S8

Question 1

If V = Span(v₁,…,v_k), then {v₁,…,v_k} is a basis of V.

Answer 1

False

To be a basis of V , the vectors would also have to be linearly independent.

Question 2

If V has dimension k, and B = {v₁, . . . , v_k} is linearly independent, then B is a basis of V .

Answer 2

True

Because V has dimension k, any set of k linearly independent vectors is a basis.

Question 3

A spanning set of maximal size is a basis.

Answer 3

False

A linearly independent set of maximal size must be a basis; a spanning set of minimal size must also be a basis.

Question 4

The pivot columns of an echelon form of a matrix A form a basis for Col(A).

Answer 4

False

The key is that two row equivalent matrices do not have to have the same column space (although they have the same row space).

For instance, think of the reduced echelon form, which always has pivot columns e₁,e₂,…. Clearly there are matrices whose column space does not have such a basis (let the first row of the matrix be all zeroes).

The confusing thing is that to find a basis for a column space of A, you find an echelon form and see what the pivot columns are, but then the corresponding columns of A form a basis.

Question 5

Row(A) = Col(A^T)

Answer 5

True

The rows of A are the columns of A^T .

Question 6

Row(A) is a subspace of ℝ^m.

Answer 6

False

The rows of A have size n, so clearly Row(A) is a subset of ℝⁿ and not (in general) of ℝ^m.

Question 7

dim Row(A) = dim Col(A)

Answer 7

True

Both are equal to the number of pivots in an echelon form of A.

Question 8

dim Row(A) + dim Nul(A) = n

Answer 8

True

This is called the Rank Theorem in the book (since dim Row(A) = rank(A) by definition). Basically, dim Row(A) equals the number of pivots in an echelon form of A, and dim Nul(A) equals the number of columns in an echelon form that do not have a pivot (because each such column gives a free variable), so the sum of these is the number of columns.

Question 9

There is a 6×9 matrix B such that dim Nul(B) = 2.

Answer 9

False

By 4), we would have dim Row(A) = 9 − 2 = 7. But Row(A) is a subspace

of ℝ⁶, which has dimension 6, and subspace cannot have higher dimension than the vector space they are in.