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Name	Status	Author	Last Modified
Find eigenvalues and eigenvectors of matrices, and power of one matrix	draft	Newcastle University Mathematics and Statistics	20/11/2019 14:50
First Order Differential Equations - Coupled FINISHED	draft	Gareth Woods	26/08/2016 14:36
Gareth's copy of Find eigenvalues and eigenvectors of matrices, and power of one matrix	draft	Gareth Woods	26/08/2016 11:21
Simon's copy of Find eigenvalues and eigenvectors of matrices, and power of one matrix	draft	Simon Vaughan	09/11/2016 16:37
Alex's copy of Simon's copy of Find eigenvalues and eigenvectors of matrices, and power of one matrix	draft	Alex Van den Hof	20/02/2017 15:52
Ch10 Q5	draft	Alex Van den Hof	20/02/2017 16:05
Harry's copy of Find eigenvalues and eigenvectors of matrices, and power of one matrix	draft	Harry Flynn	09/03/2018 14:47
Simon's copy of Find eigenvalues and eigenvectors of matrices, and power of one matrix	draft	Simon Thomas	12/06/2019 09:53
Priyanka's copy of Find eigenvalues and eigenvectors of matrices, and power of one matrix	draft	Priyanka Pandya	10/02/2020 16:21
Eigenvalues	draft	Heather Driscoll	06/05/2021 10:51
Eigenvectors	draft	Heather Driscoll	21/04/2021 15:31
Ugur's copy of Find eigenvalues and eigenvectors of matrices, and power of one matrix	Ready to use	Ugur Efem	13/10/2024 20:16
Ugur's copy of Eigenvalues	Ready to use	Ugur Efem	13/10/2024 20:15

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Question statement

Find the eigenvalues and eigenvectors for the matrices $A$ and $B$ where:

$A=\begin{pmatrix} \var{a11}&\var{a12}\\ \var{a21}&\var{a22} \end{pmatrix},\;\;\;\;\; B=\begin{pmatrix} \var{b11}&\var{b12}\\ \var{b21}&\var{b22} \end{pmatrix}$

Give any introductory information the student needs.

			Name	Type	Generated Value

test1	integer	5
a21	number	5
a22	integer	0
bn22	integer	156844
bn21	integer	13065
b22	integer	-10
b21	integer	-1
b1	integer	-7
cn11	number	-352179
mna	integer	-1
mnb	integer	-7
cn12	number	-4233060
s2	integer	1
s1	integer	1
a12	integer	1
b12	number	36
b11	integer	5
test	integer	4
cn21	number	117585
cn22	number	1411596
a11	integer	4
that	integer	-2
db	number	-14
bn12	integer	-470340
bn11	integer	-39131
da	number	-5
a1	integer	2
x2	integer	-12
c2	integer	-7
c1	integer	-1
x1	integer	-3
mxb	integer	2
mxa	integer	5
a	integer	5
b	integer	-1
f	number	9
tra	integer	4
trb	integer	-5
n	integer	6
this	integer	1
s	integer	1

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Find the eigenvalues of $A$ .

Let $a_1$ be the least eigenvalue of $A,\;\;\; a_1=\;\;$ Gap 0.

Let $a_2$ be the greatest eigenvalue of $A,\;\; a_2=\;\;$ Gap 1.

Advice

Matrix $A$

$A - \lambda I_2 = \begin{pmatrix} \var{a11}-\lambda & \var{a12}\\ \var{a21} & \var{a22}-\lambda \end{pmatrix}$
Hence the characteristic polynomial

$p(\lambda)$ is:

$\begin{eqnarray*} \mathrm{det}\left(A-\lambda I_2 \right)&=&\simplify[zeroTerm]{({a11}-lambda)({a22}-lambda)-{a12}*{a21}}\\ &=& \simplify[std]{lambda^2-{trA}*lambda+{dA}}\\ &=&\simplify[std]{(lambda-{a})(lambda-{b})} \end{eqnarray*}$
We see that on solving

$p(\lambda)=0$ we get the eigenvalues:

$\lambda_1=\var{mnA},\;\;\;\lambda_2=\var{mxA}$
Note: We could have found the characteristic polynomial by noting that for a 2 × 2 matrix

$A$ then the characteristic polynomial is

$\lambda^2-\mathrm{trace}(A)+\mathrm{det}(A)$
where

$\mathrm{trace}(A) = \var{trA},\;\;\;\mathrm{det}(A)=\var{dA}$

Finding the eigenvectors:

1. $\lambda=\var{mnA}$

We have the eigenspace is given by all $v=(x,y)^T$ such that $(\simplify{A-{mnA}}I_2)v=(0,0)^T$ i.e.

$\begin{pmatrix} \var{a11-mnA}&\var{a12}\\ \var{a21}&\var{a22-mnA} \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} =\begin{pmatrix} 0 \\ 0 \end{pmatrix}$

This gives the two equations:

$\begin{eqnarray*} \simplify[std]{{a11-mnA}x + {a12}y}&=&0\\ \simplify[std]{{a21}x + {a22-mnA}y}&=&0 \end{eqnarray*}$
There is only one equation here as we see that the equations are the same (one is a multiple of the other).

So putting $x=1$ in the first equation we get $y_1=\var{-s*(a11-mnA)}$

Hence the eigenvector we want is

$\begin{pmatrix} 1 \\ \var{-s*(a11-mnA)} \end{pmatrix}$

2. $\lambda=\var{mxA}$

In this case we have the equations:

$\begin{eqnarray*} \simplify[std]{{a11-mxA}x + {a12}y}&=&0\\ \simplify[std]{{a21}x + {a22-mxA}y}&=&0 \end{eqnarray*}$

Once again there is only one equation, so putting $x=1$ in the first equation we get $y_2=\var{-s*(a11-mxA)}$

Hence the eigenvector we want is

$\begin{pmatrix} 1 \\ \var{-s*(a11-mxA)} \end{pmatrix}$

Matrix $B$

The characteristic polynomial is given by:

$p(\lambda)=\simplify[std]{lambda^2-{b11+b22}*lambda + {dB}}$

Solving $p(\lambda)=0$ , we find the eigenvalues for $B$ are:

$\lambda_1=\var{mnB},\;\;\;\lambda_2=\var{mxB}$

Eigenvectors

1. $\lambda=\var{mnB}$

The equations are:

$\begin{eqnarray*} \simplify[std]{{b11-mnB}x + {b12}y}&=&0\\ \simplify[std]{{b21}x + {b22-mnB}y}&=&0 \end{eqnarray*}$

Putting $y=1$ in the second equation we get $x_1=\var{s*(b22-mnB)}$

Hence the eigenvector we want is

$\begin{pmatrix} \var{s*(b22-mnB)}\\1 \end{pmatrix}$

2. $\lambda=\var{mxB}$

The equations are:

$\begin{eqnarray*} \simplify[std]{{b11-mxB}x + {b12}y}&=&0\\ \simplify[std]{{b21}x + {b22-mxB}y}&=&0 \end{eqnarray*}$
Putting

$y=1$ in the second equation we get

$x_2=\var{s*(b22-mxB)}$

Hence the eigenvector we want is

$\begin{pmatrix} \var{s*(b22-mxB)}\\1 \end{pmatrix}$

For the last part we use the diagonalisation of $B$ given by the last two parts.

Thus if $x_1,\;\;x_2,\;\;\lambda_1,\;\;\lambda_2$ are as above for $B$ then we have $B=PDP^{-1} \Rightarrow B^{\var{n}}=PD^{\var{n}}P^{-1}$ where:

$\begin{eqnarray*} P &=& \begin{pmatrix} x_1 & x_2\\1&1 \end{pmatrix} = \begin{pmatrix} \var{s*(b22-mnB)} & \var{s*(b22-mxB)} \\1&1 \end{pmatrix}\Rightarrow P^{-1}= \simplify[std]{1/{x1-x2}}\begin{pmatrix} 1 & \var{-s*(b22-mxB)} \\-1&\var{s*(b22-mnB)} \end{pmatrix}\\ \\ D&=& \begin{pmatrix} \lambda_1 & 0\\0&\lambda_2 \end{pmatrix} = \begin{pmatrix} \var{mnB} & 0\\0&\var{mxB} \end{pmatrix} \Rightarrow D^{\var{n}}=\begin{pmatrix} \var{mnB^n} & 0\\0&\var{mxB^n} \end{pmatrix} \end{eqnarray*}$

Hence

$\begin{eqnarray*}B^{\var{n}}&=&PD^{\var{n}}P^{-1}\\ \\ &=&\simplify[std]{1/{x1-x2}}\begin{pmatrix} \var{s*(b22-mnB)} & \var{s*(b22-mxB)} \\1&1 \end{pmatrix}\begin{pmatrix} \var{mnB^n} & 0\\0&\var{mxB^n} \end{pmatrix}\begin{pmatrix} 1 & \var{-s*(b22-mxB)} \\-1&\var{s*(b22-mnB)} \end{pmatrix}\\ \\ &=&\begin{pmatrix} \var{bn11} & \var{bn12}\\\var{bn21}&\var{bn22} \end{pmatrix} \end{eqnarray*}$

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Linear Algebra 1 by Ugur Efem in Ugur's workspace.

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