TrueFalse 1.{ 0 } S is a subspace of every subspace S n . 2.If S is a subspace a
ID: 3115821 • Letter: T
Question
TrueFalse
1.{ 0 } S is a subspace of every subspace S n.
2.If S is a subspace and v + w S then v S.
3.If S is a subspace and v, w S then v w S.
word answer
4.Let L be a line through the origin in 3. Can L contain 2 linearly independent vectors?
5.number answer:
6.If v 0 then dim(span{v}) =?
7.If v 0 then dim(span{v} {0}) =?
8.If v 0 then dim(span{v} {0}) =?
9.If v, w are linearly independent then dim(span{v,w}) =?
10.If v, w are linearly independent then dim(span {2v; 3w}) =?
11.If v, w are linearly independent then dim(span {v w}) = ?
12.dim(4) =?
13.In the preceding why was it alright to divide by bj^2?
14.Every nonzero subspace has an orthonormal basis.(true/false)
15.The result of applying the Gram-Schmidt procedure to a basis which is orthonormal is the same basis one started with.(true/false)
16.If A, B are two matrices with compatible size, and A has rank 1 and B has rank 1, then rank(AB) 1.(true/false)
17.If A , B have the same size then rank(A + B) > rank(A) + rank(B).(true/false)
18.If A, B have the same size then rank(A + B) rank(A) + rank(B) .(true/false)
19.(true/false) If A is a nonzero n × 1 column vector then ATA is invertible.
20.(true/false) There’s a vector b such that [123] x = b is not solvable.
21.(true/false) Suppose that A^T A is invertible. Then so is A.
22.(true/false) Suppose that the square matrix A^T A has linearly independent columns. Then A must be invertible.
{ | 0 |, | |}) = ? 0 d im(spanExplanation / Answer
5. The 2 vectors are linearly independent. Hence the dimension of the spanning set is 2.
23. Let the given 2x3 matrix be denoted by A. It may be observed that { (1,5)T} is the basis for col(A). Hence dim(col(A)) = 1 so that rank(A) = 1. Now, by the rank-nullity theorem, dim(ker(A)) = nullity of A = number of columns of A – rank(A) = 3-1 = 2.
Let the given 3x3 matrix be denoted by B. The RREF of B is
1
0
-1
0
1
2
0
0
0
Hence the rank(B)= number of non-zero rows in RREF of B = 2. Then, by the rank-nullity theorem, dim(ker(B)) = nullity of B = number of columns of B – rank(B) = 3-2 = 1.
24. The 3 vectors are linearly independent. Hence the dimension of the spanning set is 3.
25. There are only 2 linearly independent vectors in the spanning set i.e. (1,2,0)T and (3,0,0)T. Therefore, the dimension of the spanning set is 2.
Please post the remaining questions again, maximum 4 at a time.
1
0
-1
0
1
2
0
0
0