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Outline
Week 10: Eigenvalues and eigenvectors
Course Notes: 6.1
Goals: Understand how to find eigenvector/eigenvalue pairs, and use themto simplify calculations involving matrix powers.
When Matrix Multiplication looks like Scalar Multiplication
[2 00 2
] [xy
]= 2
[xy
]
Recall: two vectors that are scalar multiples of one another are parallel.
When Matrix Multiplication looks like Scalar Multiplication
[2 00 2
] [xy
]= 2
[xy
]
Recall: two vectors that are scalar multiples of one another are parallel.
Reflections, Revisited
y = x
π4
Refπ4
=
[cos(2π
4 ) sin(2π4 )
sin(2π4 ) − cos(2π
4 )
]=
[0 11 0
][
0 11 0
] [xx
]= 1 ·
[xx
] [0 11 0
] [x−x
]= −1 ·
[x−x
]
Reflections, Revisited
y = x
π4
Refπ4
=
[cos(2π
4 ) sin(2π4 )
sin(2π4 ) − cos(2π
4 )
]=
[0 11 0
]
[0 11 0
] [xx
]= 1 ·
[xx
] [0 11 0
] [x−x
]= −1 ·
[x−x
]
Reflections, Revisited
y = x
π4
Refπ4
=
[cos(2π
4 ) sin(2π4 )
sin(2π4 ) − cos(2π
4 )
]=
[0 11 0
]
[0 11 0
] [xx
]= 1 ·
[xx
] [0 11 0
] [x−x
]= −1 ·
[x−x
]
Reflections, Revisited
y = x
π4
Refπ4
=
[cos(2π
4 ) sin(2π4 )
sin(2π4 ) − cos(2π
4 )
]=
[0 11 0
]
[0 11 0
] [xx
]= 1 ·
[xx
] [0 11 0
] [x−x
]= −1 ·
[x−x
]
Reflections, Revisited
y = x
π4
Refπ4
=
[cos(2π
4 ) sin(2π4 )
sin(2π4 ) − cos(2π
4 )
]=
[0 11 0
]
[0 11 0
] [xx
]= 1 ·
[xx
] [0 11 0
] [x−x
]= −1 ·
[x−x
]
Reflections, Revisited
y = x
π4
Refπ4
=
[cos(2π
4 ) sin(2π4 )
sin(2π4 ) − cos(2π
4 )
]=
[0 11 0
]
[0 11 0
] [xx
]= 1 ·
[xx
] [0 11 0
] [x−x
]= −1 ·
[x−x
]
Reflections, Revisited
y = x
π4
Refπ4
=
[cos(2π
4 ) sin(2π4 )
sin(2π4 ) − cos(2π
4 )
]=
[0 11 0
]
[0 11 0
] [xx
]= 1 ·
[xx
] [0 11 0
] [x−x
]= −1 ·
[x−x
]
Reflections, Revisited
y = x
π4
Refπ4
=
[cos(2π
4 ) sin(2π4 )
sin(2π4 ) − cos(2π
4 )
]=
[0 11 0
]
[0 11 0
] [xx
]= 1 ·
[xx
] [0 11 0
] [x−x
]= −1 ·
[x−x
]
Reflections, Revisited
y = x
π4
Refπ4
=
[cos(2π
4 ) sin(2π4 )
sin(2π4 ) − cos(2π
4 )
]=
[0 11 0
]
[0 11 0
] [xx
]= 1 ·
[xx
] [0 11 0
] [x−x
]= −1 ·
[x−x
]
Reflections, Revisited
y = x
π4
Refπ4
=
[cos(2π
4 ) sin(2π4 )
sin(2π4 ) − cos(2π
4 )
]=
[0 11 0
]
[0 11 0
] [xx
]= 1 ·
[xx
] [0 11 0
] [x−x
]= −1 ·
[x−x
]
Reflections, Revisited
y = x
π4
Refπ4
=
[cos(2π
4 ) sin(2π4 )
sin(2π4 ) − cos(2π
4 )
]=
[0 11 0
][
0 11 0
] [xx
]= 1 ·
[xx
]
[0 11 0
] [x−x
]= −1 ·
[x−x
]
Reflections, Revisited
y = x
π4
Refπ4
=
[cos(2π
4 ) sin(2π4 )
sin(2π4 ) − cos(2π
4 )
]=
[0 11 0
][
0 11 0
] [xx
]= 1 ·
[xx
]
[0 11 0
] [x−x
]= −1 ·
[x−x
]
Reflections, Revisited
y = x
π4
Refπ4
=
[cos(2π
4 ) sin(2π4 )
sin(2π4 ) − cos(2π
4 )
]=
[0 11 0
][
0 11 0
] [xx
]= 1 ·
[xx
]
[0 11 0
] [x−x
]= −1 ·
[x−x
]
Reflections, Revisited
y = x
π4
Refπ4
=
[cos(2π
4 ) sin(2π4 )
sin(2π4 ) − cos(2π
4 )
]=
[0 11 0
][
0 11 0
] [xx
]= 1 ·
[xx
]
[0 11 0
] [x−x
]= −1 ·
[x−x
]
Reflections, Revisited
y = x
π4
Refπ4
=
[cos(2π
4 ) sin(2π4 )
sin(2π4 ) − cos(2π
4 )
]=
[0 11 0
][
0 11 0
] [xx
]= 1 ·
[xx
]
[0 11 0
] [x−x
]= −1 ·
[x−x
]
Reflections, Revisited
y = x
π4
Refπ4
=
[cos(2π
4 ) sin(2π4 )
sin(2π4 ) − cos(2π
4 )
]=
[0 11 0
][
0 11 0
] [xx
]= 1 ·
[xx
] [0 11 0
] [x−x
]= −1 ·
[x−x
]
Eigenvectors and Eigenvalues
[0 11 0
] [xx
]= 1 ·
[xx
]The matrix
[0 11 0
]has eigenvector
[11
]with eigenvalue 1.[
0 11 0
] [x−x
]= −1 ·
[x−x
]
Tthe matrix
[0 11 0
]has eigenvector
[−11
]with eigenvalue −1.
Given a matrix A, a scalar λ, and a NONZERO vector x with
Ax = λx
we say x is an eigenvector of A with eigenvalue λ.
Rotation Matrix Eigenvalues
Rotθ =
[cos θ − sin θsin θ cos θ
]
Search for eigenvectors:[cos θ − sin θsin θ cos θ
] [xy
]= λ
[xy
]
θθ
θ
In general, no (real) eigenvalues of Rotθ.
Rotation Matrix Eigenvalues
Rotθ =
[cos θ − sin θsin θ cos θ
]Search for eigenvectors:[
cos θ − sin θsin θ cos θ
] [xy
]= λ
[xy
]
θθ
θ
In general, no (real) eigenvalues of Rotθ.
Rotation Matrix Eigenvalues
Rotθ =
[cos θ − sin θsin θ cos θ
]Search for eigenvectors:[
cos θ − sin θsin θ cos θ
] [xy
]= λ
[xy
]
θθ
θ
In general, no (real) eigenvalues of Rotθ.
Rotation Matrix Eigenvalues
Rotθ =
[cos θ − sin θsin θ cos θ
]Search for eigenvectors:[
cos θ − sin θsin θ cos θ
] [xy
]= λ
[xy
]
θ
θ
θ
In general, no (real) eigenvalues of Rotθ.
Rotation Matrix Eigenvalues
Rotθ =
[cos θ − sin θsin θ cos θ
]Search for eigenvectors:[
cos θ − sin θsin θ cos θ
] [xy
]= λ
[xy
]
θθ
θ
In general, no (real) eigenvalues of Rotθ.
Rotation Matrix Eigenvalues
Rotθ =
[cos θ − sin θsin θ cos θ
]Search for eigenvectors:[
cos θ − sin θsin θ cos θ
] [xy
]= λ
[xy
]
θ
θ
θ
In general, no (real) eigenvalues of Rotθ.
Rotation Matrix Eigenvalues
Rotθ =
[cos θ − sin θsin θ cos θ
]Search for eigenvectors:[
cos θ − sin θsin θ cos θ
] [xy
]= λ
[xy
]
θθ
θ
In general, no (real) eigenvalues of Rotθ.
Rotation Matrix Eigenvalues
Rotθ =
[cos θ − sin θsin θ cos θ
]Search for eigenvectors:[
cos θ − sin θsin θ cos θ
] [xy
]= λ
[xy
]
θθ
θ
In general, no (real) eigenvalues of Rotθ.
Rotation Matrix Eigenvalues
Rotθ =
[cos θ − sin θsin θ cos θ
]Search for eigenvectors:[
cos θ − sin θsin θ cos θ
] [xy
]= λ
[xy
]
θθ
θ
In general, no (real) eigenvalues of Rotθ.
Finding Eigenvectors, Given Eigenvalues
Given: 7 is an eigenvalue of the matrix
1 2 42 4 14 1 2
.
What is an eigenvector associated to that eigenvalue?
An eigenvector
xyz
with eigenvalue 7 would satisfy this equation:
1 2 42 4 14 1 2
xyz
= 7
xyz
In equation form:
x + 2y + 4x = 7x2x + 4y + z = 7y4x + y + 2z = 7z
Finding Eigenvectors, Given Eigenvalues
Given: 7 is an eigenvalue of the matrix
1 2 42 4 14 1 2
.
What is an eigenvector associated to that eigenvalue?
An eigenvector
xyz
with eigenvalue 7 would satisfy this equation:
1 2 42 4 14 1 2
xyz
= 7
xyz
In equation form:
x + 2y + 4x = 7x2x + 4y + z = 7y4x + y + 2z = 7z
Finding Eigenvectors, Given Eigenvalues
Given: 7 is an eigenvalue of the matrix
1 2 42 4 14 1 2
.
What is an eigenvector associated to that eigenvalue?
An eigenvector
xyz
with eigenvalue 7 would satisfy this equation:
1 2 42 4 14 1 2
xyz
= 7
xyz
In equation form:
x + 2y + 4x = 7x2x + 4y + z = 7y4x + y + 2z = 7z
Finding Eigenvectors, Given Eigenvalues
In equation form:
x + 2y + 4x = 7x2x + 4y + z = 7y4x + y + 2z = 7z
In better equation form:
−6x + 2y + 4x = 02x − 3y + z = 04x + y − 5z = 0
Gaussian elimination on augmented matrix:−6 2 4 02 −3 1 04 1 −5 0
→ → →
1 0 −1 00 1 −1 00 0 0 0
Finding Eigenvectors, Given Eigenvalues
In equation form:
x + 2y + 4x = 7x2x + 4y + z = 7y4x + y + 2z = 7z
In better equation form:
−6x + 2y + 4x = 02x − 3y + z = 04x + y − 5z = 0
Gaussian elimination on augmented matrix:−6 2 4 02 −3 1 04 1 −5 0
→ → →
1 0 −1 00 1 −1 00 0 0 0
Finding Eigenvectors, Given Eigenvalues
In equation form:
x + 2y + 4x = 7x2x + 4y + z = 7y4x + y + 2z = 7z
In better equation form:
−6x + 2y + 4x = 02x − 3y + z = 04x + y − 5z = 0
Gaussian elimination on augmented matrix:−6 2 4 02 −3 1 04 1 −5 0
→ → →
1 0 −1 00 1 −1 00 0 0 0
Finding Eigenvectors, Given Eigenvalues
Gaussian elimination on augmented matrix:−6 2 4 02 −3 1 04 1 −5 0
→ → →
1 0 −1 00 1 −1 00 0 0 0
Solutions: xyz
= s
111
So these are the solutions to:1 2 4
2 4 14 1 2
xyz
= 7
xyz
Finding Eigenvectors, Given Eigenvalues
Gaussian elimination on augmented matrix:−6 2 4 02 −3 1 04 1 −5 0
→ → →
1 0 −1 00 1 −1 00 0 0 0
Solutions: xy
z
= s
111
So these are the solutions to:1 2 42 4 14 1 2
xyz
= 7
xyz
Finding Eigenvectors, Given Eigenvalues
Gaussian elimination on augmented matrix:−6 2 4 02 −3 1 04 1 −5 0
→ → →
1 0 −1 00 1 −1 00 0 0 0
Solutions: xy
z
= s
111
So these are the solutions to:1 2 4
2 4 14 1 2
xyz
= 7
xyz
Finding Eigenvectors, Given Eigenvalues
The solutions to: 1 2 42 4 14 1 2
xyz
= 7
xyz
are precisely the vectors: xy
z
= s
111
So, we can choose
111
as an example of an eigenvector with eigenvalue 7.
Finding Eigenvectors, Given Eigenvalues
The solutions to: 1 2 42 4 14 1 2
xyz
= 7
xyz
are precisely the vectors: xy
z
= s
111
So, we can choose
111
as an example of an eigenvector with eigenvalue 7.
Generalized Eigenvector Finding
Suppose a matrix A has eigenvalue λ. Find an associated eigenvector x .
Ax = λx
Ax − λx = 0
(A− λI )x = 0
Generalized Eigenvector Finding
Suppose a matrix A has eigenvalue λ. Find an associated eigenvector x .
Ax = λx
Ax − λx = 0
(A− λI )x = 0
Finding Eigenvectors from Eigenvalues
The matrix
[3 66 −2
]has eigenvalues −6 and 7.
Find an eigenvector associated to each eigenvalue.
λ = −6, x =
[2−3
]λ = 7, x =
[32
]
Basis of R2:
{[2−3
],
[32
]}
Finding Eigenvectors from Eigenvalues
The matrix
[3 66 −2
]has eigenvalues −6 and 7.
Find an eigenvector associated to each eigenvalue.
λ = −6, x =
[2−3
]λ = 7, x =
[32
]
Basis of R2:
{[2−3
],
[32
]}
Finding Eigenvectors from Eigenvalues
The matrix
[3 66 −2
]has eigenvalues −6 and 7.
Find an eigenvector associated to each eigenvalue.
λ = −6, x =
[2−3
]λ = 7, x =
[32
]
Basis of R2:
{[2−3
],
[32
]}
Finding Eigenvectors from Eigenvalues
The matrix
[2 34 1
]has eigenvalues −2 and 5.
Find an eigenvector associated to each eigenvalue.
λ = −2, x =
[3−4
]λ = 5, x =
[11
]
Basis of R2:
{[3−4
],
[11
]}
Finding Eigenvectors from Eigenvalues
The matrix
[2 34 1
]has eigenvalues −2 and 5.
Find an eigenvector associated to each eigenvalue.
λ = −2, x =
[3−4
]λ = 5, x =
[11
]
Basis of R2:
{[3−4
],
[11
]}
Finding Eigenvectors from Eigenvalues
The matrix
[2 34 1
]has eigenvalues −2 and 5.
Find an eigenvector associated to each eigenvalue.
λ = −2, x =
[3−4
]λ = 5, x =
[11
]
Basis of R2:
{[3−4
],
[11
]}
How do we Find Eigenvalues, Though?
First, a reminder....
Solutions to Systems of Equations
Let A be an n-by-n matrix. The following statements are equivalent:
1) Ax = b has exactly one solution for any b .2) Ax = 0 has no nonzero solutions.3) The rank of A is n.4) The reduced form of A has no zeroes along the main diagonal.5) A is invertible6) det(A) 6= 0
statements 1-4invertiblenonzero determinant
How do we Find Eigenvalues, Though?
A matrix; λ eigenvalue, x eigenvector (so x 6= 0)
Ax = λx
Ax− λx = 0
(A− λI )x = 0
(A− λI )x = 0 has a nonzero solution
det(A− λI ) = 0
How do we Find Eigenvalues, Though?
A matrix; λ eigenvalue, x eigenvector (so x 6= 0)
Ax = λx
Ax− λx = 0
(A− λI )x = 0
(A− λI )x = 0 has a nonzero solution
det(A− λI ) = 0
Find Eigenvalues and Associated Eigenvectors
λ eigenvalue ⇔ det(A− λI ) = 0
A =
[1 03 −1
]
λ = 1,
[23
]λ = −1,
[01
] {[23
],
[01
]}is a basis of R2
B =
[1 20 1
]
λ = 1,
[10
]NOT a basis of R2!
Find Eigenvalues and Associated Eigenvectors
λ eigenvalue ⇔ det(A− λI ) = 0
A =
[1 03 −1
]λ = 1,
[23
]λ = −1,
[01
] {[23
],
[01
]}is a basis of R2
B =
[1 20 1
]
λ = 1,
[10
]NOT a basis of R2!
Find Eigenvalues and Associated Eigenvectors
λ eigenvalue ⇔ det(A− λI ) = 0
A =
[1 03 −1
]λ = 1,
[23
]λ = −1,
[01
] {[23
],
[01
]}is a basis of R2
B =
[1 20 1
]λ = 1,
[10
]NOT a basis of R2!
Find Eigenvalues and Associated Eigenvectors
A =
1 4 50 2 60 0 3
λ = 1,
100
λ = 2,
410
λ = 3,
29122
1
00
,4
10
,29
122
Basis of R3
Find Eigenvalues and Associated Eigenvectors
A =
1 4 50 2 60 0 3
λ = 1,
100
λ = 2,
410
λ = 3,
29122
1
00
,4
10
,29
122
Basis of R3
Find Eigenvalues and Associated Eigenvectors
A =
1 4 50 2 60 0 3
λ = 1,
100
λ = 2,
410
λ = 3,
29122
1
00
,4
10
,29
122
Basis of R3
Using Eigenvalues to Compute Matrix Powers
A =
1 4 50 2 60 0 3
λ = 1, k1 =
100
λ = 2, k2 =
410
λ = 3, k3 =
29122
x =
47162
= 2
100
+ 4
410
+
29122
= 2k1 + 4k2 + k3
Ax =A(2k1 + 4k2 + k3)= 2Ak1 + 4Ak2 + Ak3= 2k1 + 4(2)k2 + 3k3
Using Eigenvalues to Compute Matrix Powers
A =
1 4 50 2 60 0 3
λ = 1, k1 =
100
λ = 2, k2 =
410
λ = 3, k3 =
29122
x =
47162
= 2
100
+ 4
410
+
29122
= 2k1 + 4k2 + k3
Ax =A(2k1 + 4k2 + k3)= 2Ak1 + 4Ak2 + Ak3= 2k1 + 4(2)k2 + 3k3
Using Eigenvalues to Compute Matrix Powers
A =
1 4 50 2 60 0 3
λ = 1, k1 =
100
λ = 2, k2 =
410
λ = 3, k3 =
29122
x =
47162
= 2
100
+ 4
410
+
29122
= 2k1 + 4k2 + k3
Ax =
A(2k1 + 4k2 + k3)= 2Ak1 + 4Ak2 + Ak3= 2k1 + 4(2)k2 + 3k3
Using Eigenvalues to Compute Matrix Powers
A =
1 4 50 2 60 0 3
λ = 1, k1 =
100
λ = 2, k2 =
410
λ = 3, k3 =
29122
x =
47162
= 2
100
+ 4
410
+
29122
= 2k1 + 4k2 + k3
Ax =A(2k1 + 4k2 + k3)
= 2Ak1 + 4Ak2 + Ak3= 2k1 + 4(2)k2 + 3k3
Using Eigenvalues to Compute Matrix Powers
A =
1 4 50 2 60 0 3
λ = 1, k1 =
100
λ = 2, k2 =
410
λ = 3, k3 =
29122
x =
47162
= 2
100
+ 4
410
+
29122
= 2k1 + 4k2 + k3
Ax =A(2k1 + 4k2 + k3)= 2Ak1 + 4Ak2 + Ak3
= 2k1 + 4(2)k2 + 3k3
Using Eigenvalues to Compute Matrix Powers
A =
1 4 50 2 60 0 3
λ = 1, k1 =
100
λ = 2, k2 =
410
λ = 3, k3 =
29122
x =
47162
= 2
100
+ 4
410
+
29122
= 2k1 + 4k2 + k3
Ax =A(2k1 + 4k2 + k3)= 2Ak1 + 4Ak2 + Ak3= 2k1 + 4(2)k2 + 3k3
Using Eigenvalues to Compute Matrix Powers
A =
1 4 50 2 60 0 3
λ = 1, k1 =
100
λ = 2, k2 =
410
λ = 3, k3 =
29122
x = 2k1 + 4k2 + k3
A10x =
A10(2k1 + 4k2 + k3)= 2A10k1 + 4A10k2 + A10k3=
2k1 + 4(210)k2 + 310k3=
200
+
214
212
0
+
29 · 31012 · 3102 · 310
=
2 + 214 + 29 · 310212 + 12 · 310
2 · 310
Using Eigenvalues to Compute Matrix Powers
A =
1 4 50 2 60 0 3
λ = 1, k1 =
100
λ = 2, k2 =
410
λ = 3, k3 =
29122
x = 2k1 + 4k2 + k3
A10x =A10(2k1 + 4k2 + k3)
= 2A10k1 + 4A10k2 + A10k3=
2k1 + 4(210)k2 + 310k3=
200
+
214
212
0
+
29 · 31012 · 3102 · 310
=
2 + 214 + 29 · 310212 + 12 · 310
2 · 310
Using Eigenvalues to Compute Matrix Powers
A =
1 4 50 2 60 0 3
λ = 1, k1 =
100
λ = 2, k2 =
410
λ = 3, k3 =
29122
x = 2k1 + 4k2 + k3
A10x =A10(2k1 + 4k2 + k3)= 2A10k1 + 4A10k2 + A10k3
=
2k1 + 4(210)k2 + 310k3=
200
+
214
212
0
+
29 · 31012 · 3102 · 310
=
2 + 214 + 29 · 310212 + 12 · 310
2 · 310
Using Eigenvalues to Compute Matrix Powers
A =
1 4 50 2 60 0 3
λ = 1, k1 =
100
λ = 2, k2 =
410
λ = 3, k3 =
29122
x = 2k1 + 4k2 + k3
A10x =A10(2k1 + 4k2 + k3)= 2A10k1 + 4A10k2 + A10k3=
2k1 + 4(210)k2 + 310k3
=
200
+
214
212
0
+
29 · 31012 · 3102 · 310
=
2 + 214 + 29 · 310212 + 12 · 310
2 · 310
Using Eigenvalues to Compute Matrix Powers
A =
1 4 50 2 60 0 3
λ = 1, k1 =
100
λ = 2, k2 =
410
λ = 3, k3 =
29122
x = 2k1 + 4k2 + k3
A10x =A10(2k1 + 4k2 + k3)= 2A10k1 + 4A10k2 + A10k3=
2k1 + 4(210)k2 + 310k3=
200
+
214
212
0
+
29 · 31012 · 3102 · 310
=
2 + 214 + 29 · 310212 + 12 · 310
2 · 310
Using Eigenvalues to Compute Matrix Powers
A =
1 4 50 2 60 0 3
λ = 1, k1 =
100
λ = 2, k2 =
410
λ = 3, k3 =
29122
x = 2k1 + 4k2 + k3
A10x =A10(2k1 + 4k2 + k3)= 2A10k1 + 4A10k2 + A10k3=
2k1 + 4(210)k2 + 310k3=
200
+
214
212
0
+
29 · 31012 · 3102 · 310
=
2 + 214 + 29 · 310212 + 12 · 310
2 · 310
Random Walks and Eigenvalues
fromto
pass fail
pass 12
13
fail 12
23
λ1 = 16 , k1 =
[1−1
]; λ2 = 1, k1 =
[23
]Suppose your initial state is x0 =
[10
]. What happens after n tests?
x0 = 35k1 + 1
5k2
Pnx0 = 35P
nk1 + 15P
nk2 = 35
(16
)nk1 + 1
5k2
limn→∞
Pnx0 =1
5k2 =
[2/53/5
]
Random Walks and Eigenvalues
fromto
pass fail
pass 12
13
fail 12
23
λ1 = 16 , k1 =
[1−1
]; λ2 = 1, k1 =
[23
]
Suppose your initial state is x0 =
[10
]. What happens after n tests?
x0 = 35k1 + 1
5k2
Pnx0 = 35P
nk1 + 15P
nk2 = 35
(16
)nk1 + 1
5k2
limn→∞
Pnx0 =1
5k2 =
[2/53/5
]
Random Walks and Eigenvalues
fromto
pass fail
pass 12
13
fail 12
23
λ1 = 16 , k1 =
[1−1
]; λ2 = 1, k1 =
[23
]Suppose your initial state is x0 =
[10
]. What happens after n tests?
x0 = 35k1 + 1
5k2
Pnx0 = 35P
nk1 + 15P
nk2 = 35
(16
)nk1 + 1
5k2
limn→∞
Pnx0 =1
5k2 =
[2/53/5
]
Random Walks and Eigenvalues
fromto
pass fail
pass 12
13
fail 12
23
λ1 = 16 , k1 =
[1−1
]; λ2 = 1, k1 =
[23
]Suppose your initial state is x0 =
[10
]. What happens after n tests?
x0 = 35k1 + 1
5k2
Pnx0 = 35P
nk1 + 15P
nk2 = 35
(16
)nk1 + 1
5k2
limn→∞
Pnx0 =1
5k2 =
[2/53/5
]
Random Walks and Eigenvalues
fromto
pass fail
pass 12
13
fail 12
23
λ1 = 16 , k1 =
[1−1
]; λ2 = 1, k1 =
[23
]Suppose your initial state is x0 =
[10
]. What happens after n tests?
x0 = 35k1 + 1
5k2
Pnx0 = 35P
nk1 + 15P
nk2 = 35
(16
)nk1 + 1
5k2
limn→∞
Pnx0 =1
5k2 =
[2/53/5
]
Random Walks and Eigenvalues
fromto
pass fail
pass 12
13
fail 12
23
λ1 = 16 , k1 =
[1−1
]; λ2 = 1, k1 =
[23
]Suppose your initial state is x0 =
[10
]. What happens after n tests?
x0 = 35k1 + 1
5k2
Pnx0 = 35P
nk1 + 15P
nk2 = 35
(16
)nk1 + 1
5k2
limn→∞
Pnx0 =1
5k2 =
[2/53/5
]
Complex Eigenvalues
A =
[0 1−1 0
]x =
[23
]Calculate A90x.
λ1 = i , k1 =
[−i1
]λ2 = −i , k2 =
[i1
]
x = (1.5 + i)k1 + (1.5− i)k2
A90x = (1.5 + i)A90k1 + (1.5− i)A90k2
= (1.5 + i)(i)90k1 + (1.5− i)(−i)90k2
= (1.5 + i)(−1)k1 + (1.5− i)(−1)k2 =
[−2−3
]
Complex Eigenvalues
A =
[0 1−1 0
]x =
[23
]Calculate A90x.
λ1 = i , k1 =
[−i1
]λ2 = −i , k2 =
[i1
]
x = (1.5 + i)k1 + (1.5− i)k2
A90x = (1.5 + i)A90k1 + (1.5− i)A90k2
= (1.5 + i)(i)90k1 + (1.5− i)(−i)90k2
= (1.5 + i)(−1)k1 + (1.5− i)(−1)k2 =
[−2−3
]
Complex Eigenvalues
A =
[0 1−1 0
]x =
[23
]Calculate A90x.
λ1 = i , k1 =
[−i1
]λ2 = −i , k2 =
[i1
]
x = (1.5 + i)k1 + (1.5− i)k2
A90x = (1.5 + i)A90k1 + (1.5− i)A90k2
= (1.5 + i)(i)90k1 + (1.5− i)(−i)90k2
= (1.5 + i)(−1)k1 + (1.5− i)(−1)k2 =
[−2−3
]
Complex Eigenvalues
A =
[0 1−1 0
]x =
[23
]Calculate A90x.
λ1 = i , k1 =
[−i1
]λ2 = −i , k2 =
[i1
]
x = (1.5 + i)k1 + (1.5− i)k2
A90x = (1.5 + i)A90k1 + (1.5− i)A90k2
= (1.5 + i)(i)90k1 + (1.5− i)(−i)90k2
= (1.5 + i)(−1)k1 + (1.5− i)(−1)k2 =
[−2−3
]
Complex Eigenvalues: A Neat Trick
A =
[0 1−1 0
]x =
[23
]
λ1 = i , k1 =
[−i1
]λ2 = −i , k2 =
[i1
]
The eigenvalues and eigenvectors are complex conjugates of one another.
Ax = λx ⇔ Ax = λx ⇔ Ax = λx ⇔ Ax = λx
Complex Eigenvalues: A Neat Trick
A =
[0 1−1 0
]x =
[23
]
λ1 = i , k1 =
[−i1
]λ2 = −i , k2 =
[i1
]
The eigenvalues and eigenvectors are complex conjugates of one another.
Ax = λx ⇔ Ax = λx ⇔ Ax = λx ⇔ Ax = λx
Complex Eigenvalues: A Neat Trick
A =
[0 1−1 0
]x =
[23
]
λ1 = i , k1 =
[−i1
]λ2 = −i , k2 =
[i1
]
The eigenvalues and eigenvectors are complex conjugates of one another.
Ax = λx ⇔ Ax = λx ⇔ Ax = λx ⇔ Ax = λx
Complex Eigenvalues: A Neat Trick
Find all eigenvalues and eigenvectors of :
A =
[−3 5−2 −1
]
det
([−3 5−2 −1
]−[λ 00 λ
])= det
[−3− λ 5−2 −1− λ
]= λ2 + 4λ+ 13
Roots:−4±
√16− 4(13)
2= −2± 3i
λ1 = −2− 3i , x1 =
[1 + 3i
2
]λ2 = −2 + 3i , x2 =
[1− 3i
2
]
Complex Eigenvalues: A Neat Trick
Find all eigenvalues and eigenvectors of :
A =
[−3 5−2 −1
]
det
([−3 5−2 −1
]−[λ 00 λ
])= det
[−3− λ 5−2 −1− λ
]= λ2 + 4λ+ 13
Roots:−4±
√16− 4(13)
2= −2± 3i
λ1 = −2− 3i , x1 =
[1 + 3i
2
]λ2 = −2 + 3i , x2 =
[1− 3i
2
]
Complex Eigenvalues: A Neat Trick
Find all eigenvalues and eigenvectors of :
A =
[−3 5−2 −1
]
det
([−3 5−2 −1
]−[λ 00 λ
])= det
[−3− λ 5−2 −1− λ
]= λ2 + 4λ+ 13
Roots:−4±
√16− 4(13)
2= −2± 3i
λ1 = −2− 3i ,
x1 =
[1 + 3i
2
]λ2 = −2 + 3i , x2 =
[1− 3i
2
]
Complex Eigenvalues: A Neat Trick
Find all eigenvalues and eigenvectors of :
A =
[−3 5−2 −1
]
det
([−3 5−2 −1
]−[λ 00 λ
])= det
[−3− λ 5−2 −1− λ
]= λ2 + 4λ+ 13
Roots:−4±
√16− 4(13)
2= −2± 3i
λ1 = −2− 3i , x1 =
[1 + 3i
2
]
λ2 = −2 + 3i , x2 =
[1− 3i
2
]
Complex Eigenvalues: A Neat Trick
Find all eigenvalues and eigenvectors of :
A =
[−3 5−2 −1
]
det
([−3 5−2 −1
]−[λ 00 λ
])= det
[−3− λ 5−2 −1− λ
]= λ2 + 4λ+ 13
Roots:−4±
√16− 4(13)
2= −2± 3i
λ1 = −2− 3i , x1 =
[1 + 3i
2
]λ2 = −2 + 3i ,
x2 =
[1− 3i
2
]
Complex Eigenvalues: A Neat Trick
Find all eigenvalues and eigenvectors of :
A =
[−3 5−2 −1
]
det
([−3 5−2 −1
]−[λ 00 λ
])= det
[−3− λ 5−2 −1− λ
]= λ2 + 4λ+ 13
Roots:−4±
√16− 4(13)
2= −2± 3i
λ1 = −2− 3i , x1 =
[1 + 3i
2
]λ2 = −2 + 3i , x2 =
[1− 3i
2
]
