linear Algebra Chapter 7 Practice Problems.

Associativity of vector addition

(U+V)+W = U+(V+W). In R^4, add entry by entry and use associativity of real numbers in each entry.

Additive identity (zero vector)

A vector 0 with U+0 = U for every U. In R^4 it is (0,0,0,0).

Distributivity of scalar multiplication over vector addition

c(U+V) = cU + cV. In R^4, distribute c across each entry.

Additive inverse

For every U there is -U with U+(-U) = 0. In R^4, negate every entry.

Three conditions for a subspace

1) Contains the zero vector 2) Closed under addition 3) Closed under scalar multiplication.

Is T = {(2a, 0, -b, 0)} a subspace of R^4?

Yes. a=b=0 gives the zero vector. Sums give (2(a1+a2), 0, -(b1+b2), 0), still in T. Scalars give (2(ka), 0, -(kb), 0), still in T.

Is S = {(2a, 0, 1-b, 0)} a subspace of R^4?

Yes. Since b is any real number, 1-b takes every real value, so S = {(2a, 0, t, 0)}. The zero vector is in S (a=0, b=1). A sum has third entry 2-(b1+b2) = 1-b with b = b1+b2-1. A scalar multiple k(1-b) = 1-b' with b' = 1-k(1-b).

Is U = {(2a, 0, 1, 0)} a subspace of R^4?

No. The third entry is always 1, so the zero vector is not in U. Failing the zero vector test is enough.

Quick test to rule out a subspace

Check whether the zero vector is in the set. If it is not, the set is not a subspace.

Is S = {(x, y) in R^2 : y = 3x} a subspace?

Yes. (0,0) is in S. If y1=3x1 and y2=3x2 then y1+y2 = 3(x1+x2). Also ky = 3(kx). It is a line through the origin.

How to test if a vector is in a span

Set up a linear combination a(v1)+b(v2)+c(v3) = target, make an augmented matrix, and row reduce. If the system is consistent, the vector is in the span.

Is (2,-3,6) in span{(1,-1,3), (-2,1,0), (-1,-1,12)}?

Yes. RREF gives a=10, b=5, c=-2, a unique solution.

Do (1,-1,3), (-2,1,0), (-1,-1,12) span R^3?

Yes. There are 3 pivots (one in each column), so the vectors are linearly independent and span R^3.

Is (-2,14,1) in span{(2,1,-1), (-6,2,3), (4,7,-2)}?

Yes. The general solution is a=8-5t, b=3-t, c=t (infinitely many). With t=1 it is 3(2,1,-1)+2(-6,2,3)+1(4,7,-2).

Do (2,1,-1), (-6,2,3), (4,7,-2) span R^3?

No. There are only 2 pivots, so the vectors are linearly dependent and the span is just a plane in R^3.

Free variable in a row-reduced system means what?

Infinitely many solutions. Fewer pivots than vectors means the vectors are linearly dependent.

Is 2x^2+5x-7 in span{x^2, x+3}?

No. a(x^2)+b(x+3) forces a=2, b=5, and the constant term 3b=15, which is not -7.

How to show a set of polynomials is a basis for R2[x]

Write each polynomial as a coordinate vector relative to {1, x, x^2}, row reduce, and check that the RREF is the identity (independent and spanning).

Coordinate vector of 1+x^2 relative to {1, x, x^2}

(1, 0, 1)

Coordinate vector of -x+2x^2 relative to {1, x, x^2}

(0, -1, 2)

Coordinate vector of 1+x+x^2 relative to {1, x, x^2}

(1, 1, 1)

Is {1+x^2, -x+2x^2, 1+x+x^2} a basis for R2[x]?

Yes. The matrix of coordinate vectors row reduces to the identity, so the set is independent and spans R2[x].

How to find the coordinates of V relative to a basis B

Solve c1(b1)+c2(b2)+… = V by putting the basis vectors as columns, V as the last column, and row reducing. The solution (c1, c2, …) is V_B.

Coordinates of (3,2,-6) relative to B = {(1,0,2), (1,1,1), (0,1,3)}

V_B = (-3/2, 9/2, -5/2)

Coordinates of [3 0

1 5] relative to B = {[1 0

1 0], [0 1

0 0], [1 1

0 1], [0 1

0 1]}

c1=1, c2=-5, c3=2, c4=3, so V_B = [1 -5

2 3] (as a 2x2 arrangement) or (1, -5, 2, 3).

Coordinates in a matrix space

Match entry by entry. Each matrix entry gives one equation, so a 2x2 space gives 4 equations.

How to complete a basis of R^3 given V1 and V2

Let V3 = (a,b,c), put V1, V2, V3 as columns, and row reduce. The vectors are independent exactly when the last pivot entry is nonzero.

Completing the basis V1=(1,0,-2), V2=(3,1,2). What condition must V3=(a,b,c) satisfy?

c + 2a - 8b is not 0 (equivalently c is not 8b-2a).

Vector that completes the basis V1=(1,0,-2), V2=(3,1,2)

V3 = (0,0,1), since 1 + 0 - 0 = 1 is not 0.