 5.3.1: Find the signed area of the parallelogram formed by the following p...
 5.3.2: Find the signed volume of the parallelepiped formed by the followin...
 5.3.3: Let A = (a1, a2), B = (b1, b2), and C = (c1, c2) be points in R2. S...
 5.3.4: Suppose A, B, and C are vertices of a triangle in R2, and D is a po...
 5.3.5: Let u, v R3. Define the cross product of u and v to be the vector u...
 5.3.6: Let P be the parallelogram spanned by two vectors u, v R3. a. By in...
 5.3.7: Let a R3 be fixed. Define T : R3 R3 by T (x) = a x (see Exercise 5)...
 5.3.8: Suppose a polygon in the plane has vertices (x1, y1), (x2, y2), . ....
 5.3.9: (from the 1994 Putnam Exam) Find the value of m so that the line y ...
 5.3.10: Given any ellipse, show that there are infinitely many inscribed tr...
 5.3.11: Let x, y R3. Show that det _ x x x y y x y y _ is the square of the...
 5.3.12: Generalizing the result of Exercise 11, let v1, . . . , vk Rn. Show...
Solutions for Chapter 5.3: Signed Area in R2 and SignedVolume in R3
Full solutions for Linear Algebra: A Geometric Approach  2nd Edition
ISBN: 9781429215213
Solutions for Chapter 5.3: Signed Area in R2 and SignedVolume in R3
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Change of basis matrix M.
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.)

Cramer's Rule for Ax = b.
B j has b replacing column j of A; x j = det B j I det A

Cross product u xv in R3:
Vector perpendicular to u and v, length Ilullllvlll sin el = area of parallelogram, u x v = "determinant" of [i j k; UI U2 U3; VI V2 V3].

Diagonalizable matrix A.
Must have n independent eigenvectors (in the columns of S; automatic with n different eigenvalues). Then SI AS = A = eigenvalue matrix.

Ellipse (or ellipsoid) x T Ax = 1.
A must be positive definite; the axes of the ellipse are eigenvectors of A, with lengths 1/.JI. (For IIx II = 1 the vectors y = Ax lie on the ellipse IIA1 yll2 = Y T(AAT)1 Y = 1 displayed by eigshow; axis lengths ad

Free variable Xi.
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!).

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

Full row rank r = m.
Independent rows, at least one solution to Ax = b, column space is all of Rm. Full rank means full column rank or full row rank.

Hilbert matrix hilb(n).
Entries HU = 1/(i + j 1) = Jd X i 1 xj1dx. Positive definite but extremely small Amin and large condition number: H is illconditioned.

Iterative method.
A sequence of steps intended to approach the desired solution.

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

Linearly dependent VI, ... , Vn.
A combination other than all Ci = 0 gives L Ci Vi = O.

Multiplier eij.
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).

Norm
IIA II. The ".e 2 norm" of A is the maximum ratio II Ax II/l1x II = O"max· Then II Ax II < IIAllllxll and IIABII < IIAIIIIBII and IIA + BII < IIAII + IIBII. Frobenius norm IIAII} = L La~. The.e 1 and.e oo norms are largest column and row sums of laij I.

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.

Pivot columns of A.
Columns that contain pivots after row reduction. These are not combinations of earlier columns. The pivot columns are a basis for the column space.

Row space C (AT) = all combinations of rows of A.
Column vectors by convention.

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

Sum V + W of subs paces.
Space of all (v in V) + (w in W). Direct sum: V n W = to}.