Solutions for Chapter A.2: Introduction to Linear Algebra 5th Edition
Full solutions for Introduction to Linear Algebra  5th Edition
ISBN: 9780201658590
Solutions for Chapter A.2
Get Full SolutionsThis textbook survival guide was created for the textbook: Introduction to Linear Algebra , edition: 5. Introduction to Linear Algebra was written by and is associated to the ISBN: 9780201658590. This expansive textbook survival guide covers the following chapters and their solutions. Since 1 problems in chapter A.2 have been answered, more than 7464 students have viewed full stepbystep solutions from this chapter. Chapter A.2 includes 1 full stepbystep solutions.

Companion matrix.
Put CI, ... ,Cn in row n and put n  1 ones just above the main diagonal. Then det(A  AI) = ±(CI + c2A + C3A 2 + .•. + cnA nl  An).

Covariance matrix:E.
When random variables Xi have mean = average value = 0, their covariances "'£ ij are the averages of XiX j. With means Xi, the matrix :E = mean of (x  x) (x  x) T is positive (semi)definite; :E is diagonal if the Xi are independent.

Diagonalization
A = S1 AS. A = eigenvalue matrix and S = eigenvector matrix of A. A must have n independent eigenvectors to make S invertible. All Ak = SA k SI.

Dimension of vector space
dim(V) = number of vectors in any basis for V.

Dot product = Inner product x T y = XI Y 1 + ... + Xn Yn.
Complex dot product is x T Y . Perpendicular vectors have x T y = O. (AB)ij = (row i of A)T(column j of B).

Elimination matrix = Elementary matrix Eij.
The identity matrix with an extra eij in the i, j entry (i # j). Then Eij A subtracts eij times row j of A from row i.

Fourier matrix F.
Entries Fjk = e21Cijk/n give orthogonal columns FT F = nI. Then y = Fe is the (inverse) Discrete Fourier Transform Y j = L cke21Cijk/n.

Full column rank r = n.
Independent columns, N(A) = {O}, no free variables.

GaussJordan method.
Invert A by row operations on [A I] to reach [I AI].

Graph G.
Set of n nodes connected pairwise by m edges. A complete graph has all n(n  1)/2 edges between nodes. A tree has only n  1 edges and no closed loops.

Identity matrix I (or In).
Diagonal entries = 1, offdiagonal entries = 0.

Left nullspace N (AT).
Nullspace of AT = "left nullspace" of A because y T A = OT.

Nullspace N (A)
= All solutions to Ax = O. Dimension n  r = (# columns)  rank.

Orthogonal subspaces.
Every v in V is orthogonal to every w in W.

Orthonormal vectors q 1 , ... , q n·
Dot products are q T q j = 0 if i =1= j and q T q i = 1. The matrix Q with these orthonormal columns has Q T Q = I. If m = n then Q T = Q 1 and q 1 ' ... , q n is an orthonormal basis for Rn : every v = L (v T q j )q j •

Saddle point of I(x}, ... ,xn ).
A point where the first derivatives of I are zero and the second derivative matrix (a2 II aXi ax j = Hessian matrix) is indefinite.

Schur complement S, D  C A } B.
Appears in block elimination on [~ g ].

Spanning set.
Combinations of VI, ... ,Vm fill the space. The columns of A span C (A)!

Standard basis for Rn.
Columns of n by n identity matrix (written i ,j ,k in R3).

Tridiagonal matrix T: tij = 0 if Ii  j I > 1.
T 1 has rank 1 above and below diagonal.