Math 3191 Spring 2000 Cheat Sheet
Final

Jan Mandel

Elementary row operations: 1. add a multiple of a row to another row (does not change determinant) 2. interchange two rows (changes sign of det) 3. multiply row by a scalar (multiplies the determinant by the same scalar). Matrix of elementary row transformation: doing the transformation on A is same as replacing A by EA.
Row echelon form: has zeros under a ``staircase'': Reduced row echelon form: first entry in each row is 1; it is the only nonzero in its column (called pivot column)
A u =b is equivalent to b = $u_1 a_1 + \ldots + u_n a_n$, $A=[a_1, \ldots, a_n]$. Au=b has a solution if and only if (iff) b is in the span of the columns of A. Au=b has a solution for every b iff columns of A span R^m iff A has a pivot position in every row.
Solution of Ax=b is unique iff Au=0 has only zero solution iff there are no free variables (in the echelon form of A).
Vectors $[a_1\ldots a_m]$ are linearly dependent if there is linear combination of them that equals to zero but has some nonzero coefficients.
Solution of Ax=b is unique iff columns of A are linearly independent. Linear maps: $T: x\mapsto Ax$. A is the standard matrix for T. $A=[T(e_1), \ldots, T(e_n)]$ where $I=[e_1,\ldots,e_n]$. Matrix-matrix product: C=AB means $c_{ik} = \sum_j a_{ij}b_{k}$. Note $B[a_1,\ldots,a_n]=[Ba_1,\ldots,Ba_n]$.
Matrix inverse: B=A^-1 means AB = BA = I. A and B must be square. Only one of AB = I or BA=I is sufficient. To compute A^-1, transform [A,I] to the reduced echelon form. The following is equivalent for square A: A^-1 exists; $T: x\mapsto Ax$ is invertible; T is onto; T is one-to-one; Ax=b has a solution for every b; Ax=0 has only zero solution; $\det A \ne 0$; A has pivot in every column (in the algorithm of reduction to echelon form).
A=LU where U is the echelon form and L has ones on the diagonal and under the diagonal are minus the multipliers used in the algorithm of reduction to the echelon form (use only steps adding multiple of a row to another row). Jacobi and Gauss-Seidel iterative method: compute x_i^k+1 from equation i. Jacobi: use the values of x^k. Gauss-Seides: use the newest values available. Convergence guaranteed when A strictly diagonally dominant: sum of absolute values of offdiagonal terms in column i is less than a_ii, for all i. If C is strictly diagonally dominant, $(I-C)^{-1} = I + C + C^2 + C^3 + \ldots$.
Basis of subspace V of Rⁿ is any set $\{b_1,\ldots,b_p\}$ that is linearly independent and spans V. Each $u\in V$ can be written uniquely as $u=c_1 b_1 + \ldots c_p b_p$, with the coefficients c_i scalars; $\dim V=p$.
Nul A is the set of all x such that Ax=0. To find a basis of Nul A, solve Ax=0 by transforming A to reduced echelon form. Col A is the set of all Ax. Pivot columns of A form a basis of Col A. Rank theorem: $\dim Nul A + \dim Col A =$ number of columns of A.

Expansion of determinant by row i: $\det A = \sum_{j=1}^n (-1)^{i+j} \det A_{ij}$. Expansion by a column is similar. Determinant of triangular matrix equals the product of the diagonal terms.

Vector space H is subspace of vector space V if $H\subset V$ and $u+v\in H$ for any $u,v\in H$ and $cu\in H$ for any $u\in H$ and scalar c. Basis of a vector space V is a subset of V that spans V and is linearly independent. The coordinates of $v\in V$ relative to a basis $\{b_1,\ldots,b_p\}$ is $(x_1,\ldots,x_p) \in R^p$ such that $v=x_1 b_1 + \cdots + x_p v_p$. The mapping of vectors in V to their coordinates in R^p preserves linear dependence and independence.

The change of coordinates matrix is constructed so that $[x]_C = P_{C \leftarrow B}[x]_B$ and it is given by $P_{C \leftarrow B} = [ [b_1]_C,\ldots,[b_p]_C]$. (This matrix is also matrix of the identity operator from the basis B to C). In Rⁿ, if B and C denote also the matrices with the basis vectors as columns, $P_{C \leftarrow B}[x]_B = C^{-1}B$ and it can be found by reducing the augmented matrix $[C \ B]$ to the form $[P_{C \leftarrow B} \ I]$.

$\lambda$ and u are eigenvalue and eigenvector of A if $Au=\lambda u$and $u\ne 0$. The eigenvalues of a triangular matrix are its diagonal entries. Eigenvalues satisfy the characteristic equation $\det(A-\lambda I)=0$. If A=PBP^-1then the eigenvalues of A and B are same. A is diagonalizable if there exists a basis consisting of its eigenvectors; then A=PDP^-1, where D is a diagonal matrix with the eigenvalues on the diagonal, and the columns of P are the eigenvectors.

A basis $B=\{b_1, \ldots , b_n\}$ is orthogonal if $b_i \cdot b_j =0$ if $i\ne j$. Coefficients of x relative to orthogonal basis B are $x_i = x \cdot b_i / b_i \cdot b_i$.

A least squares solution of a rectangular system Ax=b is defined by $\Vert Ax-b\Vert^2\to \min$and it can be found by solving the normal equations A^T A x = A^T b.