Transcript Slide 1
Last lecture summary
Orthogonal matrices
• independent basis, orthogonal basis,
orthonormal vectors, normalization
• Put orthonormal vectors into a matrix
– Generally rectangular matrix – matrix with
orhonormal columns
– Square matrix with orthonormal colums –
orthogonal matrix Q
• Matrix A has non-orthogonal columns,
A → Q … Gram-Schmidt
Orthogonalization
• Q-1 = QT
• Having Q simplifies many formulas. e. g.
AT Ax AT b
QT Qxˆ QT b Ixˆ QT b xˆ QT b
• QR decomposition: A = QR
Determinants
a
b
c d
•
Properties:
–
–
–
•
ad bc
Exchange rows/cols of a matrix: reverse
sign of det
Equals to 0 if two rows/cols are same.
Det of triangular matrix is a product of
diagonal terms.
determinant is a test for invertibility
– matrix is invertible if |A| ≠ 0
– matrix is singular if |A| = 0
Eigenvalues and eigenvectors
Ax x
• Action of matrix A on vector x returns the
vector parallel (same direction) to vector x.
• x … eigenvectors, λ … eigenvalues
• spectrum, trace (sum of λ), determinant
(product of λ)
• λ = 0 … singular matrix
• Find eigendecomposition by solving
characteristic equation (leading to
characteristic polynomial)
det(A - λI) = 0
• Find λ first, then find eigenvectors by Gauss
elimination, as the solution of (A - λI)x = 0.
• algebraic multiplicity of an eigenvalue
– multiplicity of the corresponding root
• geometric multiplicity of an eigenvalue
– number of linearly independent eigenvectors
with that eigenvalue
– degenerate, deffective matrix
Diagonalization
• Square matrix A … diagonalizable if there
exists an invertible matrix P such that P
−1AP is a diagonal matrix.
• Diagonalization … process of finding a
corresponding diagonal matrix.
• Diagonalization of square matrix with
independent eigenvectors meas its
eigendecomposition
S-1AS = Λ
• Which matrices are diagonalizable (i.e.
eigendecomposed)?
– all λ’s are different (λ can be zero, the matrix is still
diagonalizable. But it is singular.)
– if some λ’s are same, I must be careful, I must check
the eigenvectors.
– If I have repeated eigenvalues, I may or may not have
n independent eigenvectors.
– If the eigenvectors are independent, than it is
diagonalizable (even if I have some repeating
eigenvalues).
• Diagonalizability is concerned with eigenvectors.
(independent)
• Invertibility is concerned with eigenvalues. (λ=0)
• diagonal matrix
– eigenvalues are on the diagonal
– eigenvectors are columns with 0 and 1
3 0
1 0
A
1 3, 2 2 x1 , x2
0 2
0
1
• triangular matrix
– eigenvalues are sitting along the diagonal
• symmetric matrix (n x n)
– real eigenvalues, real eigenvectors
– n independent eigenvectors
– they can be chosen such that they are orthonormal (A
= QΛQT)
• orthogonal matrix
– eigenvalues are +1 or -1
Symmetric matrices
• eigenvalues are real, eigenvectors are orthogonal
(may be chosen to be orthogonal)
• Symmetric matrices with all the eigenvalues positive
are called positive definite (semidefinite).
• A linear system of equations with a positive definite
matrix can be efficiently solved using the so-called
Cholesky decomposition.
– M = LLT = RTR (L – lower triangular, R – upper)
• ATA is always symmetric positive definite
provided that all columns of A are
independent.
Singular Value Decomposition
based on excelent video lectures by Gilbert Strang, MIT
http://ocw.mit.edu/OcwWeb/Mathematics/18-06Spring-2005/VideoLectures/detail/lecture29.htm
Lecture 29
SVD demonstration
• we have vectors x and y perpendicular,
find such Ax and Ay that are also
perpendicular
– If found, let’s label x as v1 and y as v2. And
Ax as u1 and Ay as u2.
– If v1 and v2 are orthonormal, u1 and u1 are
orthogonal.
– But can be normalized – multiplied by
numbers σ1 and σ2
• http://ocw.mit.edu/ans7870/18/18.06/javademo/SVD/
• So matrix A acts on vector v1 and vector
σ1u1 is obtained → Av1 = σ1u1
• And similarly Av2 = σ2u2
• Generally, this can be written in a matrix
form as
AV = UΣ
• But realize, that columns of V and U are
orthonormal. U and V are orthogonal
matrices!
• AV = UΣ → A = UΣV-1 → A = UΣVT
• Numbers σ are called singular values.
• And A = UΣVT is called singular value
decomposition (SVD).
• Factors: orthogonal matrix, diagonal matrix,
orthogonal matrix – all good matrices!
– Σ – singular values (diagonal)
– columns of U – left singular vectors
(orthogonal)
– columns of V – right singular vectors
(orthogonal)
• Any matrix has SVD!
– that’s why we actually need two orthogonal
matrices
Dimensions in SVD
A: m x n
U: m x m
V: n x n
Σ: m x n
• Consider case of square full-rank matrix
Am x m.
• Look again at Av1 = σ1u1
– σ1u1 is a linear combination of columns of A.
Where lies σ1u1?
• In C(A)
– And where lies v1?
• In C(AT)
• In SVD I’m looking for an orthogonal basis
in row space of A that gets knocked over
into an orthogonal basis in column space
of A.
• Can I construct an orthogonal basis of
C(AT)?
• Of course I can, Gram-Schmidt does this.
• However, when I take such a basis and
multiply it by A, there is no reason why the
result should also be orthogonal.
• I am looking for the special setup, where A
turns orthogonal basis vectors from row
space into orthogonal basis in the column
space.
basis vectors
in row space
of A
A v1v2
1
2
um
basis vectors
in column space
of A
vm u1u2
m
multiplying
factors
• In case A is symmetric positive definite,
the SVD becomes A = QΛQT.
– SVD of positive definite matrix directly reveals
its eigendecomposition.
• SVD of 2 x 2 nonsingular matrix
– U, Σ, V are all of 2 x 2
4 4 1 0 32
A
3 3 0 1 0
0 1 2 1
18 1 2 1
2
2
• If A is real, U and V are also real.
• The singular values σ are always real,
nonegative, and they are conventionally
arranged in descending order on the main
diagonal of Σ.
• SVD of 2 x 2 singular matrix
1 1 1 0 2
A
0 0 0 1 0
0 1
0 1
2
1
2
1
2
2
• SVD is rank revealing factorization.
rank(A) is the number of nonzero singular
values.
• Due to the experimental errors matrices
are close to singular – rank deficient
• SVD helps in this situation by revaling the
intrinsic dimension. Close to zero singular
values can then be set equal to zero.
In this example 4 x 4 matrix is generated, with 4th column almost equal to c1 – c2.
Then the 4th singular is zeroed, and matrix is reconstructed.
Matlab code
c1=[1 2 4 8]'
c2=[3 6 9 12]
c3=[12 16 3 5]'
c4 = c1-c2+0.0000001*(rand(4,1))
A = [c1 c2 c3 c4]
format long
[U,S,V]=svd(A)
U1 = U; S1=S; V1 = V;
S1(4,4)=0.0
B=U1*S1*V1'
A-B
norm(A-B)
help norm
1 3 12 1.99
2 6 16 3.99
A
4 9 3 4.99
8
12
5
3
.
99
• m x n, m > n (overedetermined system)
– with all columns independent
A = [c1 c2 c3]
[U,S,V]=svd(A)
U(1:3,1:3)*S(1:3,1:3)*V'
1
2
Av1v2 vn u1u2 unu n 1 um
n
complete this to an
orthonormal basis for the
whole column space Rm
i.e. these vectors come from
left null space, which is
orthogonal to column space
complete this by
adding zeros
• For the system m ≥ n, there exist a version
of SVD, called thin or economy SVD. The
zero block from Σ is removed, as well as
are corresponding columns from U.
1
2
Av1v2 vn u1u2 unu n 1 um
n
Dimensions of thin SVD
these are removed
A: m x n
U: m x n
V: n x n
Σ: n x n
• m x n, n > m (underdetermined system)
– with all rows independent
A = [c1(1:3) c2(1:3) c3(1:3) [15 12 63]']
[U,S,V]=svd(A)
U*S(1:3,1:3)*V(1:3,1:3)'
A v1v2
vm vm 1
vn u1u2
complete this to an
orthonormal basis for the
whole row space Rn
i.e. these vectors come from
null space, which is
orthogonal to row space
1
2
um
n
complete this by
adding zeros
Avi i ui , i 1,
,r
Basis of C(AT)
A v1
vr
Basis of C(A)
vr 1
vn u1
Basis of N(A)
ur
ur 1
um
Basis of N(AT)
1
r
• SVD chooses the right basis for the 4 subspaces
• AV=UΣ
– v1…vr: orthonormal basis in Rn for C(AT)
– vr+1…vn:
in Rn for N(A)
– u1…ur:
– ur+1…um:
in Rm for C(A)
in Rm for N(AT)
0
Uniqueness of SVD
• non-equal singular values (non-degenerate)
– U, V are unique except for the signs in columns
– Once you decide signs for U (V), signs for V (U)
are given so that A = UΣVT is guaranteed
• equal non-zero singular values (degenerate)
– not unique
– if u1 and u2 correspond to the degenerate σ, also
their any normalized linear combination is a left
singular vector corresponding to σ
– the same is true for right singular vectors
Uniqueness of SVD
• zero singular values
– Corresponding columns of U and V are added
– They form basis of N(AT) or N(A).
– They are orthonormal each to other and to the
rest of right/left singular vectors.
– And are not unique.
How do we find SVD?
•
•
•
•
A = UΣVT
I’ve got two orthogonal matrices and I
don't want to find them both at once.
I need some expression so U disappears.
Let’s do ATA (nice positive (semi)definite
symmetric). What’s the result?
ATA = VΣTUTUΣVT = VΣ2VT
This is actually a diagonalization for
positive definite ATA = QΛQT
V is found by diagonalizing ATA
AT A V
2
1
22
T
V
• How to find U?
• Look at diagonalization of AAT, so V
disappears.
– Or compute 𝑢1 =
1
𝐴𝑣1
𝜎1
• So u’s are eigenvectors of ATA, v’s are
eigenvectors of AAT.
• And eigenvalues are square roots of σ’s.
– They are same for ATA and AAT.
– They are unique.
– They have arbitrary order. However, they are
conventionally sorted in decreasing order (u, v
must be reordered accordingly)
• Example
4 4
3 3
• Rank is?
– two, invertible, nonsingular
• I'm going to look for two vectors v1 and v2
in the row space, which of course is what?
– R2
• And I'm going to look for u1, u2 in the
column space, which is also R2, and I'm
going to look for numbers σ1, σ2 so that it
all comes out right.
SVD example
4 4
A
3
3
25 7
A A
st
T
7
25
• 1 step – A A
• 2nd step – find its eigenvectors (they’ll b the vs)
and eigenvalues (squares of the σ)
T
– eigen can be guessed just by looking
– what are eigenvectors and their corresponding
eigenvalues?
•
•
•
•
[1 1]T, λ1 = 32
[-1 1]T, λ2 = 18
Actually, I made one mistake, do you see which one?
I didn’t normalize eigenvectors, they should be [1/sqrt(2),
1/sqrt(2)]T …
32
4 4
T
A
UV U
3 3
0
0 1 2 1
18 1 2 1
2
2
• 3rd step – AAT, find the u
12
T
T
2
AA U
2
U
32 0
AA
0
18
T
It’s diagonal, but just
by an accident
– What are the eigenvectors?
• [1 0]T, [0 1]T
4 4
1 0 32
T
A
UV
3 3
0 1 0
0 1 2 1
18 1 2 1
2
2
SVD expansion
• Do you remember column times row
picture of matrix multiplication?
• If you look at the UΣVT multiplication as at
the column times row problem, then you
get the SVD expansion.
T
A i ui vi
i
• This is a sum of rank-one matrices. Each
of uiviT is called mode.
• This expansion is used in data
compression application of SVD. We
investigate the spectrum of singular
values, and based on it we can
decide when to stop the expansion.
• This actually means that from Σ we
remove singular values from the end,
and that we remove coresponding
columns from U and V.
SVD image compression
load clown
colormap('gray')
image(X)
[U,S,V] = svd(X);
plot(diag(S));
size(X,1)*size(X,2)
p=1;image(U(:,1:p)*S(1:p,1:p)*V(:,1:p)');
size(U,1)*p+size(V',1)*p
To reconstruct Ap, we need to store only (m+n)·p words, compared
to m·n needed for storage of the full matrix A.
Please note, that if you put p = m (for m<n), then we need m2+m·n,
which is much more than m·n. So there is apparently some border
at which storing SVD matrices increases the storage needs!
Numerical Linear Algebra
Matrix Decompositions
based on excelent video lectures by Gilbert Strang, MIT
http://ocw.mit.edu/OcwWeb/Mathematics/18-06Spring-2005/VideoLectures/detail/lecture29.htm
Lecture 29
Conditioning and stability
• two fundamental issues in numerical
analysis
• Tell us something about the behavior after
introduction of small error.
• Conditioning – perturbation behavior of a
mathematical problem.
• Stability - perturbation behavior of an
algorithm.
Conditioning
• System’s behavior, it has nothing to do with
numerical errors (round off).
• Example: Ax = b
• well-conditioned - small change in A or b results
in a small change in the solution b
• ill-conditioned - small change in A or b results in a
large change in the solution b
• Conditioning is quantified by the condition
number, by its coupling with machine epsilon
(precission) we can then quantify how many
significant digits we can trust in the solution.
1 2 x 4
2 3 y 7
x 2
y 1
1 2 x 4.001
2 3 y 7.001
well-conditioned system
x 1.999
y 1.001
1.001 2.001 x 4
2.001 3.001 y 7
x 2.003
y 0.997
small change in coefficient matrix A
small change in result b
ill-conditioned system
2 x 4
1
2 3.999 y 7.999
x 2
y 1
2 x 4.001
1
2 3.999 y 7.998
x 3.999
y 4.000
1.001 2.001 x 4
2.001 3.998 y 7.999
x 3.994
y 0.001388
Stability
• Algorithm’s behavior, involves rounding
errors etc.
• Example:
– our computer only has two digits of precision
(i.e. only two nonzero numbers are allowed)
– we have array of 100 numbers [1.00, 0.01,
0.01, 0.01, …]
– and we want to sum them
– mathematically exact solution is 1.99
unstable
So we take the first number (1.00), then add
another (1.01), then another …. (1.10), then
another (1.11, however, this can’t be stored in
two-dogit precission, so computer rounds
down to 1.10), …, so the final result is 1.10
(compare to correct result of 1.99)
stable
Or we can sort the array first in ascending
order [0.01. 0.01, …, 1.00] and then sum:
0.01, 0.02, 0.03, …0.99, 1.99, 1.99 gets round
to 2.00 (compare to correct result of 1.99)
Matrix
factorizations/decompositions
• factorization - decomposition of an object
into a product of other objects - factors which when multiplied together give the
original .
• matrix factorization – product of some
nicer/simpler matrices that retain particular
properties of the original matrix A
– nicer matrices: orthogonal, diagonal,
symmetric, triangular
System of equations Ax = b
• System of linear equations
• It is solved by Gaussian elimination
• Gaussian elimination leads to the LU
factorization
A LU
• U is upper traiangular, L is lower triangular
with units on the diagonal
0
U
0
0
0
0
0
1
L
0
1
0
0
1
0
0
0
1
• Ax = b … LUx = b, we can say Ux = y,
and solve Ly = b and then Ux = y.
• Why is it advantageous?
1 0 y1 5
Ly
3 1 y2 6
1y1 0 y2 5
3 y1 y2 6
y1 5
3 5 y2 6 y2 9
forward substitution
• Similarly, Ux = y is solved by backward
substitution.
• LU factorization is Gauss elimination, so
when should I do LU factorization?
• If I have more bs, but A does not change. I
do LU just once, and then I repeatedly
change just RHS (b).
• LU factorization is used to calculate matrix
determinant.
– A = LU, det(A) = det(LU) = det(L)det(U) =
product of diagonal elements L (one) x product
of diagonal elements U
• Use of QR factorization for Ax = b
– A = QR, QRx = b → Rx = QTb
– Compute A = QR
– Compute y = QTb
– solve Rx = y by back-substitution
• However the standard way for Ax = b is
Gauss (LU), it requires half of the
operations than QR.
• LU factorization can also be used for
matrix inversion (however, standard way is
Gauss-Jacobi), L and U are easier to
invert.
Least squares algorithms
•
•
ATAx = ATb
Least squares via normal equations:
– Cholesky factorization ATA = RTR leads to
RTRx = ATb
1.
2.
3.
4.
form ATA and ATb
compute Cholesky ATA = RTR
solve lower triangular system RTw = ATb for w
solve upper triangular system Rx = w for x
•
Least squares with QR factorization
– ATAx = ATb
– QR factorization of A leads to
(QR)TQRx = (QR)Tb → Rx = QTb
1. compute A = QR
2. compute the vector QTb
3. solve upper triangular system Rx = QTb for x
– This is method of choice for least squares.
Eigenvalue algorithms
• Eigenvalue decomposition A = SΛS-1
• alternatively AS = SΛ
• this leads to characteristic equation polynomial (det λI – A = 0)
• However, finding roots of polynomial is
instable, it can be done for 2 x 2 or 3 x 3,
but not for 50 x 50
• Thus, this decomposition is usually not
adopted for eigenproblems.
• Eigenproblem via iterative QR algorithm
• Will be shown without proof:
A(0) = A
for k = 1, 2, …
Q(k)R(k) = A(k-1)
A(k) = R(k)Q(k)
QR factorization of A(k-1)
Recombine factors in reverse order
• Other iterative algorithms exist for special
purposes (get first 30 eigenvalues, get
eigenvalues from the given interval, nonsymmetric matrices)