Solutions for Chapter 8: Algebra

Tv = Av + Vo = linear transformation plus shift.

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 •

If diagonalizable, they share n eigenvectors.

Row n + 1 counts corners, edges, faces, ... of a cube in Rn.

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

A sequence of steps intended to approach the desired solution.

If A has s independent eigenvectors, its "generalized" eigenvector matrix M gives 1 = diag(lt, ... , 1s). The block his Akh +Nk where Nk has 1 's on diagonall. Each block has one eigenvalue Ak and one eigenvector.

Current Law: net current (in minus out) is zero at each node. Voltage Law: Potential differences (voltage drops) add to zero around any closed loop.

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

The pivot row j is multiplied by eij and subtracted from row i to eliminate the i, j entry: eij = (entry to eliminate) / (jth pivot).

= column times row = rank one matrix.

There are n! orders of 1, ... , n. The n! P 's have the rows of I in those orders. P A puts the rows of A in the same order. P is even or odd (det P = 1 or -1) based on the number of row exchanges to reach I.

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

P = aaT laTa has rank l.

Unit vector u is reflected to Qu = -u. All x intheplanemirroruTx = o have Qx = x. Notice QT = Q-1 = Q.

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

Entries AL = Ajj. AT is n by In, AT A is square, symmetric, positive semidefinite. The transposes of AB and A-I are BT AT and (AT)-I.

Orthonormal columns (complex analog of Q).

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