- 11.4.1: In Exercises 14, a and b are the legs of a right triangle and c is ...
- 11.4.2: In Exercises 14, a and b are the legs of a right triangle and c is ...
- 11.4.3: In Exercises 14, a and b are the legs of a right triangle and c is ...
- 11.4.4: In Exercises 14, a and b are the legs of a right triangle and c is ...
- 11.4.5: APPLICATION Triangles that are similar to a right triangle with sid...
- 11.4.6: Cal and Al are trying to solve the problem . Cal says that . Al dis...
- 11.4.7: You will need a centimeter ruler for this problem. a. Measure the l...
- 11.4.8: Miya was trying to solve the problem x2 + 42 = 52. She took the squ...
- 11.4.10: Mini-Investigation Strips of graph paper may help in 10a. a. Draw o...
- 11.4.11: A 27-inch TV has a screen that measures 27 inches on its diagonal. ...
- 11.4.12: In Exercise 10, you showed that a triangle with side lengths of 5, ...
- 11.4.13: APPLICATION When objects block sunlight, they cast shadows, and sim...
- 11.4.14: Ibrahim Patterson is planning to expand his square deck. He will ad...
Solutions for Chapter 11.4: The Pythagorean Theorem
Full solutions for Discovering Algebra: An Investigative Approach | 2nd Edition
Associative Law (AB)C = A(BC).
Parentheses can be removed to leave ABC.
Circulant matrix C.
Constant diagonals wrap around as in cyclic shift S. Every circulant is Col + CIS + ... + Cn_lSn - l . Cx = convolution c * x. Eigenvectors in F.
Column space C (A) =
space of all combinations of the columns of A.
Complete solution x = x p + Xn to Ax = b.
(Particular x p) + (x n in nullspace).
Dimension of vector space
dim(V) = number of vectors in any basis for V.
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 IIA-1 yll2 = Y T(AAT)-1 Y = 1 displayed by eigshow; axis lengths ad
Exponential eAt = I + At + (At)2 12! + ...
has derivative AeAt; eAt u(O) solves u' = Au.
A = L U. If elimination takes A to U without row exchanges, then the lower triangular L with multipliers eij (and eii = 1) brings U back to A.
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.
Gram-Schmidt orthogonalization A = QR.
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.
Linear combination cv + d w or L C jV j.
Vector addition and scalar multiplication.
Ln = 2,J, 3, 4, ... satisfy Ln = L n- l +Ln- 2 = A1 +A~, with AI, A2 = (1 ± -/5)/2 from the Fibonacci matrix U~]' Compare Lo = 2 with Fo = O.
Minimal polynomial of A.
The lowest degree polynomial with meA) = zero matrix. This is peA) = det(A - AI) if no eigenvalues are repeated; always meA) divides peA).
If N NT = NT N, then N has orthonormal (complex) eigenvectors.
Particular solution x p.
Any solution to Ax = b; often x p has free variables = o.
Positive definite matrix A.
Symmetric matrix with positive eigenvalues and positive pivots. Definition: x T Ax > 0 unless x = O. Then A = LDLT with diag(D» O.
Row picture of Ax = b.
Each equation gives a plane in Rn; the planes intersect at x.
Singular Value Decomposition
(SVD) A = U:E VT = (orthogonal) ( diag)( orthogonal) First r columns of U and V are orthonormal bases of C (A) and C (AT), AVi = O'iUi with singular value O'i > O. Last columns are orthonormal bases of nullspaces.
Spectral Theorem A = QAQT.
Real symmetric A has real A'S and orthonormal q's.
If x gives the movements of the nodes, K x gives the internal forces. K = ATe A where C has spring constants from Hooke's Law and Ax = stretching.