LINEAR ALGEBRA TEST 2 Q&A

LINEAR ALGEBRA TEST 2 Q&A
Theorem 4.1.1 let vector v = [x,y,z] - Answer-

Definition 4.1 - Answer-Two vectors u & v are parallel if there is some non-zero constant
c such that u = cv

Theorem (4.2.1): Let u, v, and w denote vectors in R3 - Answer-1. v ·w is a real number
2. v ·w = w ·v
3. v ·0 = 0 = 0 ·v
4. v ·v = ∥v∥^2
5. (kv) ·w = k(w ·v) = v ·(kw) for all scalars k
6. u ·(v ±w) = u ·v ±u ·w

Theorem (4.2.4): Let u and d ̸= 0 be vectors - Answer-1. The projection of u on d is
given by
projdu = (u ·d/(∥d∥)^2 )d.
2. The vector u −projdu is orthogonal to d.

Theorem (4.2.2) Let v&w be non zero vectors - Answer-Then v · w = ||v|| * ||w|| *
cos(theta)

Defn (4.2.1) - Answer-Two vectors v and w are orthogonal if v ·w = 0.

Defn (4.2.2) Normal Vector - Answer-A nonzero vector n is called a normal vector for a
plane if it is orthogonal to every vector in the plane.

Theorem (4.2.5): Let v and w be vectors in R3. - Answer-1. v ×w is a vector orthogonal
to both v and w.
2. If v and w are nonzero, then v ×w = 0 if and only if v and w are parallel.

Theorem (4.3.1): u·(v×w) = - Answer-u·(v×w) = det(v x w) but i j k are replaced with the
parts of vector u

Theorem (4.3.2): Let u, v, and w denote vectors in R3. - Answer-1. u ×v is a vector.
2. u ×v is orthogonal to both u and v.
3. u ×0 = 0 = 0 ×u
4. u ×u = 0
5. u ×v = −(v ×u)
6. (ku) ×v = k(u ×v) = u ×(kv) for any scalar k.
7. u ×(v + w) = (u ×v) + (u ×w)8. (v + w) ×u = (v ×u) + (w + u)

Theorem (4.3.3):If u and v are vectors in R3, then ||u ×v||^2 = - Answer-||u||^2 *||v||^2 −
(u ·v)^2.

, Theorem 4.3.4 If u and v are two nonzero vectors and θ is the angle between u and v,
then ||u×v|| = - Answer-1. ||u×v|| = ||u|| * ||v|| * sinθ = the area of the parallelogram
determined by u and v.
2. u and v are parallel if and only if u×v = 0.

Theorem 4.3.5 - Answer-The volume of the parallelepiped determined by three vectors
w, u, and v is given by |w ·(u×v)|.

Defn 5.1.1 A set U of vectors in Rn is called a subspace of Rn if it satisfies: - Answer-
1)The zero vector 0 ∈U.
2)If x ∈U and y ∈U then x + y ∈U.
3)If x ∈U, then ax ∈U for every real number a.

Null Space - Answer-The null space of an m x n matrix A is the set of all solutions to Ax
=0

null A = { x e R^n | Ax = 0}

Image space - Answer-im A = { Ax | x e R^n }

Span - Answer-The set of all linear combinations of a set of vectors

span{x1,x2,...,xk} =.{t1x1 + t2x2 + ... + tkxk | ti e R^n}

Theorem (5.1.1): Let U = span {x1,x2,...,xk} in Rn, then - Answer-1. U is a subspace of
Rn containing each xi.
2. If W is a subspace of Rn and each xi ∈W, then U ⊆W.

Defn 5.2.1 - Answer-A set of vectors {x1,x2,...,xn} is called linearly independent t1x1 +
t2x2 +... + tkxk = 0 implies t1 = t2 = tk = 0

If not linearly independent you can take a vector and solve it as a linear combination of
the others

Thm 5.2.1 If {x1,x2,...,xk} are an independent set of vectors in Rn - Answer-Every vector
in span {x1,x2,...,xk} has a unique representation as a linear combination of xis

Thm 5.2.2 If A = [a1, a2, ... ,an] is mxn - Answer-1) {a1, a2, ..., an} is independent in Rn
IFF Ax=0 implies x=0
2) Rm = span{a1, a2, ..., an} IFF Ax = b has a solution for every b in Rm

Thm (5.2.3) A is nxn TFAE - Answer-1) A is invertible
2) The columns of A are linearly independent
3) The columns of A span Rn
4) The rows of A are linearly independent
5) The rows of A span Rn