Eigenspaces
Change of Basis
CS 130: Mathematical Methods in Computer Science Vector Spaces and Linear Systems Nestine Hope S. Hernandez Algorithms and Complexity Laboratory Department of Computer Science University of the Philippines, Diliman nshernandez@dcs.upd.edu.ph Day 6
Eigenspaces
Change of Basis
Vector Spaces and Linear Systems
Eigenspaces Eigenvalues and Eigenvectors Computing for the Eigenvalues and Eigenvectors Diagonalization
Change of Basis Coordinate Vector with respect to a Basis Transition Matrices
Eigenspaces
Change of Basis
Vector Spaces and Linear Systems
Eigenspaces Eigenvalues and Eigenvectors Computing for the Eigenvalues and Eigenvectors Diagonalization
Change of Basis Coordinate Vector with respect to a Basis Transition Matrices
Eigenspaces
Change of Basis
Given A =
1 −2
1 4
and b =
1 1
.
Eigenspaces
Change of Basis
Given A =
1 −2
1 4
1 −2
We have Ab =
and b = 1 4
1 1
1 1
=
.
Eigenspaces
Change of Basis
Given A =
1 −2
1 4
1 −2
We have Ab =
and b = 1 4
1 1
1 1
.
=
2 2
=
Eigenspaces
Change of Basis
Given A =
1 −2
1 4
1 −2
We have Ab =
and b = 1 4
1 1
1 1
.
=
2 2
=2
1 1
=
Eigenspaces
Change of Basis
Given A =
1 −2
1 4
1 −2
We have Ab =
and b = 1 4
1 1
1 1
.
=
2 2
=2
1 1
= 2b.
Eigenspaces
Change of Basis
Given A =
1 −2
1 4
1 −2
We have Ab =
Now consider A
3 2
and b = 1 4
=
1 1
1 −2
1 1
.
=
1 4
2 2
3 2
=2
=
1 1
= 2b.
Eigenspaces
Change of Basis
Given A =
1 −2
1 4
1 −2
We have Ab =
Now consider A
3 2
and b = 1 4
=
1 1
1 −2
1 1
.
=
1 4
2 2
3 2
=2
=
1 1
5 2
.
= 2b.
Eigenspaces
Change of Basis
Given A =
1 −2
1 4
1 −2
We have Ab =
Now consider A Also consider A
and b = 1 4
3 2
1 2
= =
1 1
1 1
.
=
1 −2
1 4
1 −2
1 4
2 2
=2
3 2
1 2
=
=
1 1
5 2
.
= 2b.
Eigenspaces
Change of Basis
Given A =
1 −2
1 4
1 −2
We have Ab =
Now consider A Also consider A
and b = 1 4
3 2
1 2
= =
1 1
1 1
.
=
1 −2
1 4
1 −2
1 4
2 2
1 1
5 2
.
3 6
=2
3 2
1 2
=
=
= 2b.
=
Eigenspaces
Change of Basis
Given A =
1 −2
1 4
1 −2
We have Ab =
Now consider A Also consider A
and b = 1 4
3 2
1 2
= =
1 1
1 1
.
=
1 −2
1 4
1 −2
1 4
2 2
1 1
5 2
.
3 6
=2
3 2
1 2
=
=
= 2b.
=3
1 2
.
Eigenspaces
Change of Basis
Eigenvalues and Eigenvectors
Let A be an n × n matrix. The real number λ is called an eigenvalue of A if there exists a nonzero n-vector ~x such that A~x = λ~x.
Eigenspaces
Change of Basis
Eigenvalues and Eigenvectors
Let A be an n × n matrix. The real number λ is called an eigenvalue of A if there exists a nonzero n-vector ~x such that A~x = λ~x. Every nonzero n-vector ~x satisfying the above equation is called an eigenvector of A associated with the eigenvalue λ.
Eigenspaces
Change of Basis
Eigenvalues and Eigenvectors
Let A be an n × n matrix. The real number λ is called an eigenvalue of A if there exists a nonzero n-vector ~x such that A~x = λ~x. Every nonzero n-vector ~x satisfying the above equation is called an eigenvector of A associated with the eigenvalue λ. Eigenvalues are also called proper values, characteristic values, and latent values.
Eigenspaces
Change of Basis
Eigenvalues and Eigenvectors
Let A be an n × n matrix. The real number λ is called an eigenvalue of A if there exists a nonzero n-vector ~x such that A~x = λ~x. Every nonzero n-vector ~x satisfying the above equation is called an eigenvector of A associated with the eigenvalue λ. Eigenvalues are also called proper values, characteristic values, and latent values.
Example 1. The only eigenvalue of In is λ = 1 and every nonzero vector in Rn is an eigenvector of In associated with the eigenvalue λ = 1.
Eigenspaces
Change of Basis
Eigenvalues and Eigenvectors
Let A be an n × n matrix. The real number λ is called an eigenvalue of A if there exists a nonzero n-vector ~x such that A~x = λ~x. Every nonzero n-vector ~x satisfying the above equation is called an eigenvector of A associated with the eigenvalue λ. Eigenvalues are also called proper values, characteristic values, and latent values.
Example 1. The only eigenvalue of In is λ = 1 and every nonzero vector in Rn is an eigenvector of In associated with the eigenvalue λ = 1. » – 0 12 2. Let A = . 1 0 2
Eigenspaces
Change of Basis
Eigenvalues and Eigenvectors
Let A be an n × n matrix. The real number λ is called an eigenvalue of A if there exists a nonzero n-vector ~x such that A~x = λ~x. Every nonzero n-vector ~x satisfying the above equation is called an eigenvector of A associated with the eigenvalue λ. Eigenvalues are also called proper values, characteristic values, and latent values.
Example 1. The only eigenvalue of In is λ = 1 and every nonzero vector in Rn is an eigenvector of In associated with the eigenvalue λ = 1. » – 0 12 2. Let A = . 1 0 » 2– » – 1 1 Note, A = 21 . 1 1
Eigenspaces
Change of Basis
Eigenvalues and Eigenvectors
Let A be an n × n matrix. The real number λ is called an eigenvalue of A if there exists a nonzero n-vector ~x such that A~x = λ~x. Every nonzero n-vector ~x satisfying the above equation is called an eigenvector of A associated with the eigenvalue λ. Eigenvalues are also called proper values, characteristic values, and latent values.
Example 1. The only eigenvalue of In is λ = 1 and every nonzero vector in Rn is an eigenvector of In associated with the eigenvalue λ = 1. » – 0 12 2. Let A = . 1 0 » 2– » – » – » – 1 1 1 1 Note, A = 21 . Also, A = − 12 . 1 1 −1 −1
Eigenspaces
Change of Basis
Vector Spaces and Linear Systems
Eigenspaces Eigenvalues and Eigenvectors Computing for the Eigenvalues and Eigenvectors Diagonalization
Change of Basis Coordinate Vector with respect to a Basis Transition Matrices
Eigenspaces
Change of Basis
Characteristic Polynomial of a Matrix
Let A = [aij ] be an n × n matrix. The determinant
λ − a11 −a12 ···
−a21 λ − a22 · · ·
p(λ) = det(λIn − A) =
: :
−an1 −an2 ··· is called the characteristic polynomial of A.
a1n −a2n : λ − ann
Eigenspaces
Change of Basis
Characteristic Polynomial of a Matrix
Let A = [aij ] be an n × n matrix. The determinant
λ − a11 −a12 ···
−a21 λ − a22 · · ·
p(λ) = det(λIn − A) =
: :
−an1 −an2 ··· is called the characteristic polynomial of A. The equation p(λ) = det(λIn − A) = 0 is called the characteristic equation of A.
a1n −a2n : λ − ann
Eigenspaces
Change of Basis
Computing for the Eigenvalues and Eigenvectors
The procedure for finding the eigenvalues and associated eigenvectors of a matrix is as follows.
Eigenspaces
Change of Basis
Computing for the Eigenvalues and Eigenvectors
The procedure for finding the eigenvalues and associated eigenvectors of a matrix is as follows. Step 1. Determine the real roots of the characteristic polynomial p(λ) = det(λIn − A). These are the eigenvalues of A.
Eigenspaces
Change of Basis
Computing for the Eigenvalues and Eigenvectors
The procedure for finding the eigenvalues and associated eigenvectors of a matrix is as follows. Step 1. Determine the real roots of the characteristic polynomial p(λ) = det(λIn − A). These are the eigenvalues of A. Step 2. For each eigenvalue λ, find all the nontrivial solutions to the system (λIn − A)~x = ~0. These are the eigenvectors of A associated with the eigenvalue λ.
Eigenspaces
Change of Basis
Computing for the Eigenvalues and Eigenvectors
The procedure for finding the eigenvalues and associated eigenvectors of a matrix is as follows. Step 1. Determine the real roots of the characteristic polynomial p(λ) = det(λIn − A). These are the eigenvalues of A. Step 2. For each eigenvalue λ, find all the nontrivial solutions to the system (λIn − A)~x = ~0. These are the eigenvectors of A associated with the eigenvalue λ.
Example A=
0 1 2
1 2
0
Eigenspaces
Change of Basis
List of Nonsingular Equivalences Suppose A is an n Ă— n matrix. Then the following statements are equivalent: 1. A is nonsingular. 2. A is row equivalent to In . 3. Ax = 0 has only the trivial solution. 4. The linear system Ax = b has a unique solution for every n Ă— 1 matrix b. 5. det(A) 6= 0. 6. A has rank n. 7. A has nullity 0. 8. The rows of A form a linearly independent set of n vectors in lRn . 9. The columns of A form a linearly independent set of n vectors in lRn . 10. Zero is NOT an eigenvalue of A.
Eigenspaces
Change of Basis
More Examples Find the characteristic polynomial, eigenvalues and eigenvectors of the following matrices: 2 3 1 0 0 1 0 5 1. 4 0 0 −1 1 2 3 1 −1 −2 1 0 5 2. 4 0 0 1 3 2 3 5 1 1 5 −1 5 3. 4 1 1 −1 5 2 3 1 0 2 −1 6 0 1 4 −2 7 7 4. 6 4 2 −1 0 1 5 2 −1 −1 2
Eigenspaces
Change of Basis
More Examples Find the characteristic polynomial, eigenvalues and eigenvectors of the following matrices: 2 3 1 0 0 1 0 5 1. 4 0 0 −1 1 2 3 1 −1 −2 1 0 5 2. 4 0 0 1 3 2 3 5 1 1 5 −1 5 3. 4 1 1 −1 5 2 3 1 0 2 −1 6 0 1 4 −2 7 7 4. 6 4 2 −1 0 1 5 2 −1 −1 2 The set S consisting of all eigenvectors of A associated with λ as well as the zero vector is called the eigenspace associated with λ.
Eigenspaces
Change of Basis
Vector Spaces and Linear Systems
Eigenspaces Eigenvalues and Eigenvectors Computing for the Eigenvalues and Eigenvectors Diagonalization
Change of Basis Coordinate Vector with respect to a Basis Transition Matrices
Eigenspaces
Change of Basis
Example Given A =
1 1 . −2 4
Eigenspaces
Change of Basis
Example Given A = Let P =
1 1
1 1 . −2 4 1 2
.
Eigenspaces
Change of Basis
Example Given A = Let P =
1 1
1 1 . −2 4 1 2
. Then P −1 =
2 −1
−1 1
.
Eigenspaces
Change of Basis
Example Given A = Let P =
1 1
1 1 . −2 4 1 2
. Then P −1 =
Compute for B = P −1 AP .
2 −1
−1 1
.
Eigenspaces
Change of Basis
Example Given A = Let P =
1 1
1 1 . −2 4 1 2
. Then P −1 =
Compute for B = P −1 AP . 2 −1 1 We have B = −1 1 −2
2 −1
1 4
−1 1
.
1 1
1 2
=
2 0
0 3
.
Eigenspaces
Change of Basis
Example Given A = Let P =
1 1
1 1 . −2 4 1 2
. Then P −1 =
Compute for B = P −1 AP . 2 −1 1 We have B = −1 1 −2 We say that B is similar to A.
2 −1
1 4
−1 1
.
1 1
1 2
=
2 0
0 3
.
Eigenspaces
Change of Basis
Example Given A = Let P =
1 1
1 1 . −2 4 1 2
. Then P −1 =
Compute for B = P −1 AP . 2 −1 1 We have B = −1 1 −2
2 −1
1 4
−1 1
.
1 1
1 2
=
2 0
0 3
.
We say that B is similar to A. A matrix B is said to be similar to a matrix A if there is a nonsingular matrix P such that B = P −1 AP .
Eigenspaces
Change of Basis
Diagonalizable Matrix A matrix A is diagonalizable if it is similar to a diagonal matrix. In this case we also say that A can be diagonalized.
Eigenspaces
Change of Basis
Diagonalizable Matrix A matrix A is diagonalizable if it is similar to a diagonal matrix. In this case we also say that A can be diagonalized.
Theorem An n Ă— n matrix A is diagonalizable if and only if it has n linearly independent eigenvectors.
Eigenspaces
Change of Basis
Diagonalizable Matrix A matrix A is diagonalizable if it is similar to a diagonal matrix. In this case we also say that A can be diagonalized.
Theorem An n Ă— n matrix A is diagonalizable if and only if it has n linearly independent eigenvectors. In this case A is similar to a diagonal matrix D, with P −1 AP = D, whose diagonal elements are the eigenvalues of A, while P is a matrix whose columns are respectively the n linearly independent eigenvectors of A.
Eigenspaces
Change of Basis
Vector Spaces and Linear Systems
Eigenspaces Eigenvalues and Eigenvectors Computing for the Eigenvalues and Eigenvectors Diagonalization
Change of Basis Coordinate Vector with respect to a Basis Transition Matrices
Eigenspaces
Change of Basis
If S = {v1 , v2 , · · · , vn } is an ordered basis for the n-dimensional vector space V , then every vector in V can be uniquely expressed in the form v = c1 v1 + c2 v2 + · · · + cn vn , where c1 , c2 , · · · , cn are real numbers.
Eigenspaces
Change of Basis
If S = {v1 , v2 , · · · , vn } is an ordered basis for the n-dimensional vector space V , then every vector in V can be uniquely expressed in the form v = c1 v1 + c2 v2 + · · · + cn vn , where c1 , c2 , · · · , cn are real numbers. We shall refer to
c1 c2 [v]S = : cn
as the coordinate vector of v with respect to the ordered basis of S. The entries of [v]S are called the coordinates of v with respect to S.
Eigenspaces
Change of Basis
Example Let S = {v1 , v2 , v3 , v4 } be a basis for lR4 , where v1 = (1, 1, 0, 0), v2 = (2, 0, 1, 0), v3 = (0, 1, 2, −1), v4 = (0, 1, −1, 0).
Eigenspaces
Change of Basis
Example Let S = {v1 , v2 , v3 , v4 } be a basis for lR4 , where v1 = (1, 1, 0, 0), v2 = (2, 0, 1, 0), v3 = (0, 1, 2, −1), v4 = (0, 1, −1, 0). If v = (1, 2, −6, 2), compute for [v]S .
Eigenspaces
Change of Basis
Example Let S = {v1 , v2 , v3 , v4 } be a basis for lR4 , where v1 = (1, 1, 0, 0), v2 = (2, 0, 1, 0), v3 = (0, 1, 2, −1), v4 = (0, 1, −1, 0). If v = (1, 2, −6, 2), compute for [v]S . We need to compute constants c1 , c2 , c3 , c4 s.t. v = c1 v1 + c2 v2 + c3 v3 + c4 v4 .
Eigenspaces
Change of Basis
Example Let S = {v1 , v2 , v3 , v4 } be a basis for lR4 , where v1 = (1, 1, 0, 0), v2 = (2, 0, 1, 0), v3 = (0, 1, 2, −1), v4 = (0, 1, −1, 0). If v = (1, 2, −6, 2), compute for [v]S . We need to compute constants c1 , c2 , c3 , c4 s.t. v = c1 v1 + c2 v2 + c3 v3 + c4 v4 . This leads us to the linear system whose augmented matrix is 3 2 1 2 0 0 1 6 1 0 1 1 2 7 7 6 4 0 1 2 −1 −6 5 0 0 −1 0 2 or equivalently, ˆ
v1T
v2T
v3T
v4T
vT
˜
Eigenspaces
Change of Basis
Example Let S = {v1 , v2 , v3 , v4 } be a basis for lR4 , where v1 = (1, 1, 0, 0), v2 = (2, 0, 1, 0), v3 = (0, 1, 2, −1), v4 = (0, 1, −1, 0). If v = (1, 2, −6, 2), compute for [v]S . We need to compute constants c1 , c2 , c3 , c4 s.t. v = c1 v1 + c2 v2 + c3 v3 + c4 v4 . This leads us to the linear system whose augmented matrix is 3 2 1 2 0 0 1 6 1 0 1 1 2 7 7 6 4 0 1 2 −1 −6 5 0 0 −1 0 2 or equivalently, v1T v2T v3T v4T We obtain c1 = 3, c2 = −1, c3 = −2, c4 = 1 ˆ
vT
˜
Eigenspaces
Change of Basis
Example Let S = {v1 , v2 , v3 , v4 } be a basis for lR4 , where v1 = (1, 1, 0, 0), v2 = (2, 0, 1, 0), v3 = (0, 1, 2, −1), v4 = (0, 1, −1, 0). If v = (1, 2, −6, 2), compute for [v]S . We need to compute constants c1 , c2 , c3 , c4 s.t. v = c1 v1 + c2 v2 + c3 v3 + c4 v4 . This leads us to the linear system whose augmented matrix is 3 2 1 2 0 0 1 6 1 0 1 1 2 7 7 6 4 0 1 2 −1 −6 5 0 0 −1 0 2 or equivalently, ˜ v1T v2T v3T v4T v T We obtain c1 = 3, c2 = −1, c3 = −2, c4 = 1 so the coordinate vector of v wrt the basis S is 2 3 3 6 −1 7 7 [v]S = 6 4 −2 5 1 ˆ
Eigenspaces
Change of Basis
Example Now suppose we use T = {w1 , w2 , w3 , w4 } as a basis for lR4 , where w1 = (1, 0, 1, 2), w2 = (0, 1, 1, 2), w3 = (1, 1, 1, 3), w4 = (1, 0, −1, −1).
Eigenspaces
Change of Basis
Example Now suppose we use T = {w1 , w2 , w3 , w4 } as a basis for lR4 , where w1 = (1, 0, 1, 2), w2 = (0, 1, 1, 2), w3 = (1, 1, 1, 3), w4 = (1, 0, −1, −1). Again, let v = (1, 2, −6, 2). Compute for [v]T .
Eigenspaces
Change of Basis
Example Now suppose we use T = {w1 , w2 , w3 , w4 } as a basis for lR4 , where w1 = (1, 0, 1, 2), w2 = (0, 1, 1, 2), w3 = (1, 1, 1, 3), w4 = (1, 0, −1, −1). Again, let v = (1, 2, −6, 2). Compute for [v]T . We have ˆ
w1T w2T 2 1 0 6 0 1 6 4 1 1 2 2
w3T 1 1 1 3
w4T 1 0 −1 −1
vT
˜ 3
1 2 7 7 −6 5 2
=
Eigenspaces
Change of Basis
Example Now suppose we use T = {w1 , w2 , w3 , w4 } as a basis for lR4 , where w1 = (1, 0, 1, 2), w2 = (0, 1, 1, 2), w3 = (1, 1, 1, 3), w4 = (1, 0, −1, −1). Again, let v = (1, 2, −6, 2). Compute for [v]T . We have ˆ
w1T w2T 2 1 0 6 0 1 6 4 1 1 2 2
w3T 1 1 1 3
w4T 1 0 −1 −1
vT
˜ 3
1 2 7 7 −6 5 2
The coordinate vector of v wrt the basis T is 2 3 −13 6 −17 7 7 [v]T = 6 4 19 5 −5
=
Eigenspaces
Change of Basis
Example Now suppose we use U = {e1 , e2 , e3 , e4 } as a basis for lR4 , where e1 = (1, 0, 0, 0), e2 = (0, 1, 0, 0), e3 = (0, 0, 1, 0), e4 = (0, 0, 0, 1).
Eigenspaces
Change of Basis
Example Now suppose we use U = {e1 , e2 , e3 , e4 } as a basis for lR4 , where e1 = (1, 0, 0, 0), e2 = (0, 1, 0, 0), e3 = (0, 0, 1, 0), e4 = (0, 0, 0, 1). Again, let v = (1, 2, −6, 2). Compute for [v]U .
Eigenspaces
Change of Basis
Example Now suppose we use U = {e1 , e2 , e3 , e4 } as a basis for lR4 , where e1 = (1, 0, 0, 0), e2 = (0, 1, 0, 0), e3 = (0, 0, 1, 0), e4 = (0, 0, 0, 1). Again, let v = (1, 2, −6, 2). Compute for [v]U . We have eT1 2 1 6 0 6 4 0 0
ˆ
eT2
eT3
eT4
0 1 0 0
0 0 1 0
0 0 0 1
vT
˜
= 3
1 2 7 7 −6 5 2
Eigenspaces
Change of Basis
Example Now suppose we use U = {e1 , e2 , e3 , e4 } as a basis for lR4 , where e1 = (1, 0, 0, 0), e2 = (0, 1, 0, 0), e3 = (0, 0, 1, 0), e4 = (0, 0, 0, 1). Again, let v = (1, 2, −6, 2). Compute for [v]U . We have eT1 2 1 6 0 6 4 0 0
ˆ
eT2
eT3
eT4
0 1 0 0
0 0 1 0
0 0 0 1
vT
The coordinate vector of v wrt the basis U is 2 3 1 6 2 7 7 [v]U = 6 4 −6 5 2
˜
= 3
1 2 7 7 −6 5 2
Eigenspaces
Change of Basis
Vector Spaces and Linear Systems
Eigenspaces Eigenvalues and Eigenvectors Computing for the Eigenvalues and Eigenvectors Diagonalization
Change of Basis Coordinate Vector with respect to a Basis Transition Matrices
Eigenspaces
Change of Basis
Suppose S = {v1 , v2 , · · · , vn } and T = {w1 , w2 , · · · , wn } are bases for the n-dimensional vector space V .
Eigenspaces
Change of Basis
Suppose S = {v1 , v2 , · · · , vn } and T = {w1 , w2 , · · · , wn } are bases for the n-dimensional vector space V . If v is any vector in V , then v = c1 w1 + c2 w2 + · · · + cn wn so that
Eigenspaces
Change of Basis
Suppose S = {v1 , v2 , · · · , vn } and T = {w1 , w2 , · · · , wn } are bases for the n-dimensional vector space V . If v is any vector in V , then v = c1 w1 + c2 w2 + · · · + cn wn so that c1 c2 [v]T = : . cn
Eigenspaces
Change of Basis
Suppose S = {v1 , v2 , · · · , vn } and T = {w1 , w2 , · · · , wn } are bases for the n-dimensional vector space V . If v is any vector in V , then v = c1 w1 + c2 w2 + · · · + cn wn so that c1 c2 [v]T = : . cn Then [v]S
=
[c1 w1 + c2 w2 + · · · + cn wn ]S
Eigenspaces
Change of Basis
Suppose S = {v1 , v2 , · · · , vn } and T = {w1 , w2 , · · · , wn } are bases for the n-dimensional vector space V . If v is any vector in V , then v = c1 w1 + c2 w2 + · · · + cn wn so that c1 c2 [v]T = : . cn Then [v]S
= [c1 w1 + c2 w2 + · · · + cn wn ]S = [c1 w1 ]S + [c2 w2 ]S + · · · + [cn wn ]S
Eigenspaces
Change of Basis
Suppose S = {v1 , v2 , · · · , vn } and T = {w1 , w2 , · · · , wn } are bases for the n-dimensional vector space V . If v is any vector in V , then v = c1 w1 + c2 w2 + · · · + cn wn so that c1 c2 [v]T = : . cn Then [v]S
= [c1 w1 + c2 w2 + · · · + cn wn ]S = [c1 w1 ]S + [c2 w2 ]S + · · · + [cn wn ]S = c1 [w1 ]S + c2 [w2 ]S + · · · + cn [wn ]S
Eigenspaces
Change of Basis
Suppose S = {v1 , v2 , · · · , vn } and T = {w1 , w2 , · · · , wn } are bases for the n-dimensional vector space V . If v is any vector in V , then v = c1 w1 + c2 w2 + · · · + cn wn so that c1 c2 [v]T = : . cn Then [v]S
= [c1 w1 + c2 w2 + · · · + cn wn ]S = [c1 w1 ]S + [c2 w2 ]S + · · · + [cn wn ]S = c1 [w1 ]S + c2 [w2 ]S + · · · + cn [wn ]S " # c1 c2 = [w1 ]S [w2 ]S · · · [wn ]S : cn
Eigenspaces
Change of Basis
Suppose S = {v1 , v2 , · · · , vn } and T = {w1 , w2 , · · · , wn } are bases for the n-dimensional vector space V . If v is any vector in V , then v = c1 w1 + c2 w2 + · · · + cn wn so that c1 c2 [v]T = : . cn Then [v]S
PS←T = [w1 ]S
= [c1 w1 + c2 w2 + · · · + cn wn ]S = [c1 w1 ]S + [c2 w2 ]S + · · · + [cn wn ]S = c1 [w1 ]S + c2 [w2 ]S + · · · + cn [wn ]S " # c1 c2 = [w1 ]S [w2 ]S · · · [wn ]S : cn [w2 ]S · · · [wn ]S is called the transition
matrix from the T -basis to the S-basis.
Eigenspaces
Change of Basis
Example Let S = {v1 , v2 , v3 , v4 } and T = {w1 , w2 , w3 , w4 } be bases for lR4 , where v1 = (1, 1, 0, 0), v2 = (2, 0, 1, 0), v3 = (0, 1, 2, −1), v4 = (0, 1, −1, 0) w1 = (1, 0, 1, 2), w2 = (0, 1, 1, 2), w3 = (1, 1, 1, 3), w4 = (1, 0, −1, −1).
Eigenspaces
Change of Basis
Example Let S = {v1 , v2 , v3 , v4 } and T = {w1 , w2 , w3 , w4 } be bases for lR4 , where v1 = (1, 1, 0, 0), v2 = (2, 0, 1, 0), v3 = (0, 1, 2, −1), v4 = (0, 1, −1, 0) w1 = (1, 0, 1, 2), w2 = (0, 1, 1, 2), w3 = (1, 1, 1, 3), w4 = (1, 0, −1, −1). Compute the transition matrix PS←T from the T -basis to the S-basis.
Eigenspaces
Change of Basis
Example Let S = {v1 , v2 , v3 , v4 } and T = {w1 , w2 , w3 , w4 } be bases for lR4 , where v1 = (1, 1, 0, 0), v2 = (2, 0, 1, 0), v3 = (0, 1, 2, −1), v4 = (0, 1, −1, 0) w1 = (1, 0, 1, 2), w2 = (0, 1, 1, 2), w3 = (1, 1, 1, 3), w4 = (1, 0, −1, −1). Compute the transition matrix PS←T from the T -basis to the S-basis. To compute PS←T ,
Eigenspaces
Change of Basis
Example Let S = {v1 , v2 , v3 , v4 } and T = {w1 , w2 , w3 , w4 } be bases for lR4 , where v1 = (1, 1, 0, 0), v2 = (2, 0, 1, 0), v3 = (0, 1, 2, −1), v4 = (0, 1, −1, 0) w1 = (1, 0, 1, 2), w2 = (0, 1, 1, 2), w3 = (1, 1, 1, 3), w4 = (1, 0, −1, −1). Compute the transition matrix PS←T from the T -basis to the S-basis. To compute PS←T , we need to find a1 , a2 , a3 , a4 such that a1 v1 + a2 v2 + a3 v3 + a4 v4 = w1 .
Eigenspaces
Change of Basis
Example Let S = {v1 , v2 , v3 , v4 } and T = {w1 , w2 , w3 , w4 } be bases for lR4 , where v1 = (1, 1, 0, 0), v2 = (2, 0, 1, 0), v3 = (0, 1, 2, −1), v4 = (0, 1, −1, 0) w1 = (1, 0, 1, 2), w2 = (0, 1, 1, 2), w3 = (1, 1, 1, 3), w4 = (1, 0, −1, −1). Compute the transition matrix PS←T from the T -basis to the S-basis. To compute PS←T , we need to find a1 , a2 , a3 , a4 such that a1 v1 + a2 v2 + a3 v3 + a4 v4 = w1 . Similarly, we need b1 , b2 , b3 , b4 such that b1 v1 + b2 v2 + b3 v3 + b4 v4 = w2 ,
Eigenspaces
Change of Basis
Example Let S = {v1 , v2 , v3 , v4 } and T = {w1 , w2 , w3 , w4 } be bases for lR4 , where v1 = (1, 1, 0, 0), v2 = (2, 0, 1, 0), v3 = (0, 1, 2, −1), v4 = (0, 1, −1, 0) w1 = (1, 0, 1, 2), w2 = (0, 1, 1, 2), w3 = (1, 1, 1, 3), w4 = (1, 0, −1, −1). Compute the transition matrix PS←T from the T -basis to the S-basis. To compute PS←T , we need to find a1 , a2 , a3 , a4 such that a1 v1 + a2 v2 + a3 v3 + a4 v4 = w1 . Similarly, we need b1 , b2 , b3 , b4 such that b1 v1 + b2 v2 + b3 v3 + b4 v4 = w2 , c1 , c2 , c3 , c4 where c1 v1 + c2 v2 + c3 v3 + c4 v4 = w3 ,
Eigenspaces
Change of Basis
Example Let S = {v1 , v2 , v3 , v4 } and T = {w1 , w2 , w3 , w4 } be bases for lR4 , where v1 = (1, 1, 0, 0), v2 = (2, 0, 1, 0), v3 = (0, 1, 2, −1), v4 = (0, 1, −1, 0) w1 = (1, 0, 1, 2), w2 = (0, 1, 1, 2), w3 = (1, 1, 1, 3), w4 = (1, 0, −1, −1). Compute the transition matrix PS←T from the T -basis to the S-basis. To compute PS←T , we need to find a1 , a2 , a3 , a4 such that a1 v1 + a2 v2 + a3 v3 + a4 v4 = w1 . Similarly, we need b1 , b2 , b3 , b4 such that b1 v1 + b2 v2 + b3 v3 + b4 v4 = w2 , c1 , c2 , c3 , c4 where c1 v1 + c2 v2 + c3 v3 + c4 v4 = w3 , and d1 , d2 , d3 , d4 where d1 v1 + d2 v2 + d3 v3 + d4 v4 = w4 . Note that the coefficient matrix of all four linear systems is the same.
Eigenspaces
Change of Basis
Example Let S = {v1 , v2 , v3 , v4 } and T = {w1 , w2 , w3 , w4 } be bases for lR4 , where v1 = (1, 1, 0, 0), v2 = (2, 0, 1, 0), v3 = (0, 1, 2, −1), v4 = (0, 1, −1, 0) w1 = (1, 0, 1, 2), w2 = (0, 1, 1, 2), w3 = (1, 1, 1, 3), w4 = (1, 0, −1, −1). Compute the transition matrix PS←T from the T -basis to the S-basis. To compute PS←T , we need to find a1 , a2 , a3 , a4 such that a1 v1 + a2 v2 + a3 v3 + a4 v4 = w1 . Similarly, we need b1 , b2 , b3 , b4 such that b1 v1 + b2 v2 + b3 v3 + b4 v4 = w2 , c1 , c2 , c3 , c4 where c1 v1 + c2 v2 + c3 v3 + c4 v4 = w3 , and d1 , d2 , d3 , d4 where d1 v1 + d2 v2 + d3 v3 + d4 v4 = w4 . Note that the coefficient matrix of all four linear systems is the same. Hence, we compute for the RREF of – » T T v1 v2 v3T v4T : w1T : w2T : w3T : w4T .
Eigenspaces
Change of Basis
Example Let S = {v1 , v2 , v3 , v4 } and T = {w1 , w2 , w3 , w4 } be bases for lR4 , where v1 = (1, 1, 0, 0), v2 = (2, 0, 1, 0), v3 = (0, 1, 2, −1), v4 = (0, 1, −1, 0) w1 = (1, 0, 1, 2), w2 = (0, 1, 1, 2), w3 = (1, 1, 1, 3), w4 = (1, 0, −1, −1). Compute the transition matrix PS←T from the T -basis to the S-basis. To compute PS←T , we need to find a1 , a2 , a3 , a4 such that a1 v1 + a2 v2 + a3 v3 + a4 v4 = w1 . Similarly, we need b1 , b2 , b3 , b4 such that b1 v1 + b2 v2 + b3 v3 + b4 v4 = w2 , c1 , c2 , c3 , c4 where c1 v1 + c2 v2 + c3 v3 + c4 v4 = w3 , and d1 , d2 , d3 , d4 where d1 v1 + d2 v2 + d3 v3 + d4 v4 = w4 . Note that the coefficient matrix of all four linear systems is the same. Hence, we compute for the RREF of – » T T v1 v2 v3T v4T : w1T : w2T : w3T : w4T . 2 3 13 16 21 −9 6 −6 −8 −10 5 7 7. We have PS←T = 6 4 −2 −2 −3 1 5 −11 −13 −17 8
Eigenspaces
Change of Basis
Example From a previous exercise, if v = (1, 2, −6, 2), we obtained −13 −17 [v]T = 19 . −5
Eigenspaces
Change of Basis
Example From a previous exercise, if v = (1, 2, −6, 2), we obtained −13 −17 [v]T = 19 . −5 We can find [v]S by computing for PS←T [v]T .
Eigenspaces
Change of Basis
Example From a previous exercise, if v = (1, 2, −6, 2), we obtained −13 −17 [v]T = 19 . −5 We can find [v]S by computing for PS←T [v]T . 13 16 21 −9 −13 −6 −17 −8 −10 5 [v]S = PS←T [v]T = −2 −2 −3 1 19 −11 −13 −17 8 −5
3 −1 = −2 1
Eigenspaces
Change of Basis
Theorem Let S = {v1 , v2 , · · · , vn } and T = {w1 , w2 , · · · , wn } are bases for the n-dimensional vector space V . Let PS←T be the transition matrix from the T -basis to the S-basis. −1 Then PS←T is nonsingular and PS←T is the transition matrix from the S-basis to the T -basis. In our previous example, verify that 1 4 5 −2 3 5 10 −1 −1 PT ←S = PS←T = −2 −5 −9 2 2 3 4 0
Eigenspaces
Change of Basis
For Self-Study
Eigenspaces
Change of Basis
Properties of Coordinate Vectors If S is a basis for an n-dimensional vector space V , v and w are vectors in V , and k is a scalar, then (a) v + w S = v S + w S (b) kv S = k v S Moreover, the results in (a) and (b) can be generalized to show that for vectors vi in V and scalars ki (c) k1 v1 + k2 v2 + · · · + kn vn S = k1 v1 S + k2 v2 S + · · · + kn vn S
Eigenspaces
Change of Basis
Questions? See you next meeting!