Example 2
Consider the matrix
1 2
A=
0 1
which has characteristic polynomial
1 2
=( 1)2
0 1
What are its eigenvalues and eigenspaces?
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Multiplicity of eigenvalue
The algebraic multiplicity of a particular eigenvalue k is m if
det(A I) = · · · (k )m · · ·
The geometric multiplicity of a particular eigenvalue k is the
dimension of the eigenspace Ek .
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Back to Example 2
Consider the matrix
1 2
A=
0 1
Algebraic multiplicity=
Geometric multiplicity=
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Algebraic vs Geometric
For any eigenvalue:
1 6 geometric multiplicity 6 algebraic multiplicity
I The algebra tells us the maximum possible multiplicity
I The geometry tells us the actual multiplicity
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Diagonalisable
A linear transformation f : Rn ! Rn is diagonalisable if there is
a basis B for Rn such that [f ]BB is a diagonal matrix.
A matrix A is called diagonalisable if the corresponding linear
transformation is diagonalisable.
This happens exactly if we can find a basis for Rn consisting of
eigenvectors for A.
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Diagonalisability test
A matrix A is diagonalisable if and only if
I It has all real eigenvalues (no complex)
I Every eigenvalue has the maximum possible geometric
multiplicity
In other words, each eigenvalue has geometric multiplicity equal
to its algebraic multiplicity.
Special case: if the matrix has n distinct eigenvalues, then it is
diagonalisable.
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Example
What are the eigenvalues and eigenvectors of
2 3
1 1 1
A = 4 0 1 1 5?
0 1 1
Find characteristic polynomial:
1 1 1
det(A I) = 0 1 1
0 1 1
= ( 1)( 2)
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Working
Solve three systems of linear equations:
I Ax = 0x has solutions S = {t(0, 1, 1) : t 2 R}
I Ax = 1x has solutions S = {t(1, 0, 0) : t 2 R}
I Ax = 2x has solutions S = {t(2, 1, 1) : t 2 R}
So B = {(0, 1, 1), (1, 0, 0), (2, 1, 1)} is a basis for Rn of
eigenvectors.
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Working: Change of basis
The matrix 2 3
0 1 2
P = PSB = 4 1 0 1 5
1 0 1
and 2 3
0 0 0
P 1 AP = 4 0 1 0 5
0 0 2
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Summary
To diagonalise a matrix A
I Find the eigenvalues 1 , 2, . . ., k and their multiplicities
If some has geometric multiplicity too low then fail.
I Find a basis for each eigenspace E 1 , E 2 , . . . , E k
.
I Put the basis vectors into the columns of P
Note that P is a square matrix.
Then P 1 AP = D where D is the diagonal matrix of
eigenvalues.
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High powers
What is A5 , if A is diagonalisable?
A5 = (P DP 1
)(P DP 1
)(P DP 1
)(P DP 1
)(P DP 1
)
1 1 1 1 1
= P D(P P )D(P P )D(P P )D(P P )DP
1
= P DIDIDIDIDP
= P D5 P 1
If 2 3 2 3
0 0 5 0 0
1 1
D=4 0 2 0 5 D5 = 4 0 5
2 0 5
0 0 0 0 5
3 3
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Special Case
A matrix is symmetric if A = AT .
If A is a symmetric matrix, then
I Each eigenvalue of A is real
I Each eigenvalue has geometric multiplicity equal to
algebraic multiplicity
I Eigenvectors from distinct eigenspaces are orthogonal
In other words, symmetric matrices are diagonalisable.
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Applications
There are lots of applications of eigenvalues/eigenspaces to
I engineering (stress tensors)
I physics (mechanics)
I study of graphs (Google Page Rank)
I data analysis
I computer graphics (image compression)
This concludes our study of Linear Algebra.
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