Characteristic poly, eigenvalues and eigenvectors 3x3, digonailsability (non-randomised)

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Created this as a copy of Characteristic poly, eigenvalues and eigenvectors 3x3 non-randomised.

Name	Status	Author	Last Modified
Characteristic poly, eigenvalues and eigenvectors 3x3 non-randomised	Ready to use	Julia Goedecke	24/05/2022 20:45
Characteristic poly, eigenvalues and eigenvectors 3x3, digonailsability (non-randomised)	Ready to use	Ugur Efem	13/10/2024 20:15

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Question statement

Find the characteristic polynomial, eigenvalues and eigenvectors (bases for eigenspaces) for the matrix $\var{matrixA}$ . Decide whether the matrix is diagonalisable.

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			Name	Type	Generated Value

matrixA	matrix	Matrix of size 3×3
a	integer	-1
b	integer	2
c	integer	5
chiA	expression	(t + 1)(t - 2)(t - 5)
v1	matrix	matrix([0],[-3],[1])
v2	matrix	matrix([-1],[0],[1])
v3	matrix	matrix([1],[0],[0])
RREFa	matrix	Matrix of size 3×3
RREFb	matrix	Matrix of size 3×3
RREFc	matrix	Matrix of size 3×3
n	integer	3
chiAexpanded	expression	t^3 - 6*t^2 + 3t + 10

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matrix([5,1,3],[0,-1,0],[0,1,2])

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Generated value: `matrix`

matrix([5,1,3],[0,-1,0],[0,1,2])

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RREFa
RREFb
RREFc

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Char poly Gap-fill
1. Gap char_A Mathematical expression
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2. Gap evals List of numbers
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evectors Gap-fill
1. Gap Gap 0. Number entry
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2. Gap Gap 1. Matrix entry
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Diagonalisability Choose one from a list

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The characteristic polynomial of $A$ is defined as $\chi_A(t)=\det(tI-A)$. So in this case you should calculate

$\begin{vmatrix} \simplify{t-{matrixA[0][0]}} & \var{-matrixA[0][1]} & \var{-matrixA[0][2]} \\ \var{-matrixA[1][0]} & \simplify{t-{matrixA[1][1]}} & \var{-matrixA[1][2]}\\ \var{-matrixA[2][0]} & \var{-matrixA[2][1]} & \simplify{t-{matrixA[2][2]}} \end{vmatrix}=(\simplify{t-{matrixA[0][0]}})\begin{vmatrix} \simplify{t-{matrixA[1][1]}} & \var{-matrixA[1][2]}\\ \var{-matrixA[2][1]} & \simplify{t-{matrixA[2][2]}} \end{vmatrix}=\simplify[]{(t-{matrixA[0][0]})((t-{matrixA[1][1]})(t-{matrixA[2][2]})-{matrixA[1][2]*matrixA[2][1]})}= \var{chiA}=\var{chiAexpanded}$.

Here we first expanded the determinant in the first column, and then we used the formula for the determinant a $2\times 2$ matrix in the last step: $ \begin{vmatrix}a&b\\c&d\end{vmatrix}=ad-bc$.

The eigenvalues are exactly the roots of the characteristic polynomial. So the eigenvalues of $A$ are $\var{a}$, $\var{b}$ and $\var{c}$.

To calculate the eigenvectors for a given eigenvalues $\lambda$, we calculate the kernel of $A-\lambda I$. So, for

$\lambda=\var{a}$, we calculate the kernel of $\var{matrixA-a*id(n)}$. The reduced row echelon form is $\var[fractionNumbers]{RREFa}$. So the eigenvector for $\lambda=\var{a}$ is

$\var[fractionNumbers]{v1}$ (or any multiple of this).

And for $\lambda=\var{b}$, the reduced row echelon form of $\var{matrixA-b*id(n)}$ is $\var[fractionNumbers]{RREFb}$, and the eigenvector is $\var[fractionNumbers]{v2}$ (or any multiple).

And for $\lambda=\var{c}$, the reduced row echelon form of $\var{matrixA-c*id(n)}$ is $\var[fractionNumbers]{RREFc}$, and the eigenvector is $\var[fractionNumbers]{v3}$ (or any multiple).

Recall that the algebraic multiplicity of an eigenvalue is it multiplicity as a root of the characteristic polynomial. Then, it is clear that all the eigenvalues of $A$ has albegraic multiplicity $1$.

Also, recall that the geometric multiplicity of an eigenvalue $\lambda$ is the dimension of the subspace genereted by eigenvectors associated to $\lambda$. Observe that here every eigenvalue has geometric multiplicty $1$. Therefore, for every eigenvalue, its algebraic and geoemtric multiplicities are eqaul. That implies $A$ is diagonalisable.

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This question is used in the following exam:

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

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Characteristic poly, eigenvalues and eigenvectors 3x3, digonailsability (non-randomised)

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Characteristic poly, eigenvalues and eigenvectors 3x3, digonailsability (non-randomised)

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Ugur Efem 5 months, 1 week ago

Ugur Efem 5 months, 1 week ago

Ugur Efem 5 months, 1 week ago

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Generated value: `matrix`