Invariant subspaces, Eigenvalues, Eigenvectors

Andrew Thangaraj

Aug-Nov 2020

Recap

  • Vector space \(V\) over a scalar field \(F\)
    • \(F\): real field \(\mathbb{R}\) or complex field \(\mathbb{C}\) in this course
  • \(m\times n\) matrix A represents a linear map \(T:F^n\to F^m\)
    • dim null \(T\) + dim range \(T\) = dim \(V\)
  • Linear equation: \(Ax=b\)
    • Solution (if it exists): \(u+\) null\((A)\)
  • Four fundamental subspaces of a matrix
    • Column space, row space, null space, left null space
  • Determinant of a square matrix
    • Function with many interesting properties
  • Change of basis for linear map
    • Can result in simpler matrix representation

Invariant subspace

\(T:V\to V\) 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?

Why invariant subspaces?

\(T\leftrightarrow\begin{bmatrix} 1&2\\ -1&4 \end{bmatrix}\)

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

Eigenvalues

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\((A-\lambda I)\)

Eigenvectors

\(T:V\to V\) 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 \(3\), \((0,1)\)

Examples and multiplicities

\(A=\begin{bmatrix} 1&2\\ -1&4 \end{bmatrix}\)

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 \(T-\lambda I\)

Examples

\(A=\begin{bmatrix} 1&2\\ 0&1 \end{bmatrix}\)

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 \(A\)

  1. Modern numerical methods can compute eigenvalues and eigenvectors numerically.

  2. In most cases, det\((A-\lambda I)\) is not a preferred method.

  3. Determinant approach is useful for small cases and hand calculations in quizzes!

Linear independence of eigenvectors

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