Transcript Chapter 4
Eigenvalues and Eigenvectors
Computations
And
Geometry
Eigenvalues and Eigenvectors
Let A be a π × π matrix if there exists π β β a scalar,
and x a nonzero vector π± β βπ such that π΄π± = ππ±. We
call π an eigenvalue of the matrix A and x an eigenvector
of the matrix A.
π is a Eigenvalue for A
π± β π½ is a Eigenvector for A
π΄π± = ππ±
Numerical Interpretation of Eigenvalues
In terms of matrix arithmetic eigenvalues turn matrix multiplication into scalar multiplication.
Numerically knowing an eigenvalue tremendously lowers the number of operations required
to get the result of a matrix multiplication. This optimizes the performance of certain
algorithms that are used in the areas of computer graphics and engineering.
The question that arises in this chapter is how can we determine what the eigenvalues and
eigenvectors of a matrix are? What properties do eigenvalues of a matrix have?
Eigenvalues and Singular Matrices
Remember a square matrix A is singular if and only if there exists
a nonzero solution x to the matrix equation π΄π± = π½. In order to
apply this we subtract ππ± from both sides of the equation, but
we can not immediately factor out x since A is a matrix and π a
scalar the operation π΄ β π is not defined. Multiply x by the
identity matrix I, we get that π is a eigenvalue for A when π΄ β ππΌ
is singular, or the eigenvalues of A are the values π that make the
matrix π΄ β ππΌ singular.
π΄π± = ππ±
π΄π± β ππ± = π½
π΄π± β ππΌπ± = π½
π΄ β ππΌ π± = π½
If and only if:
π΄ β ππΌ ππ π ππππ’πππ
Eigenvectors and Null Spaces
If the matrix π΄ β ππΌ is singular then all nonzero vectors x
such that π΄ β ππΌ π± = π½ are the eigenvectors of A. All
vectors x such that π΄ β ππΌ π± = π½ is the null space of A. This
we call the eigenspace of the eigenvalue π.
π± ππ ππππππ£πππ‘ππ ππ π΄
Then,
π± β π© π΄ β ππΌ
Remember a 2 × 2 matrix ππ ππ is singular if and only if ππ β ππ = 0. This gives a method
for how to find the eigenvalues of 2 × 2 matrices.
Finding Eigenvalues of 2 × 2 Matrices
To find the eigenvalues of a 2 × 2 matrix π΄ =
π
π
π
π
we do the following:
1) Form the matrix π΄ β ππΌ this amounts to subtracting π
down the main diagonal of the matrix.
2) Solve the equation π β π π β π β ππ = 0 for π.
Treat π as the variable and either factor this or use
the quadratic equation if it can not be factored.
Finding Eigenvectors of 2 × 2 Matrices
After you have found each eigenvalue π, to find the
eigenvectors associated with an eigenvalue π find a basis
for π© π΄ β ππΌ since basis vectors are never zero.
1 0
π π
βπ
0 1
π π
π π
π 0
=
β
π π
0 π
πβπ
π
=
π
πβπ
Singular, if and only if,
π β π π β π β ππ = 0
π± ππ ππππππ£πππ‘ππ
x a basis vector for:
π© π΄ β ππΌ
Example
Find the eigenvalues and associated eigenvectors for the 2 × 2
matrix A given to the right.
π΄=
7
β5
10
β8
7βπ
10
, which is singular if 7 β π β8 β π β β5 10 = 0
β5 β8 β π
Solve this equation for π by first multiplying it out then
7 β π β8 β π β β5 10
trying to factor it or else use the quadratic equation if it
= π2 + π β 56 + 50
will not factor.
= π2 + π β 6
From this we see that π = β3 or π = 2.
= π+3 πβ2 =0
π΄ β ππΌ =
For the eigenvalue -3,
10 10
π΄ β β3 πΌ = π΄ + 3πΌ =
β5 β5
1 1
Row reducing, we get:
0 0
βπ₯2
β1
Vector form of solution: π₯ = π₯2
2
1
For the eigenvalue 2,
5
10
π΄ β 2πΌ =
β5 β10
1 2
Row reducing, we get:
0 0
β2π₯2
β2
Vector form of solution:
= π₯2
π₯2
1
Check,
7 10 β1
3
β1
π΄
=
=
β5 β8 1
β3
1
3
β1
β3
=
β3
1
β1
Eigenvalue -3 with eigenvector
.
1
Check,
7 10 β2
β4
β2
π΄
=
=
β5 β8 1
2
1
β4
β2
2
=
2
1
β4
Eigenvalue 2 with eigenvector
.
2
Geometric Interpretation of Real Eigenvectors
If π: βπ β βπ is a linear transformation given by
π π― = ππ―. If x is an eigenvector for M with eigenvalue
π then the result of applying T to x changes the length
of x (maybe points in opposite direction π < 0).
Example
Let π: β2 β β2 be a reflection through the y-axis and M the
matrix so that π π― = ππ―. Find the eigenvalues and
eigenvectors for M.
βπ₯1
β1 0 π₯1
From the picture we get ππ₯ = π₯ =
0 1 π₯2
2
β1 β π
0
π β ππΌ =
, singular if β1 β π 1 β π = 0
0
1βπ
0 0
β2 0
π β β1 πΌ = π + πΌ =
, πβπΌ =
,
0 2
0 0
0 1
1 0
Row reduces to:
Row reduces to:
0 0
0 0
π₯
1
0
0
Vector solution: 1 = π₯1
Vector
solution:
=
π₯
2
0
0
π₯2
1
1
0
Eigenvector for -1 is:
Eigenvector
for
1
is:
0
1
π π± = ππ± = ππ±
π±
π₯1
π₯2
y
βπ₯1
π₯2
x
Eigenvalues: π = β1, π = 1
π
π
0
1
1
0
=1
= β1
0
1
1
0
0
1
1
0
Vectors on the positive x-axis are sent to negative x-axis and visa versa. While vectors on the
y-axis remain fixed.
Example
Let π: β2 β β2 be a projection onto the vector π° = 11 (i.e. π π― = πππππ° π―). Find the matrix
M such that π π― = ππ― and compute its eigenvalues and eigenvectors.
π π1 = πππππ°
π π2 = πππππ°
1
=
0
0
=
1
1
2
1
2
1
2
1
2
π π π =
1
2
1
2
1
2
1
2
Factoring this we get π π β 1 = 0 this gives
eigenvalues π = 0 and π = 1.
π°=
β1
1
π
1
1
π π° =1
β1
1
=0
π β ππΌ =
1
2
1
2
Singular if, 12 β π
Or, π2 β π = 0
βπ
β π β 14 = 0
1
2
1
2
πβ 0 πΌ=π=
1
2
1
2
,
1 1
0 0
βπ₯2
β1
Vector solution: π₯ = π₯2
2
1
β1
Eigenvector for 0 is:
1
1
1
β1
2
1
2
1
2
β1
2
For a projection,
πβπΌ =
πππππ° π° = π° π π° π° = 1π°, 1 is an eigenvalue
Row reduces to:
π°ππ°
If v is orthogonal to w,
πππππ° π― =
1
2
1
2
Row reduces to:
β1
1
π―π π°
π°
π°ππ°
1
2
βπ
= 0π°, 0 is an eigenvalue
,
1 β1
0 0
π₯2
1
Vector solution: π₯ = π₯2
2
1
1
Eigenvector for 0 is:
1
Example
Let π: β2 β β2 be a counterclockwise rotation of π2 (i.e. 90°). Find the matrix M such that
π π― = ππ― and compute its eigenvalues and eigenvectors.
0
π π1 =
1
β1
π π2 =
0
π π π =
0 β1
1 0
π β ππΌ =
βπ
1
β1
,
βπ
Singular if,
π2 + 1 = 0
The equation π2 + 1 = 0 has no real solutions, so this matrix has no real eigenvalues. In fact
we can see geometrically that a 90° rotation do not fix or reverse the direction of any vector
in the plane. This matrix has no real eigenvalues.
The eigenvalues for this matrix are π = π and π = βπ which are known as imaginary
numbers. More advanced course in linear algebra give an interpretation to these values but
for right now we will say this matrix has no real eigenvalues.
Because the matrix has no eigenvalues it has no corresponding eigenvectors also.