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# Typical Problems

• Linear System: $Ax = b$
• Least Squares Problem $\lVert Ax - b\lVert_{2}$
• Eigenvalue problem $Ax = \lambda x$, where $x$ - vector $\lambda$ - scalar
• Singular Value problem $A^T Ax = \lambda x$

# Definitions

## Nonsingular matrix

A square matrix that is not singular, i.e. one that has a matrix inverse. Nonsingular matrices are sometimes also called regular matrices. A square matrix is nonsingular if its determinant is nonzero

## Inversing Triangular Matrix

Solve directly:

$\left(\begin{array}{lll} {a} & {b} & {c} \\ {0} & {d} & {e} \\ {0} & {0} & {f} \end{array}\right) \left(\begin{array}{lll} {x} & {y} & {z} \\ {0} & {y} & {v} \\ {0} & {0} & {w} \end{array}\right) = \left(\begin{array}{lll} {1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1} \end{array}\right)$ $\left(\begin{array}{lll} {1 / a} & {-b /(a d)} & {(b e-c d) /(a f d)} \\ {0} & {1 / d} & {-e /(f d)} \\ {0} & {0} & {1 / f} \end{array}\right)$

from which we see directly that the matrix is invertible if all $a,d$ and $f$ are different from zero

## Positive Definite matrix

$A$ $n \times n$ symmetric real matrix $M$ is said to be positive definite if $x^{\top} M x>0$ for all non-zero $x$ in $\mathbb{R}^{n}$.

$M \text { positive definite } \Longleftrightarrow x^{\top} M x>0 \text { for all } x \in \mathbb{R}^{n} \backslash {0}$

## Positive Semidefinite matrix

$A$ $n \times n$ symmetric real matrix $M$ is said to be positive semidefinite or non-negative definite if $x^{\top} M x \geq 0$ for all $x$ in $\mathbb{R}^{n} .$ Formally,

$M \text { positive definite } \Longleftrightarrow x^{\top} M x>0 \text { for all } x \in \mathbb{R}^{n}$

## Rank of a matrix

The rank of a matrix is defined as

• the maximum number of linearly independent column vectors in the matrix or
• the maximum number of linearly independent row vectors in the matrix.

Both definitions are equivalent.

## Eigenvalues and eigenvectors

Eigenvalues (characteristic roots, proper values, latent roots) are a special set of scalars associated with a linear system of equations. Each eigenvalue is paired with a corresponding so-called eigenvector.

Let $A$ be a linear transformation represented by a matrix A. If there is a vector $\mathbf{X} \in \mathbb{R}^{n} \neq 0$ such that:

$A X=\lambda X$

for some scalar $\lambda,$ then $\lambda$ is called the eigenvalue of A with corresponding (right) eigenvector $\mathbf{X}$.

Letting A be a $k \times k$ square matrix

$\left[\begin{array}{cccc} {a_{11}} & {a_{12}} & {\cdots} & {a_{1 k}} \\ {a_{21}} & {a_{22}} & {\cdots} & {a_{2 k}} \\ {\vdots} & {\vdots} & {\ddots} & {\vdots} \\ {a_{k 1}} & {a_{k 2}} & {\cdots} & {a_{k k}} \end{array}\right]$

with eigenvalue $\lambda,$ then the corresponding eigenvectors satisfy

$\left[\begin{array}{cccc} {a_{11}} & {a_{12}} & {\cdots} & {a_{1 k}} \\ {a_{21}} & {a_{22}} & {\cdots} & {a_{2 k}} \\ {\vdots} & {\vdots} & {\ddots} & {\vdots} \\ {a_{k 1}} & {a_{k 2}} & {\cdots} & {a_{k k}} \end{array}\right]\left[\begin{array}{c} {x_{1}} \\ {x_{2}} \\ {\vdots} \\ {x_{k}} \end{array}\right]=\lambda\left[\begin{array}{c} {x_{1}} \\ {x_{2}} \\ {\vdots} \\ {x_{k}} \end{array}\right]$

which is equivalent to the homogeneous system

$\left[\begin{array}{cccc} {a_{11}-\lambda} & {a_{12}} & {\cdots} & {a_{1 k}} \\ {a_{21}} & {a_{22}-\lambda} & {\cdots} & {a_{2 k}} \\ {\vdots} & {\vdots} & {\ddots} & {\vdots} \\ {a_{k 1}} & {a_{k 2}} & {\cdots} & {a_{k k}-\lambda} \end{array}\right]\left[\begin{array}{c} {x_{1}} \\ {x_{2}} \\ {\vdots} \\ {x_{k}} \end{array}\right]=\left[\begin{array}{c} {0} \\ {0} \\ {\vdots} \\ {0} \end{array}\right]$

Equation ( 4) can be written compactly as

$(A-\lambda I) X=0$

## Condition number

Condition number determines to what degree an error in the numerical representation of the input data of a problem affects the error in the result. If condition number is $10^k$ , you lose roughly $k$ decimal digits of precision

$k(A) = ||A−1|| · ||A||$

## Norms

\begin{aligned} &|\mathbf{x}|_{\infty} \equiv \max \left|x_{i}\right| \\ &|\mathbf{x}|_{p} \equiv\left(\sum_{i}\left|x_{i}\right|^{p}\right)^{1 / p} \end{aligned} $\begin{array}{llll} {\text { name }} & {\text { symbol}} & {\text { value }} & {\text { approx. }} \\ {L^{1}-\text { norm }} & {|\mathbf{x}|_{1}} & {6} & {6.000} \\ {L^{2}-\text { norm }} & {|\mathbf{x}|_{2}} & {\sqrt{14}} & {3.742} \\ {L^{3}-\text { norm }} & {|\mathbf{x}|_{3}} & {6^{2 / 3}} & {3.302} \\ {L^{4}-\text { norm }} & {|\mathbf{x}|_{4}} & {2^{1 / 4} \sqrt{7}} &{3.146} \\ {L^{\infty}-\text { norm }} & {|\mathbf{x}|_{\infty}} & {3} & {3.000} \end{array}$

# Matrix Factorization:

A factorization of the matrix $A$ is a representation of $A$ as a product of several “simpler” matrices, which make the problem at hand easier to solve. We give two examples.

$\left[\begin{array}{cccc} {a_{11}} & {} & {} & {} \\ {a_{21}} & {a_{22}} & {} & {} \\ {\vdots} & {\vdots} & {\ddots} & {} \\ {a_{n 1}} & {a_{n 2}} & {\ldots} & {a_{n n}} \end{array}\right] \left[\begin{array}{c} {x_{1}} \\ {x_{2}} \\ {\vdots} \\ {x_{n}} \end{array}\right] =\left[\begin{array}{c} {b_{1}} \\ {b_{2}} \\ {\vdots} \\ {b_{n}} \end{array}\right]$

## Forward Substitution

$\text{for } i=1 \text{ to } n \\ \quad x_{i}=\left(b_{i}-\sum_{k=1}^{i-1} a_{i k} x_{k}\right) / a_{i i} \\ \text{end for}$

## Backward Substitution

An analogous idea, back substitution, works if $A$ is upper triangular. To use this to solve a general system $A x=b$ we need the following matrix factorization, which is just a restatement of Gaussian elimination.

## LU Factorization

LU on example:

• rows operations
$A = \left[ \begin{array} { r r r } { 1 } & { 4 } & { - 3 } \\ { - 2 } & { 8 } & { 5 } \\ { 3 } & { 4 } & { 7 } \end{array} \right] \begin{matrix}\\\textbf{2}R_{1}+R_{2}\\\textbf{-3}R_{1}+R_{2}\end{matrix} \quad L = \left[ \begin{array} { c c c } { 1 } & { 0 } & { 0 } \\ { ? } & { 1 } & { 0 } \\ { ? } & { ? } & { 1 } \end{array} \right]$
• 2 and -3 goes down to $L$ with changed sign
$U = \left[ \begin{array} { r r r } { 1 } & { 4 } & { - 3 } \\ {0 } & { 16 } & { -1 } \\ { 0 } & { -8 } & { 16 } \end{array} \right] \begin{matrix}\\ \\\textbf{0.5}R_{2}+R_{3}\end{matrix} \quad L = \left[ \begin{array} { c c c } { 1 } & { 0 } & { 0 } \\ { - 2 } & { 1 } & { 0 } \\ { 3 } & { ? } & { 1 } \end{array} \right]$
• 0.5 goes to $L$ with minus sign.
$U = \left[ \begin{array} { r r r } { 1 } & { 4 } & { - 3 } \\ {0 } & { 16 } & { -1 } \\ { 0 } & { 0 } & { 15.5 } \end{array} \right] \quad L = \left[ \begin{array} { c c c } { 1 } & { 0 } & { 0 } \\ { - 2 } & { 1 } & { 0 } \\ { 3 } & { -0.5 } & { 1 } \end{array} \right]$

## Gaussian Elimination

Partial Pivoting - Sort rows by its first elements absolute value. Choose $a_{i,k}$ and swap that $k$-th row with $i$-th row for i in (1,n) .

THEOREM If the $n-b y-n$ matrix $A$ is nonsingular, there exists

• a permutation matrix $P$ (the identity matrix with its rows permuted),
• a nonsingular lower triangular matrix $L,$ and
• a nonsingular upper triangular matrix $U$

such that $A=P \cdot L \cdot U .$ To solve $A x=b,$ we solve the equivalent system $P L U x=b$ as follows:

• $\left.L U x=P^{-1} b=P^{T} b \quad \text { (permute entries of } b\right)$
• $U x=L^{-1}\left(P^{T} b\right) \quad \text{(forward substitution),}$
• $x=U^{-1}\left(L^{-1} P^{T} b\right) \quad \text{(back substitution).}$

Important thing $L^{-1} \neq L^{T}$ and $U^{-1} \neq U^{T}$ - check inversion of triangular matrix in section definitions.

## Cholesky Factorization

The Cholesky decomposition is roughly twice as efficient as the LU decomposition for solving systems of linear equations. $\bf{A}$ needs to be symmetric. Every symmetric, positive definite matrix A can be decomposed into a product of a unique lower triangular matrix L and its transpose. $A = LL^{T}$, where $L$ is a lower triangular matrix with real and positive diagonal entries.

### Algorithm:

\begin{aligned} l_{1,1} &=\sqrt{a_{11}} \\ l_{j, 1} &=\frac{a_{j 1}}{l_{11}}, \quad j \in[2, n] \\ l_{i, i} &=\sqrt{a_{i i}-\sum_{p=1}^{i-1} l_{i p}^{2}}, \quad i \in[2, n] \\ l_{j, i} &=\left(a_{j i}-\sum_{p=1}^{i-1} l_{i p} l_{j p}\right) / l_{i i}, \quad i \in[2, n-1], j \in[i+1, n] \end{aligned}

### Example

$A = \left[\begin{array}{ccc} {4} & {12} & {-16} \\ {12} & {37} & {-43} \\ {-16} & {-43} & {98} \end{array}\right]$
• $i = 1:$
\begin{aligned} & l_{1,1} = \sqrt{a_{1,1}} = 2 \\ & l_{2,1} = \frac{a_{2,1}}{l_{1,1}} = 6 \\ & l_{3,1} = \frac{a_{3,1}}{l_{1,1}} = -8 \end{aligned} \quad L = \left[ \begin{array} { r r r } { 2 } & { 0 } & { 0 } \\ { 6 } & { ? } & { 0 } \\ { - 8 } & { ? } & { ? } \end{array} \right]
• $i = 2:$
\begin{aligned} & { l _ { 2,2 } = \sqrt { a _ { 2,2 } - l _ { 2,1 } ^ { 2 }} } = \sqrt{37 - 6^{2}} = 1\\ & { l_ { 3,2 } = \left( a _ { 3,2 } - l _ { 2 , 1 } \cdot l _ { 3,1 } \right) / l _ { 2,2 } } = (- 43 - (6 \cdot (- 8))/1 = 5 \end{aligned} \quad L = \left[ \begin{array} { r r r } { 2 } & { 0 } & { 0 } \\ { 6 } & { 1 } & { 0 } \\ { - 8 } & { 5 } & { ? } \end{array} \right]
• $i = 3$
\begin{aligned} l _ { 3,3 } & = \sqrt { a _ { 3,3 } - \left( l _ { 3,1 } ^ { 2 } + l _ { 3,2 } ^ { 2 } \right) } = \\ & = \sqrt { 98 - \left( ( - 8 ) ^ { 2 } + 5 ^ { 2 } \right) } = \\ & = \sqrt { 98 - ( 64 + 25 ) } = \\ & = \sqrt { 98 - 89 } = \sqrt { 9 } = 3 \end{aligned} $L = \left[ \begin{array} { r r r } { 2 } & { 0 } & { 0 } \\ { 6 } & { 1 } & { 0 } \\ { - 8 } & { 5 } & { 3 } \end{array} \right] \quad L^{T} = \left[ \begin{array} { r r r } { 2 } & { 6 } & { -8 } \\ { 0 } & { 1 } & { 5 } \\ { 0 } & { 0 } & { 3 } \end{array} \right]$

## QR factorization:

THEOREM $3.1 .$ QR decomposition. Let $A$ be $m-b y-n$ with $m \geq n .$ Suppose that $A$ has full column rank. Then there exists a unique $m-b y-n$ orthogonal matrix $Q\left(Q^{T} Q=I_{n}\right)$ and a unique $n-b y-n$ upper triangular matrix $R$ with positive diagonals $r_{i i}>0$ such that $A=Q R$

### QR Factorization Solves:

• linear equations
• least squares problems
• constrained least squares problems

### Algorithms for QR factorization:

• Gram-Schmidt algorithm
• complexity is $2 m n^{2}$ flops
• not recommended in practice (sensitive to rounding errors)
• Householder algorithm
• complexity is $2 m n^{2}-(2 / 3) n^{3}$ flops
• represents $Q$ as a product of elementary orthogonal matrices
• the most widely used algorithm

### Gram-Schmidt algorithm

Given: $m \times n$ matrix $A$ with linearly independent columns $a_{1}, \ldots, a_{n}$ \text{Algorithm} \\ \quad \text{for } k=1 \text{ to } n: \quad \begin{aligned} \tilde{q}_{1} &= a_{1} \\ R_{1 k} &=\left\|\tilde{q}_{1}\right\| \\ q_{1} &=\frac{1}{R_{1 k}}\tilde{q}_{1} \\ & \vdots \\ R_{k-1, k} &=q_{k-1}^{T} a_{k} \\ \tilde{q}_{k} &=a_{k}-\left(R_{1 k} q_{1}+R_{2 k} q_{2}+\cdots+R_{k-1, k} q_{k-1}\right) \\ R_{k k} &=\left\|\tilde{q}_{k}\right\| \\ q_{k} &=\frac{1}{R_{k k}} \tilde{q}_{k} \end{aligned} In MATLAB:

[m, n] = size(A);
Q = zeros(m,n);
R = zeros(n,n);
for k = 1:n
R(1:k-1,k) = Q(:,1:k-1)’ * A(:,k);
v = A(:,k) - Q(:,1:k-1) * R(1:k-1,k);
R(k,k) = norm(v);
Q(:,k) = v / R(k,k);
end;


## Hessenberg Factorization

The Hessenberg form of $A$ is a factorization $A=Q_{0} \cdot H \cdot Q_{0}^{H}$ where $Q_{0}$ is an orthogonal matrix and $H$ is upper Hessenberg. How can you compute it in the case when $A$ is a $3 \times 3$ matrix?

A matrix $H=\left(h_{i, j}\right): i, j \in \mathbb{R}^{n \times n}$ is upper Hessenberg if all its coefficients below the lower secondary diagonal are zero, that is, if $h_{i, j}=0$ whenever $i \geq j+2$

### Algorithm

• Choose $Q_{1}$ so
$Q_{1} A=\left[\begin{array}{ccccc} {x} & {x} & {x} & {x} & {x} \\ {x} & {x} & {x} & {x} & {x} \\ {0} & {x} & {x} & {x} & {x} \\ {0} & {x} & {x} & {x} & {x} \\ {0} & {x} & {x} & {x} & {x} \end{array}\right] \quad \text { and } \quad A_{1} \equiv Q_{1} A Q_{1}^{T}=\left[\begin{array}{ccccc} {x} & {x} & {x} & {x} & {x} \\ {x} & {x} & {x} & {x} & {x} \\ {0} & {x} & {x} & {x} & {x} \\ {0} & {x} & {x} & {x} & {x} \\ {0} & {x} & {x} & {x} & {x} \end{array}\right]$

$Q_{1}$ leaves the first row of $Q_{1} A$ unchanged, and $Q_{1}^{T}$ leaves the first column of $Q_{1} A Q_{1}^{T}$ unchanged, including the zeros.

• Choose $Q_{2}$ so
$Q_{2} A_{1}=\left[\begin{array}{ccccc} {x} & {x} & {x} & {x} & {x} \\ {x} & {x} & {x} & {x} & {x} \\ {0} & {x} & {x} & {x} & {x} \\ {0} & {0} & {x} & {x} & {x} \\ {0} & {0} & {x} & {x} & {x} \end{array}\right] \text { and } A_{2} \equiv Q_{2} A_{1} Q_{2}^{T}=\left[\begin{array}{ccccc} {x} & {x} & {x} & {x} & {x} \\ {x} & {x} & {x} & {x} & {x} \\ {0} & {x} & {x} & {x} & {x} \\ {0} & {0} & {x} & {x} & {x} \\ {0} & {0} & {x} & {x} & {x} \end{array}\right]$

$Q_{2}$ changes only the last three rows of $A_{1},$ and $Q_{2}^{T}$ leaves the first two columns of $Q_{2} A_{1} Q_{2}^{T}$ unchanged, including the zeros.

• Choose $Q_{3}$ so
$Q_{3} A_{2}=\left[\begin{array}{ccccc} {x} & {x} & {x} & {x} & {x} \\ {x} & {x} & {x} & {x} & {x} \\ {o} & {x} & {x} & {x} & {x} \\ {0} & {o} & {x} & {x} & {x} \\ {o} & {o} & {o} & {x} & {x} \end{array}\right] \quad \text { and } \quad A_{3}=Q_{3} A_{2} Q_{3}^{T}=\left[\begin{array}{ccccc} {x} & {x} & {x} & {x} & {x} \\ {x} & {x} & {x} & {x} & {x} \\ {0} & {x} & {x} & {x} & {x} \\ {0} & {0} & {x} & {x} & {x} \\ {0} & {0} & {0} & {x} & {x} \end{array}\right]$

which is upper Hessenberg. Altogether $A_{3}=\left(Q_{3} Q_{2} Q_{1}\right) \cdot A\left(Q_{3} Q_{2} Q_{1}\right)^{T} \equiv$ $Q A Q^{T}$

ALGORITHM $4.6$ Reduction to upper Hessenberg form: $\text{if } Q \text{ is desired, set }Q=I\\$

\begin{aligned} \text {for } i &=1: n-2 \\ u_{i} &= \text { House }(A(i+1: n, i)) \\ P_{i} &=I-2 u_{i} u_{i}^{T} & /^{*} Q_{i}=\operatorname{diag}\left(I^{i \times i}, P_{i}\right) * / \\ A(i+1: n, i: n)&=P_{i} \cdot A(i+1: n, i: n) \\ A(1: n, i+1: n)&=A(1: n, i+1: n) \cdot P_{i} \\ \text{if } Q \text{ is desired} & \\ Q(i+1: n, i: n) &=P_{i} \cdot Q(i+1: n, i: n) \quad /^{*} Q=Q_{i} \cdot Q^{*} \\ \quad\text{end if} & \\ \text{end for} & \\ \end{aligned}

## Schur Factorization

Schur decomposition allows one to write an arbitrary matrix as unitarily equivalent to an upper triangular matrix whose diagonal elements are the eigenvalues of the original matrix. The Schur decomposition reads as follows: if $A$ is a $n \times n$ square matrix with complex entries, then $A$ can be expressed as: $A=Q U Q^{-1}$

• $Q$ is a unitary matrix (so that its inverse $Q^{-1}$ is also the conjugate transpose $Q^{*}$ of $Q$ ), and

• $U$ is an upper triangular matrix, which is called a Schur form of $A$.

• If $A$ is real, its Schur factorization is $A =QUQ^{T}$

• columns of $Q$ are called the Schur vectors
• he eigenvalues of $A$ appear on the diagonal of $U$

The first step in a Schur decomposition is a Hessenberg decomposition.

For example, the Schur decomposition of the matrix $A=\left[\begin{array}{lll}{3} & {2} & {1} \\{4} & {2} & {1} \\{4} & {4} & {0}\end{array}\right]$ is $Q=\left[\begin{array}{ccc} {0.49857} & {0.76469} & {0.40825} \\ {0.57405} & {0.061628} & {-0.81650} \\ {0.64953} & {-0.64144} & {0.40825} \end{array}\right]\\ T=\left[\begin{array}{ccc} {6.6056} & {4.4907} & {-0.82632} \\ {0.00000} & {-0.60555} & {1.0726} \\ {0.00000} & {0.00000} & {-1.00000} \end{array}\right]$ and the eigenvalues of $T$ are: $\lambda_{1}=-1, \lambda_{2}=3-\sqrt{13}=-0.60555 \ldots . \lambda_{3}=3+\sqrt{13}=6.6055 \ldots$

# TODO

• Singular value decomposition (SVD)

• Jacobi method

• Gauss-Seidel method (iterative scheme)

• SOR(w) method

# Bibliography

[BL] K. Bryan and T. Leise, The \$25,000,000,000 eigenvector: The linear algebra behind Google, SIAM Review 48 (2006) 569–581.

[D] J. Demmel, Applied Numerical Linear Algebra, SIAM 1997.

[E] L. Elden, A note on the eigenvalues of the Google matrix, e-print arXiv:math/0401177 (2004), 2 pages.

[S] J. Shlens, A tutorial on principal components analysis, e-print arXiv:1404:1100 (2014), 12 pages

23 Mar 2020 Mateusz Dorobek

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