 3.C.1: Suppose V and W are finitedimensional and T 2 L.V; W /. Show thatw...
 3.C.2: Suppose D 2 LP3.R/;P2.R/is the differentiation map defined byDp D p...
 3.C.3: Suppose V and W are finitedimensional and T 2 L.V; W /. Provethat ...
 3.C.4: Suppose v1;:::; vm is a basis of V and W is finitedimensional. Sup...
 3.C.5: Suppose v1;:::; vm is a basis of V and W is finitedimensional. Sup...
 3.C.6: Suppose V and W are finitedimensional and T 2 L.V; W /. Prove that...
 3.C.7: Verify 3.36
 3.C.8: Verify 3.38
 3.C.9: Prove 3.52
 3.C.10: Suppose A is an mbyn matrix and C is an nbyp matrix. Prove that...
 3.C.11: Suppose a D a1 anis a 1byn matrix and C is an nbypmatrix. Prove...
 3.C.12: Give an example with 2by2 matrices to show that matrix multiplica...
 3.C.13: Prove that the distributive property holds for matrix addition and ...
 3.C.14: Prove that matrix multiplication is associative. In other words, su...
 3.C.15: Suppose A is an nbyn matrix and 1 j; k n. Show that the entry inr...
Solutions for Chapter 3.C: Matrices
Full solutions for Linear Algebra Done Right (Undergraduate Texts in Mathematics)  3rd Edition
ISBN: 9783319110790
Solutions for Chapter 3.C: Matrices
Get Full SolutionsThis expansive textbook survival guide covers the following chapters and their solutions. Linear Algebra Done Right (Undergraduate Texts in Mathematics) was written by and is associated to the ISBN: 9783319110790. Chapter 3.C: Matrices includes 15 full stepbystep solutions. Since 15 problems in chapter 3.C: Matrices have been answered, more than 6693 students have viewed full stepbystep solutions from this chapter. This textbook survival guide was created for the textbook: Linear Algebra Done Right (Undergraduate Texts in Mathematics), edition: 3.

Associative Law (AB)C = A(BC).
Parentheses can be removed to leave ABC.

Basis for V.
Independent vectors VI, ... , v d whose linear combinations give each vector in V as v = CIVI + ... + CdVd. V has many bases, each basis gives unique c's. A vector space has many bases!

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.)

Column picture of Ax = b.
The vector b becomes a combination of the columns of A. The system is solvable only when b is in the column space C (A).

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.

Cyclic shift
S. Permutation with S21 = 1, S32 = 1, ... , finally SIn = 1. Its eigenvalues are the nth roots e2lrik/n of 1; eigenvectors are columns of the Fourier matrix F.

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

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.

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

Factorization
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.

Incidence matrix of a directed graph.
The m by n edgenode incidence matrix has a row for each edge (node i to node j), with entries 1 and 1 in columns i and j .

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

Least squares solution X.
The vector x that minimizes the error lie 112 solves AT Ax = ATb. Then e = b  Ax is orthogonal to all columns of A.

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

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

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 •

Polar decomposition A = Q H.
Orthogonal Q times positive (semi)definite H.

Semidefinite matrix A.
(Positive) semidefinite: all x T Ax > 0, all A > 0; A = any RT R.

Singular matrix A.
A square matrix that has no inverse: det(A) = o.

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