# 9231. FP2. Matrices 2

Eigen-things

If Ae = ๐e, then e is an eigenvector of matrix A and lambda is the corresponding eigenvalue of matrix A.

|A-๐I| = 0 (remember that I is the identity matrix) gives us the characteristic equation of a matrix. Solving this gives us the eigenvalues of matrix A.

Thinking geometrically, the eigenvectors are the invariant vectors under transformation by A, and the eigenvalues are the scale factor by which A enlarges them.

Worked Examples

• Find the characteristic equation, the eigenvalues and the eigenvectors of $\begin{pmatrix} -2 & 1 \\ 6 & 3 \end{pmatrix}$
• Find the characteristic equation, the eigenvalues and the eigenvectors of $\begin{pmatrix} 3 & 2 & 4 \\ 1 & 2 & 0 \\ 1 & 2 & -1 \end{pmatrix}$

N.B. In exam questions you will also sometimes be given an eigenvector and asked to find the corresponding eigenvector, and sometimes the converse.

Exercise

Further Matrix Algebra

• If matrix A has eigenvector e and eigenvalue ๐, and
• matrix B has eigenvector e and eigenvalue ฮผ, then:
• Ae = ๐e and Be = ฮผe, so
• (A+B)e = (๐+ฮผ)e and
• ABe = Aฮผe = ฮผAe = ฮผ๐e.
• So matrix A+B has eigenvalue ฮผ+๐; and
• matrix AB has eigenvalue ฮผ๐
• Each has the same eigenvectors.
• Further, we can conclude that Ane has eigenvalue ๐n.
• N.B. Adding kI to a matrix increases the eigenvalues by k.

Cayley Hamilton Theorem (another way to find an inverse matrix)

For a square matrix, A, consider the characteristic equation: PA(๐)=|A-๐I| = 0.

Clearly PA(A)=0

So for A๐3 + b๐2 + c๐ + d = 0, we have aA3 + bA2 + cA + d = 0. We can rearrange this to give:

• aA3 + bA2 + cA = -dI
• A(aA2 + bA + c) = -dI
• $\frac{A(aA^2 + bA + c)}{-d} = I$

So we have a new formula for the inverse of A, i.e. $A^{-1} = \frac{-(aA^2 + bA + c)}{d}$, or in general for an nxn matrix, $A^{-1} = \frac{aA^{n-1} + a_{n-1}A^{n-2} + ... + a_1}{-a_0}$

Worked Examples

• The matrix A is given by $\begin{pmatrix} -6 & 2 & 3 \\ -14 & 3 & 10 \\ -4 & 2 & 1 \end{pmatrix}$. The matrix B is given by A + 3I. Two eigenvalues of a are ๐1=-2 and ๐2=1 and their corresponding eigenvectors are $\begin{pmatrix} 5 \\ -2 \\ 8 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$. Find the eigenvalues and corresponding eigenvectors for B.
• The matrix A is given as $\begin{pmatrix} -3 & 0 \\ 5 & 2 \end{pmatrix}$. Determine the eigenvalues and corresponding eigenvectors of the matrix A6.
• The matrix A is given as $\begin{pmatrix} 1 & 2 & -1 \\ 0 & 2 & 4 \\ 0 & 0 & -3 \end{pmatrix}$. Determine the characteristic equation of A and use it to determine the matrix A-1.

Exercise

Diagonalisation (Matrices raised to a power)

An efficient method for calculating powers of matrices is diagonalisation.

We can rewrite matrix A as PDP-1, where P is the matrix formed from the eigenvectors of A and D is the matrix formed from the corresponding eigenvalues of A.

(PDP-1)n = PDP-1PDP-1โฆPDP-1 = PDnP-1

So (PDP-1)n = PDnP-1

Dn is easy to calculate, because D is a diagonal matrix: Dn = $\begin{pmatrix} \lambda_1 & 0 & 0 & 0 \\ 0 & \lambda_2 & 0 & 0 \\ 0 & 0 & \lambda_3 & 0 \\ 0 & 0 & 0 & \lambda_4 \end{pmatrix}^n = \begin{pmatrix} (\lambda_1)^n & 0 & 0 & 0 \\ 0 & (\lambda_2)^n & 0 & 0 \\ 0 & 0 & (\lambda_3)^n & 0 \\ 0 & 0 & 0 & (\lambda_4)^n \end{pmatrix}$.

Only non-singular matrices can be diagonalised, and they must have distinct eigenvalues and eigenvectors.

Worked Examples

• Find Mn, where M = $\begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix}$
• B = $\begin{pmatrix} 1 & 2 & 0 \\ 0 & 3 & 1 \\ 0 & 0 & 2 \end{pmatrix}$ has eigenvalues 1,2 and 3, with corresponding eigenvectors $\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$, $\begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$. Find B6

Exercise

Solving Systems of Equations

We can form an augmented matrix with the coefficients of the variables on the left and the RHS on the right, and use row reduction to reduce to triangular form.

We can then either back-solve or reduce to reduced row echelon form and read off the solution.

Worked Example

Find the unique solution for the system of equations:

• 2x + 5z = -3
• x + y + 2z = 0
• x – y + 4z = -4

Equations without a unique solution

Some systems of equations will have no solutions and some will have infinitely many.  We can visualise the equations as three planes that can meet at a point, at a line, or not at all.

If they have no solutions, the row reduction will lead to a contradiction, e.g. 0z=7.  If they have infinitely many solutions, one row will reduce to zeroes, in which case the solution can be stated as a vector line, as in the following example.

Find a solution set for the system of equations:

• 2x + 3y + z = 1
• 4x + 10y + 5z = 3
• 2x + 11y + 7z = 3

Exercise & General Exercise: