 1&0\\ 0&1 \end{bmatrix}\)
\(\mathbb{R}^2\): invariant
\(A=\begin{bmatrix} 2&0\\ 0&3 \end{bmatrix}\)
span\(\{(1,0)\}\): invariant, span\(\{(0,1)\}\): invariant
\(A=\begin{bmatrix} 1&2\\ -1&4 \end{bmatrix}\)
Do invariant subspaces exist?
\(T\leftrightarrow\begin{bmatrix} 1&2\\ -1&4 \end{bmatrix}\)
span\(\{(2,1)\}\): invariant, span\(\{(1,1)\}\): invariant
Change of basis
\(B=\{(2,1),(1,1)\}\)
\(T\leftrightarrow\begin{bmatrix} 2&0\\ 0&3 \end{bmatrix}\)
Found a basis under which \(T\) is a diagonal matrix
one-dimensional invariant subspace
\(T:V\to V\) is an operator, dim \(V=n\), \(T\leftrightarrow A\) in some basis
span\(\{v\}\) is a one-dimensional invariant subspace iff \(Tv=\lambda v\)
\(\lambda\in F\) is an eigenvalue of \(T\) if there exists a \(v\) such that \(v\ne 0\) and \(Tv=\lambda v\).
Do eigenvalues exist?
\(Tv=\lambda v\), \(v\ne0\) iff \((T-\lambda I)v=0\), \(v\ne0\) iff \(T-\lambda I\): non-invertible
\(T-\lambda I\) non-invertible iff det\((A-\lambda I)\) = 0
det\((A-\lambda I)\): polynomial of degree \(n\) in \(\lambda\)
Roots of det\((A-\lambda I)\) are eigenvalues of \(T\).
Eigenvalues are for the operator \(T\) and are the same under any basis
det\((S^{-1}AS-\lambda I)=\) det\((S^{-1}(A-\lambda I)S)=\) det\((A-\lambda I)\)
\(T:V\to V\) is an operator
\(v\in V\) is an eigenvector corresponding to an eigenvalue \(\lambda\in F\) of \(T\) if \(v\ne 0\) and \(Tv=\lambda v\).
If \(v\) is an eigenvector, then so is \(\alpha v\) for \(\alpha\ne 0\). The scaled versions are not considered new.
For the same eigenvalue \(\lambda\), linearly independent eigenvectors, if they exist, need to be found.
Eigenvectors of \(\lambda\): basis of null \(T-\lambda I\)
Example 1
\(A=\begin{bmatrix} 2&0\\ 0&3 \end{bmatrix}\)
det\((A-\lambda I)=(2-\lambda)(3-\lambda)=0\)
Eigenvalues: \(2\) and \(3\)
Eigenvectors: for \(2\), \((1,0)\); for \(3\), \((0,1)\)
\(A=\begin{bmatrix} 1&2\\ -1&4 \end{bmatrix}\)
det\((A-\lambda I)=(1-\lambda)(4-\lambda)+2=\lambda^2-5\lambda+6=0\)
Eigenvalues: \(2\) and \(3\)
Eigenvectors: for \(2\), \((2,1)\); for \(3\), \((1,1)\)
\(A=\begin{bmatrix} 1&0\\ 0&1 \end{bmatrix}\)
det\((A-\lambda I)=(1-\lambda)^2=0\)
Eigenvalues: \(1\) repeated twice
Eigenvectors: for \(1\), \((1,0)\) and \((0,1)\)
algebraic multiplicity of eigenvalue \(\lambda\): mutliplicity of root \(\lambda\) in det\((A-\lambda I)\)
geometric multiplicity of eigenvalue \(\lambda\): dim null \(T-\lambda I\)
\(A=\begin{bmatrix} 1&2\\ 0&1 \end{bmatrix}\)
det\((A-\lambda I)=(1-\lambda)^2=0\)
Eigenvalues: \(1\) repeated twice
Eigenvectors: for \(1\), \((1,0)\)
Algebraic multiplicity = 2 and Geometric multiplicity = 1
General case: \(n\times n\) matrix \(A\)
Modern numerical methods can compute eigenvalues and eigenvectors numerically.
In most cases, det\((A-\lambda I)\) is not a preferred method.
Determinant approach is useful for small cases and hand calculations in quizzes!
Let \(v_1,\ldots,v_m\) be eigenvectors of distinct eigenvalues \(\lambda_1,\ldots,\lambda_m\). Then, \(v_1,\ldots,v_m\) are linearly independent.
Proof
Let \(k\) be the least value such that \(v_k\in\) span\(\{v_1,\ldots,v_{k-1}\}\) \[v_k=a_1v_1+\cdots+a_{k-1}v_{k-1}\]
Applying \(T-\lambda_k I\), \[0=a_1(\lambda_1-\lambda_k)v_1+\cdots+a_{k-1}(\lambda_{k-1}-\lambda_k)v_{k-1}\]
Contradicts with \(k\) being least value chosen above