Linear Algebra

Intro

Basic idea behind linear algebra: solving large number of linear algebraic equations simultaneously.

$a_{i,1} x_1 + a_{i,2} x_2 + a_{i,3} x_3 + \dots + a_{i,N} x_{N} = b_{N} \qquad i=1 \dots M$

There are N unknowns and M equations, forming a linear system from matrices:

$\mathbf{A \cdot x} = \mathbf{b}$

where lowercase letters denote vectors, and the dot represents multiplication of two matrices or a matrix and a vector, or dot product of two vectors.

Nonsingular vs Singular

We are not guaranteed to find a solution to a given linear system Ax=b

row degeneracy - one or more of the M equations is a linear combination of other equations
column degeneracy - one or more of the N variables are interchangeable and cannot be distinguished (are non-unique)
for square matrices, these two conditions are equivalent

Having singular system of equations means we can't find good solutions.

Tasks of Computational Linear Algebra

Beyond simply solving Ax=b for the unknowns x, we may want to do other things:

Solve more than one matrix equation, $\mathbf{A \cdot x}_j = \mathbf{b}_j$ , where the matrix A does not change but unknown/RHS vectors do change
Calculating the inverse of the matrix
Calculating the determinant of a square matrix
Calculating the condition number of a matrix

If M < N (row degeneracy), or if M = N and equations are degenerate, there are fewer equations than unknowns. If fewer equations than unknowns:

Either no solution vector, or
More than one solution vector
If more than one solution, solution space consists of particular solution added to linear combination of N-M vectors (which are in the nullspace of the matrix A)
SVD can be used to solve this problem, by find these vectors (SVD finds the nullspace of a matrix A)

If M > N (column degeneracy), there are fewer unknowns than equations. If fewer unknowns than equations:

Think least squares - more data points than unknowns
Looking for best fit/compromise solution
Linear least-squares problem can be used to solve overdetermined linear system

Linear Algebra Software

Software:

Blas is short for "basic linear algebra subroutines" and provides the linear algebra operations that compose Lapack
Lapack is the most common linear algebra package (Linpack is old and was replaced by Lapack)
ScaLapack is version of Lapack for parallel architectures
Lapack routines divided by precision, algorithm, simplifications (tridiag., sparse, dense, etc.)
Atlas is version of Lapack that automatically tunes itself to run as fast as possible for your system's architecture when it is compiled

Direct vs Iterative:

Major dividing line between routines
Direct - algorithms that execute in a predictable number of operations
Iterative - attempt to converge on the desired answer to within a desired tolerance, taking as many steps as necessary

Gauss-Jordan Elimination

Classic technique for solving linear equations: combine equations to cancel out unknowns, turning A into the identity matrix

Gauss-Jordan elimination produces the solution of the equations and the matrix inverse $\mathbf{A}^{-1}$ - this is an expensive extra thing to produce, not always desirable to have it

Deficiencies:

All right hand side values must be stored and manipulated at once (not feasible for very large systems of equations)
Gauss-Jordan is three times slower than alternative techniques that do not produce the matrix inverse

Advantages:

Straightforward
Solid
Pivoting

Pivoting:

Pivoting is just interchanging rows (partial pivoting) or rows and columns (full pivoting) to put desired elements on matrix diagonals
Partial pivoting is easier, full pivoting requires keeping track of permutations, and un-doing these when solution vector achieved

Storage requirements of method:

The matrix inverse of A is gradually built up in A, as original A is overwritten
Likewise solution vector x can gradually replace right-hand side vector b (solution vector is overwritten one row element at a time, each time an element is overwritten it will not be used again)

On the subject of row vs. column elimination strategy:

if we perform row operations, row transformations build right to left, so right hand side is transformed at each stage from one vector to another
if we perform column operations, the column operations build from left to right, so we have to remember each transformation at each stage, and apply them all in reverse order at the end
this means column operations are much less useful

Gaussian Elimination with Back-Substitution

Like Gauss-Jordan, Gaussian elimination utilizes pivoting to eliminate elements from the matrix A and simplify it, but doesn't go as far as simplifying it to the identity matrix - simplifies it to an upper-triangular matrix

Triangular matrices make back-substitution easy - solving first equation involves one unknown, solving second equation involves only one new unknown, etc.

Represents midpoint between full elimination (Gauss-Jordan) and triangular decomposition (LU, QR)

Advantage of Gaussian elimination with back-substitution over Gauss-Jordan elimination is faster in terms of number of operations:

Inner loops of Gauss-Jordan elimination have one subtraction and one multiplication, executed $$ N^3 $$ and $$ N^2 m $$ times (where m is number of right hand sides)
Inner loops in Gaussian elimination are only executed $\dfrac{1}{3} N^3$ times and $\dfrac{1}{2} N^2 m$ times
Back-substitution of RHS is $\dfrac{1}{2} N^2$ executions of similar loop (one mult plus one subt)
Gaussian elimination has a factor of three advantage over Gauss-Jordan, which erodes as m approaches N (number of RHS approaches number of equations/unknowns)

Calculation of inverse matrix is equivalent to the case of m = N (basically solving Ax=b, and letting b be each column of an identity matrix)

Gaussian elimination and back-subst. requires $\dfrac{1}{3} N^3$ operations for matrix reduction, $\dfrac{1}{2}N^3$ for RHS manipulation, and $\dfrac{1]{2} N^3$ for N back-substitutions, for a total of $\dfrac{4}{3} N^3$ operations
Gauss-Jordan only requires $$ N^3 $$ operations for matrix inversion, so it seems to win out
However, because all but one of RHS vector elements are 0, the operations count for Gaussian elimination is actually lower
Gauss-Jordan matrix inversion costs $\dfrac{1}{6} N^3$ operations

Note: the 1/6 comes from the fact that the matrix multiplication step normally requires 8 operations, but due to the sparseness of the RHS, we are only performing 1 of the 8 operations. Hence, $\drac{1}{8} \times \dfrac{4}{3} = \dfrac{1}{6}$

LU Decomposition