Introduction to Linear Algebra 4th Edition - Solutions by Chapter

The old basis vectors v j are combinations L mij Wi of the new basis vectors. The coordinates of CI VI + ... + cnvn = dl wI + ... + dn Wn are related by d = M c. (For n = 2 set VI = mll WI +m21 W2, V2 = m12WI +m22w2.)

Remove row i and column j; multiply the determinant by (-I)i + j •

(Particular x p) + (x n in nullspace).

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

A(B + C) = AB + AC. Add then multiply, or mUltiply then add.

has derivative AeAt; eAt u(O) solves u' = Au.

Column i has no pivot in elimination. We can give the n - r free variables any values, then Ax = b determines the r pivot variables (if solvable!).

Independent columns in A, orthonormal columns in Q. Each column q j of Q is a combination of the first j columns of A (and conversely, so R is upper triangular). Convention: diag(R) > o.

A symmetric matrix with eigenvalues of both signs (+ and - ).

Vector addition and scalar multiplication.

A combination other than all Ci = 0 gives L Ci Vi = O.

= Xl (column 1) + ... + xn(column n) = combination of columns.

Some power of N is the zero matrix, N k = o. The only eigenvalue is A = 0 (repeated n times). Examples: triangular matrices with zero diagonal.

In each column, choose the largest available pivot to control roundoff; all multipliers have leij I < 1. See condition number.

Vectors x with aT x = O. Plane is perpendicular to a =1= O.

Orthogonal Q times positive (semi)definite H.

The minimum cost vector x * is found by moving from comer to lower cost comer along the edges of the feasible set (where the constraints Ax = b and x > 0 are satisfied). Minimum cost at a comer!

The transpose is -K, since Kij = -Kji. Eigenvalues are pure imaginary, eigenvectors are orthogonal, eKt is an orthogonal matrix.

The transpose is AT = A, and aU = a ji. A-I is also symmetric.

Sequence of n real numbers v = (VI, ... , Vn) = point in Rn.