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Name	Status	Author	Last Modified
Find eigenvalues, characteristic polynomial and a normalised eigenvector of a 3x3 matrix	draft	Newcastle University Mathematics and Statistics	20/11/2019 14:50
Harry's copy of Find eigenvalues, characteristic polynomial and a normalised eigenvector of a 3x3 matrix	draft	Harry Flynn	09/03/2018 14:48
Find eigenvalues, characteristic polynomial and a normalised eigenvector of a 3x3 matrix	draft	Simon Thomas	07/02/2019 10:47
Simon's copy of Find eigenvalues, characteristic polynomial and a normalised eigenvector of a 3x3 matrix	draft	Simon Thomas	03/04/2019 09:26
Ugur's copy of Find eigenvalues, characteristic polynomial and a normalised eigenvector of a 3x3 matrix	Ready to use	Ugur Efem	13/10/2024 20:16

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Question statement

Let $M=\var{M}$ . Answer the following questions.

Give any introductory information the student needs.

			Name	Type	Generated Value

a	integer	-2
b	integer	7
D	matrix	Matrix of size 3×3
g	integer	3
det	integer	-2
M	matrix	Matrix of size 3×3
coefflamsq	integer	2
invp	matrix	Matrix of size 3×3
p	matrix	Matrix of size 3×3
coefflam	integer	1
eigen2	integer	1
eigen3	integer	-1
eigen1	integer	2

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random(-2..2 except 0)

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Description

Describe what this variable represents, and list any assumptions made about its value.

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Generated value: `integer`

-2

→ Used by:

b
invp
p

Parts

Part a) Gap-fill
1. Gap Gap 0. Number entry
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2. Gap Gap 1. Number entry
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Part b) Gap-fill
1. Gap Gap 0. Mathematical expression
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2. Gap Gap 1. Number entry
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3. Gap Gap 2. Mathematical expression
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4. Gap Gap 3. Mathematical expression
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Ask the student a question, and give any hints about how they should answer this part.

You are given that $\var{vector(1,-b,-a)}$ is an eigenvector of $M$ .

Find the corresponding eigenvalue: Gap 0.

Also $\var{vector(1,1-b,-a)}$ is an eigenvector of $M$ .

Find the corresponding eigenvalue: Gap 1.

Advice

a)

For the eigenvector $\var{vector(1,-b,-a)}$ we have:

$\var{M}\var{vector(1,-b,-a)}=\var{vector(eigen1,-eigen1*b,-eigen1*a)}=\var{eigen1}\var{vector(1,-b,-a)}$ .

Hence the corresponding eigenvalue is $\var{eigen1}$ .

Similarly, for the eigenvector $\var{vector(1,1-b,-a)}$ we have:

$\var{M}\var{vector(1,1-b,-a)}=\var{vector(eigen3,eigen3*(1-b),-eigen3*a)}=\var{eigen3}\var{vector(1,1-b,-a)}$ .

Hence the corresponding eigenvalue for this eigenvector is $\var{eigen3}$ .

b)

The characteristic polynomial is given by $P_M(\lambda)=\operatorname{det}(M-\lambda I_3)$ . The roots of this are the eigenvalues.

We find that in this case:

$\begin{align}\operatorname{det}(M-\lambda I_3)&=\operatorname{det}\begin{pmatrix}\var{m[0][0]}-\lambda &\var{m[0][1] }&\var{m[0][2]}\\ \var{m[1][0]}&\var{m[1][1] }-\lambda &\var{m[1][2]}\\ \var{m[2][0]}&\var{m[2][1] } &\var{m[2][2]}-\lambda \end{pmatrix}\\&=\simplify{-lambda^3+{coefflamsq}*lambda^2+{coefflam}*lambda+{det}}\end{align}$

Now we know that $\simplify{lambda-{eigen1}}$ and $\simplify{lambda-{eigen3}}$ are both factors of the characteristic polynomial.

Hence we have:

$\simplify{-lambda^3+{coefflamsq}*lambda^2+{coefflam}*lambda+{det}=-(lambda-{eigen1})(lambda-{eigen3})(lambda-r)}$ for some number $r$ .

Since the constant term in the characteristic polynomial is the product of the three eigenvalues, $r=\simplify[!basic]{{det}/({eigen1}*{eigen3})={eigen2}}$ and this is the third eigenvalue.

To find an eigenvector corresponding to this eigenvalue we solve the equation:

$\simplify{M*vector(x,y,z)={M}*vector(x,y,z)={eigen2}*vector(x,y,z)}$

This gives the equations:

$\begin{align} \simplify{{m[0][0]}*x+{m[0][1]}*y+{m[0][2]}*z}&=\simplify{{eigen2}*x}\\ \simplify{{m[1][0]}*x+{m[1][1]}*y+{m[1][2]}*z}&=\simplify{{eigen2}*y}\\ \simplify{{m[2][0]}*x+{m[2][1]}*y+{m[2][2]}*z}&=\simplify{{eigen2}*z}\end{align}$

The question asked you to find an eigenvector with $x=1$ and if we substitute this into the equations we find (only need to use two of the equations) that $y=\var{-b}$ and $z=\simplify[all,fractionNumbers]{{b}/{g}}$ .

Hence the eigenvector we want is $\simplify[all,fractionNumbers]{vector(1,{-b},{b}/{g})}$ .

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Ugur's copy of Find eigenvalues, characteristic polynomial and a normalised eigenvector of a 3x3 matrix

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Ugur's copy of Find eigenvalues, characteristic polynomial and a normalised eigenvector of a 3x3 matrix

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Taxonomy: mathcentre

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Ugur Efem 5 months, 1 week ago

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Generated value: integer

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b)

Generated value: `integer`