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Lesson 14Eigenvalues and Eigenvectors
Math 20
October 22, 2007
Announcements
I Midterm almost done
I Problem Set 5 is on the WS. Due October 24
I OH: Mondays 1–2, Tuesdays 3–4, Wednesdays 1–3 (SC 323)
I Prob. Sess.: Sundays 6–7 (SC B-10), Tuesdays 1–2 (SC 116)
Math 20 - October 22, 2007.GWBMonday, Oct 22, 2007
Page1of16
Geometric effect of a diagonal linear transformation
Example
Let A =
(2 00 −1
). Draw the effect of the linear transformation
which is multiplication by A.
x
y
e1 Ae1
e2
Ae2
v
Av
Geometric effect of a diagonal linear transformation
Example
Let A =
(2 00 −1
). Draw the effect of the linear transformation
which is multiplication by A.
x
y
e1 Ae1
e2
Ae2
v
Av
Geometric effect of a diagonal linear transformation
Example
Let A =
(2 00 −1
). Draw the effect of the linear transformation
which is multiplication by A.
x
y
e1
Ae1
e2
Ae2
v
Av
Math 20 - October 22, 2007.GWBMonday, Oct 22, 2007
Page2of16
Geometric effect of a diagonal linear transformation
Example
Let A =
(2 00 −1
). Draw the effect of the linear transformation
which is multiplication by A.
x
y
e1 Ae1
e2
Ae2
v
Av
Geometric effect of a diagonal linear transformation
Example
Let A =
(2 00 −1
). Draw the effect of the linear transformation
which is multiplication by A.
x
y
e1 Ae1
e2
Ae2
v
Av
Geometric effect of a diagonal linear transformation
Example
Let A =
(2 00 −1
). Draw the effect of the linear transformation
which is multiplication by A.
x
y
e1 Ae1
e2
Ae2
v
Av
Math 20 - October 22, 2007.GWBMonday, Oct 22, 2007
Page4of16
Geometric effect of a diagonal linear transformation
Example
Let A =
(2 00 −1
). Draw the effect of the linear transformation
which is multiplication by A.
x
y
e1 Ae1
e2
Ae2
v
Av
Math 20 - October 22, 2007.GWBMonday, Oct 22, 2007
Page4of16
Math 20 - October 22, 2007.GWBMonday, Oct 22, 2007
Page5of16
Geometric effect of a diagonal linear transformation
Example
Let A =
(2 00 −1
). Draw the effect of the linear transformation
which is multiplication by A.
x
y
e1 Ae1
e2
Ae2
v
Av
Geometric effect of a diagonal linear transformation
Example
Let A =
(2 00 −1
). Draw the effect of the linear transformation
which is multiplication by A.
x
y
e1 Ae1
e2
Ae2
v
Av
Geometric effect of a diagonal linear transformation
Example
Let A =
(2 00 −1
). Draw the effect of the linear transformation
which is multiplication by A.
x
y
e1 Ae1
e2
Ae2
v
Av
Geometric effect of a diagonal linear transformation
Example
Let A =
(2 00 −1
). Draw the effect of the linear transformation
which is multiplication by A.
x
y
e1 Ae1
e2
Ae2
v
Av
Example
Draw the effect of the linear transformation which is multiplicationby A2.
x
y
Example
Draw the effect of the linear transformation which is multiplicationby A2.
x
y
Example
Draw the effect of the linear transformation which is multiplicationby A2.
x
y
Geometric effect of a non-diagonal linear transformation
Example
Let B =
(1/2 3/23/2 1/2
). Draw the effect of the linear transformation
which is multiplication by B.
x
y
e1
Be1e2
Be2
v
Bv
Geometric effect of a non-diagonal linear transformation
Example
Let B =
(1/2 3/23/2 1/2
). Draw the effect of the linear transformation
which is multiplication by B.
x
y
e1
Be1e2
Be2
v
Bv
Geometric effect of a non-diagonal linear transformation
Example
Let B =
(1/2 3/23/2 1/2
). Draw the effect of the linear transformation
which is multiplication by B.
x
y
e1
Be1
e2
Be2
v
Bv
Geometric effect of a non-diagonal linear transformation
Example
Let B =
(1/2 3/23/2 1/2
). Draw the effect of the linear transformation
which is multiplication by B.
x
y
e1
Be1e2
Be2
v
Bv
Geometric effect of a non-diagonal linear transformation
Example
Let B =
(1/2 3/23/2 1/2
). Draw the effect of the linear transformation
which is multiplication by B.
x
y
e1
Be1e2
Be2
v
Bv
Geometric effect of a non-diagonal linear transformation
Example
Let B =
(1/2 3/23/2 1/2
). Draw the effect of the linear transformation
which is multiplication by B.
x
y
e1
Be1e2
Be2
v
Bv
Geometric effect of a non-diagonal linear transformation
Example
Let B =
(1/2 3/23/2 1/2
). Draw the effect of the linear transformation
which is multiplication by B.
x
y
e1
Be1e2
Be2
v
Bv
Math 20 - October 22, 2007.GWBMonday, Oct 22, 2007
Page6of16
Geometric effect of a non-diagonal linear transformation
Example
Let B =
(1/2 3/23/2 1/2
). Draw the effect of the linear transformation
which is multiplication by B.
x
y
e1
Be1e2
Be2
v
Bv
Geometric effect of a non-diagonal linear transformation
Example
Let B =
(1/2 3/23/2 1/2
). Draw the effect of the linear transformation
which is multiplication by B.
x
y
e1
Be1e2
Be2
v
Bv
Geometric effect of a non-diagonal linear transformation
Example
Let B =
(1/2 3/23/2 1/2
). Draw the effect of the linear transformation
which is multiplication by B.
x
y
e1
Be1e2
Be2
v
Bv
Geometric effect of a non-diagonal linear transformation
Example
Let B =
(1/2 3/23/2 1/2
). Draw the effect of the linear transformation
which is multiplication by B.
x
y
v1
Bv1
v2
Bv2
2e2
2Be2
Math 20 - October 22, 2007.GWBMonday, Oct 22, 2007
Page8of16
Geometric effect of a non-diagonal linear transformation
Example
Let B =
(1/2 3/23/2 1/2
). Draw the effect of the linear transformation
which is multiplication by B.
x
y
v1
Bv1
v2
Bv2
2e2
2Be2
Geometric effect of a non-diagonal linear transformation
Example
Let B =
(1/2 3/23/2 1/2
). Draw the effect of the linear transformation
which is multiplication by B.
x
y
v1
Bv1
v2
Bv2
2e2
2Be2
Geometric effect of a non-diagonal linear transformation
Example
Let B =
(1/2 3/23/2 1/2
). Draw the effect of the linear transformation
which is multiplication by B.
x
y
v1
Bv1
v2
Bv2
2e2
2Be2
Math 20 - October 22, 2007.GWBMonday, Oct 22, 2007
Page9of16
Geometric effect of a non-diagonal linear transformation
Example
Let B =
(1/2 3/23/2 1/2
). Draw the effect of the linear transformation
which is multiplication by B.
x
y
v1
Bv1
v2
Bv2
2e2
2Be2
Geometric effect of a non-diagonal linear transformation
Example
Let B =
(1/2 3/23/2 1/2
). Draw the effect of the linear transformation
which is multiplication by B.
x
y
v1
Bv1
v2
Bv2
2e2
2Be2
Geometric effect of a non-diagonal linear transformation
Example
Let B =
(1/2 3/23/2 1/2
). Draw the effect of the linear transformation
which is multiplication by B.
x
y
v1
Bv1
v2
Bv2
2e2
2Be2
Geometric effect of a non-diagonal linear transformation
Example
Let B =
(1/2 3/23/2 1/2
). Draw the effect of the linear transformation
which is multiplication by B.
x
y
v1
Bv1
v2
Bv2
2e2
2Be2
Geometric effect of a non-diagonal linear transformation
Example
Let B =
(1/2 3/23/2 1/2
). Draw the effect of the linear transformation
which is multiplication by B.
x
y
v1
Bv1
v2
Bv2
2e2
2Be2
Geometric effect of a non-diagonal linear transformation
Example
Let B =
(1/2 3/23/2 1/2
). Draw the effect of the linear transformation
which is multiplication by B.
x
y
v1
Bv1
v2
Bv2
2e2
2Be2
Geometric effect of a non-diagonal linear transformation
Example
Let B =
(1/2 3/23/2 1/2
). Draw the effect of the linear transformation
which is multiplication by B.
x
y
v1
Bv1
v2
Bv2
2e2
2Be2
Geometric effect of a non-diagonal linear transformation
Example
Let B =
(1/2 3/23/2 1/2
). Draw the effect of the linear transformation
which is multiplication by B.
x
y
v1
Bv1
v2
Bv2
2e2
2Be2
The big concept
DefinitionLet A be an n × n matrix. The number λ is called an eigenvalueof A if there exists a nonzero vector x ∈ Rn such that
Ax = λx. (1)
Every nonzero vector satisfying (1) is called an eigenvector of Aassociated with the eigenvalue λ.
Math 20 - October 22, 2007.GWBMonday, Oct 22, 2007
Page10of16
Finding the eigenvector(s) corresponding to an eigenvalue
Example (Worksheet Problem 1)
Let A =
(0 −2−3 1
). The number 3 is an eigenvalue for A. Find
an eigenvector corresponding to this eigenvalue.
SolutionWe seek x such that
Ax = 3x =⇒ (A− 3I)x = 0.
[A− 3I 0
]=
[−3 −2 0−3 −2 0
]
[1 2/3 00 0 0
]So
(−2/3
1
), or
(−23
), are possible eigenvectors.
Math 20 - October 22, 2007.GWBMonday, Oct 22, 2007
Page11of16
Finding the eigenvector(s) corresponding to an eigenvalue
Example (Worksheet Problem 1)
Let A =
(0 −2−3 1
). The number 3 is an eigenvalue for A. Find
an eigenvector corresponding to this eigenvalue.
SolutionWe seek x such that
Ax = 3x =⇒ (A− 3I)x = 0.
[A− 3I 0
]=
[−3 −2 0−3 −2 0
]
[1 2/3 00 0 0
]So
(−2/3
1
), or
(−23
), are possible eigenvectors.
Finding the eigenvector(s) corresponding to an eigenvalue
Example (Worksheet Problem 1)
Let A =
(0 −2−3 1
). The number 3 is an eigenvalue for A. Find
an eigenvector corresponding to this eigenvalue.
SolutionWe seek x such that
Ax = 3x =⇒ (A− 3I)x = 0.
[A− 3I 0
]=
[−3 −2 0−3 −2 0
]
[1 2/3 00 0 0
]
So
(−2/3
1
), or
(−23
), are possible eigenvectors.
Math 20 - October 22, 2007.GWBMonday, Oct 22, 2007
Page12of16
Finding the eigenvector(s) corresponding to an eigenvalue
Example (Worksheet Problem 1)
Let A =
(0 −2−3 1
). The number 3 is an eigenvalue for A. Find
an eigenvector corresponding to this eigenvalue.
SolutionWe seek x such that
Ax = 3x =⇒ (A− 3I)x = 0.
[A− 3I 0
]=
[−3 −2 0−3 −2 0
]
[1 2/3 00 0 0
]So
(−2/3
1
), or
(−23
), are possible eigenvectors.
Math 20 - October 22, 2007.GWBMonday, Oct 22, 2007
Page13of16
Example (Worksheet Problem 2)
The number −2 is also an eigenvalue for A. Find an eigenvector.
SolutionWe have
A + 2I =
(2 −2−3 3
)
(1 −10 0
).
So
(11
)is an eigenvector for the eigenvalue −2.
Example (Worksheet Problem 2)
The number −2 is also an eigenvalue for A. Find an eigenvector.
SolutionWe have
A + 2I =
(2 −2−3 3
)
(1 −10 0
).
So
(11
)is an eigenvector for the eigenvalue −2.
Summary
To find the eigenvector(s) of a matrix corresponding to aneigenvalue λ, do Gaussian Elimination on A− λI.
Find the eigenvalues of a matrix
Example (Worksheet Problem 3)
Determine the eigenvalues of
A =
(−4 −33 6
).
SolutionAx = λx has a nonzero solution if and only if (A− λI)x = 0 has anonzero solution, if and only if A− λI is not invertible. Whatmagic number determines whether a matrix is invertible? Thedeterminant! So to find the possible values λ for which this couldbe true, we need to find the determinant of A− λI.
Find the eigenvalues of a matrix
Example (Worksheet Problem 3)
Determine the eigenvalues of
A =
(−4 −33 6
).
SolutionAx = λx has a nonzero solution if and only if (A− λI)x = 0 has anonzero solution, if and only if A− λI is not invertible.
Whatmagic number determines whether a matrix is invertible? Thedeterminant! So to find the possible values λ for which this couldbe true, we need to find the determinant of A− λI.
Math 20 - October 22, 2007.GWBMonday, Oct 22, 2007
Page14of16
Math 20 - October 22, 2007.GWBMonday, Oct 22, 2007
Page15of16
Find the eigenvalues of a matrix
Example (Worksheet Problem 3)
Determine the eigenvalues of
A =
(−4 −33 6
).
SolutionAx = λx has a nonzero solution if and only if (A− λI)x = 0 has anonzero solution, if and only if A− λI is not invertible. Whatmagic number determines whether a matrix is invertible?
Thedeterminant! So to find the possible values λ for which this couldbe true, we need to find the determinant of A− λI.
Find the eigenvalues of a matrix
Example (Worksheet Problem 3)
Determine the eigenvalues of
A =
(−4 −33 6
).
SolutionAx = λx has a nonzero solution if and only if (A− λI)x = 0 has anonzero solution, if and only if A− λI is not invertible. Whatmagic number determines whether a matrix is invertible? Thedeterminant!
So to find the possible values λ for which this couldbe true, we need to find the determinant of A− λI.
Find the eigenvalues of a matrix
Example (Worksheet Problem 3)
Determine the eigenvalues of
A =
(−4 −33 6
).
SolutionAx = λx has a nonzero solution if and only if (A− λI)x = 0 has anonzero solution, if and only if A− λI is not invertible. Whatmagic number determines whether a matrix is invertible? Thedeterminant! So to find the possible values λ for which this couldbe true, we need to find the determinant of A− λI.
det(A− λI) =
∣∣∣∣−4− λ −33 6− λ
∣∣∣∣= (−4− λ)(6− λ)− (−3)(3)
= (−24− 2λ+ λ2) + 9 = λ2 − 2λ− 15
= (λ+ 3)(λ− 5)
So λ = −3 and λ = 5 are the eigenvalues for A.
We can find the eigenvectors now, based on what we did before.We have
A + 3I =
(−1 −33 9
)
(1 30 0
)So
(−31
)is an eigenvector for this eigenvalue. Also,
A− 5I =
(−9 −33 1
)
(1 1/30 0
)
So
(−1/3
1
)or
(−13
)would be an eigenvector for this eigenvalue.
We can find the eigenvectors now, based on what we did before.We have
A + 3I =
(−1 −33 9
)
(1 30 0
)So
(−31
)is an eigenvector for this eigenvalue. Also,
A− 5I =
(−9 −33 1
)
(1 1/30 0
)
So
(−1/3
1
)or
(−13
)would be an eigenvector for this eigenvalue.
We can find the eigenvectors now, based on what we did before.We have
A + 3I =
(−1 −33 9
)
(1 30 0
)
So
(−31
)is an eigenvector for this eigenvalue. Also,
A− 5I =
(−9 −33 1
)
(1 1/30 0
)
So
(−1/3
1
)or
(−13
)would be an eigenvector for this eigenvalue.
We can find the eigenvectors now, based on what we did before.We have
A + 3I =
(−1 −33 9
)
(1 30 0
)So
(−31
)is an eigenvector for this eigenvalue.
Also,
A− 5I =
(−9 −33 1
)
(1 1/30 0
)
So
(−1/3
1
)or
(−13
)would be an eigenvector for this eigenvalue.
We can find the eigenvectors now, based on what we did before.We have
A + 3I =
(−1 −33 9
)
(1 30 0
)So
(−31
)is an eigenvector for this eigenvalue. Also,
A− 5I =
(−9 −33 1
)
(1 1/30 0
)
So
(−1/3
1
)or
(−13
)would be an eigenvector for this eigenvalue.
We can find the eigenvectors now, based on what we did before.We have
A + 3I =
(−1 −33 9
)
(1 30 0
)So
(−31
)is an eigenvector for this eigenvalue. Also,
A− 5I =
(−9 −33 1
)
(1 1/30 0
)
So
(−1/3
1
)or
(−13
)would be an eigenvector for this eigenvalue.
We can find the eigenvectors now, based on what we did before.We have
A + 3I =
(−1 −33 9
)
(1 30 0
)So
(−31
)is an eigenvector for this eigenvalue. Also,
A− 5I =
(−9 −33 1
)
(1 1/30 0
)
So
(−1/3
1
)or
(−13
)would be an eigenvector for this eigenvalue.
We can find the eigenvectors now, based on what we did before.We have
A + 3I =
(−1 −33 9
)
(1 30 0
)So
(−31
)is an eigenvector for this eigenvalue. Also,
A− 5I =
(−9 −33 1
)
(1 1/30 0
)
So
(−1/3
1
)or
(−13
)would be an eigenvector for this eigenvalue.
Summary
To find the eigenvalues of a matrix A, find the determinant ofA− λI. This will be a polynomial in λ, and its roots are theeigenvalues.
Example (Worksheet Problem 4)
Find the eigenvalues of
A =
(−10 6−15 9
).
SolutionThe characteristic polynomial is∣∣∣∣−10− λ 6−15 9− λ
∣∣∣∣ = (−10−λ)(9−λ)−(−15)(6) = λ2+λ = λ(λ+1).
The eigenvalues are 0 and −1. A set of eigenvectors are
(35
)and(
23
).
Example (Worksheet Problem 4)
Find the eigenvalues of
A =
(−10 6−15 9
).
SolutionThe characteristic polynomial is∣∣∣∣−10− λ 6−15 9− λ
∣∣∣∣ = (−10−λ)(9−λ)−(−15)(6) = λ2+λ = λ(λ+1).
The eigenvalues are 0 and −1. A set of eigenvectors are
(35
)and(
23
).
Example (Worksheet Problem 4)
Find the eigenvalues of
A =
(−10 6−15 9
).
SolutionThe characteristic polynomial is∣∣∣∣−10− λ 6−15 9− λ
∣∣∣∣ = (−10−λ)(9−λ)−(−15)(6) = λ2+λ = λ(λ+1).
The eigenvalues are 0 and −1.
A set of eigenvectors are
(35
)and(
23
).
Example (Worksheet Problem 4)
Find the eigenvalues of
A =
(−10 6−15 9
).
SolutionThe characteristic polynomial is∣∣∣∣−10− λ 6−15 9− λ
∣∣∣∣ = (−10−λ)(9−λ)−(−15)(6) = λ2+λ = λ(λ+1).
The eigenvalues are 0 and −1. A set of eigenvectors are
(35
)and(
23
).
Applications
I In a Markov Chain, the steady-state vector is an eigenvectorcorresponding to the eigenvalue 1.
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