MATH1012 - Lecture 13

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MATH1012 - Lecture 12
Skeleton Slides

Systematic Approach to Eigenvalues

• If $Av=\lambda v$ then: $(A-\lambda I)v=0$
  • $I$ is the identity matrix
  • When does $\det (A-\lambda I)=0$ have non-zero solutions?
    • These are the only eigenvalues that will work for $A$
    • Remember, $\lambda$ is the eigenvalue
• We will solve this equation and then find all values for $\lambda$ rather than manually checking if each value of $\lambda$ works
• Remember, $v$ is an eigenvector
  • i.e. a solution to the equation $Av=\lambda v$ for a given $A$ and $\lambda$

You do not need to know $v$, this process takes a vector $A$, finds all possible eigenvalues $\lambda$ it could produce and then allows you to solve for any possible $v$. Once you have $\lambda$, create a homogenous system of equations (by moving the variables to the right-hand side) and solve for $v$ (do this for each potential $\lambda$). Worked solution in the lecture recording.

Using Eigenvalues to Solve High Powers

• Eigenvalues makes high power linear algebra analysis easy
• Suppose $Av=\lambda v$:
  • $AAv=\lambda \lambda v={\lambda }^{2}v$
  • $A^{3}v={\lambda }^{3}v$
• Typically, you’ll express a vector in the form of two eigenvectors
  • $A^{500}[\genfrac{}{}{0ex}{}{1}{3}]$ → $[\genfrac{}{}{0ex}{}{1}{3}]=\alpha [\genfrac{}{}{0ex}{}{1}{1}]+\beta [\genfrac{}{}{0ex}{}{1}{-1}]$ → $[\genfrac{}{}{0ex}{}{1}{3}]=2[\genfrac{}{}{0ex}{}{1}{1}]-1[\genfrac{}{}{0ex}{}{1}{-1}]$
    • The eigenvectors are given from solution to lecture example
  • $A^{500}[\genfrac{}{}{0ex}{}{1}{3}]=[\genfrac{}{}{0ex}{}{1}{3}]\to A^{500}[\genfrac{}{}{0ex}{}{1}{3}]=2A^{500}[\genfrac{}{}{0ex}{}{1}{1}]-A^{500}[\genfrac{}{}{0ex}{}{1}{-1}]$
  • Full example in lecture recording

Characteristic Polynomial

• If $A$ is an $n\times n$ matrix, then $\det (A-\lambda I)$ is a polynomial of degree $n$ in $\lambda$, called the characteristic polynomial of $A$
• $\det (A-\lambda I)=0$ is also known as the characteristic equation
  • Its solutions are the eigenvalues of $A$
  • There are always $n$ complex solutions
    • But they may not be real and may be repeated

Eigenspaces

• For a given eigenvalue $\lambda$, the eigenspace is the set of all eigenvectors corresponding to $\lambda$, together with the zero vector ($v$ cannot be $\varnothing$)
  • Represented by $E_{\lambda }$
• $E_{\lambda }=\{v\in {\mathbb{R}}^n|Av=\lambda v\}$ which also equals $\ker (A-\lambda I)$
  • When describing an eigenspace, you should find a basis for it

Multiplicity of an Eigenvalue

• There can be repeated solutions for a characteristic equation
  • e.g. $(\lambda -\alpha {)}^n=0$ has $n$ solutions with the same $\lambda =\alpha$
• The algebraic multiplicity of a particular eigenvalue $k$ is $m$, if: $\det (A-\lambda I)=...(k-\lambda {)}^m...$
  Remember, this is referring to an eigenvalue, not eigenspace!
• The geometric multiplicity of a particular eigenvalue $k$ is the dimension of the eigenspace $E_k$
  • i.e. how many vectors is in the basis of the eigenspace
  • It will always be less than or equal to the algebraic multiplicity $(m)$

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MATH1012 - Lecture 14
Annotated Slides