Today we’ll further investigate the
Row Space and Column Space of a Matrix


Row Space = the set spanned by the row vectors of a matrix
Column Space = the set spanned by the column vectors of a matrix
The rank of a matrix is equal to the number of pivots of the row reduced echelon form.
You can think of each pivot as being in a row or in a column. It is an important result in
linear algebra that the rank of a matrix is equal to the rank of it’s transpose… the linear
independence of row vectors and column vectors are ‘in some sense’ interchangeable.
(read Theorem 4.28 on page 126.)
Finding The Basis For the Span of a Set of Vectors
A basis of a vector space V is a set of vectors B  V such that
1. B spans V
2. B is linearly independent
Even though there are many possible bases for B, they will all have the same number of
vectors: n vectors corresponding to the dimension of V.
Common Problem (Quiz Wednesday):
Find a basis for the span of a set of vectors T=v1, v2, v3 and v4.
“Theory”
S= Span(T) = a v1 +b v2 +c v3 +d v4 where a,b,c,d 2.
What is a basis for S?
The basis is a set B that spans S.
We already know that S is spanned by v1, v2, v3, v4.
… And B is linearly independent.
We need to check that v1, v2, v3, v4 are lin indep, and take out the dependent
vectors.
“Solution”
Check linear dependence of v1, v2, v3, v4.
There are two ways to do this: using the row space or the column space.
Column space – see practice quiz 7.
The test for linear dependence will result in a matrix with vectors forming the
columns. The solution will indicate which columns are linearly dependent on the others.
A pivot column corresponds to a linearly independent vector and a non-pivot column
corresponds to a linearly dependent vector.
Row space – elementary row operations preserve the row space. Therefore, the span of a
reduced matrix is the same as the original matrix.
How would you complete the two problems in practice quiz 7 using the row space
method?
Comparing Row Space and Column Space
The row space doesn’t change when you reduce a matrix so the span of the rows
of M is equal to the span of the rows of RREF(M). However, since all pivots are collected
at the top, you lose information about which row vectors of M generated pivots.
The column space does change when you reduce a matrix. However, since all
columns are connected with coefficients that are tied with each vector, you preserve
information about which column vectors generate pivots.
Dimension
A basis of a vector space V is a set of vectors B  V such that
1. B spans V
2. B is linearly independent
Related:
A basis is a minimal spanning spanning set. (fewest # of vectors needed to span)
A basis is a maximal linear independent set. (greatest # of lin independent vectors)
The number of elements in any basis of V is invariant and is called the dimension of U.
Examples of Vector Spaces with their Dimension
Vector Space
Standard Basis
Dimension
R2
(M(2,1))
1 0
0 1
  
2
R3
(M(3,1))
1 
0 
 
0 
0 
1 
 
0 
0 
0 
 
1 
3
n
M(2,2)
n
1 0 0 1 0 0 0 0
0 0 0 0 1 0 0 1





M(m,n)
P2
Pn
4
mxn
{1, x, x2}
1, x, x ,..., x 
2
n
3
n+1
Why invariant? Suppose there is a basis for V with n elements.
(1) If a set has more than n vectors it is linearly dependent. (can’t be a basis) THM 1
(2) If a set has fewer than n elements, it does not span V. (can’t be a basis) THM 2
Proof (1):
Basis for V is B = {b1, b2, …, bn}.
Another set S = {s1, s2, …, sn}
By definition of basis, every vector in S is in the span of B, so every vector si can be
written as a linear combination of b1, b2, …, bn.
Thus, every vector in S is linearly dependent upon B and has at most n degrees of
freedom  cannot all be independent.
Proof (2):
Assume that a set S = {s1, s2, …, sm}with fewer than n elements does span V. Then every
element in B can be expressed as a linear combination of s1, s2, …, sm. Therefore, the
elements in B have at most m degrees of freedom, so you cannot have more than m
independent vectors. However, all elements of B are independent because B was a basis
 contradiction.
What Dimension “n” Means:
 the smallest # of elements needed to span V is n (def)
 no set of more then n vectors of V can be independent (Thm 1)
(Proof: Using a basis for V with n vectors,
show the n+1th one must be dependent.)
 any basis of V contains exactly n vectors (Thm 2)
(Proof: follows from Theorem 1)
 any set of n vectors that spans V must be lin indep (Thm 3)
 any set of n independent vectors spans V (Thm 3)
Let V be a vector space V with dimension
n and let S be a set of n vectors v1, …, vn 
V . Then the following are equivalent:
 S is linearly independent
 S is a basis of V
 S spans V
“TFAE” means:
Given that any 1 of the 3 is true, they are
all are true.
Given that any 1 of the 3 are false, they are
all false.
Subspace Dimension Theorem
For U and W are finite-dimensional subspaces of V then”
 If UW then dim(U)≤dim(W)
 If UW and dim(U)=dim(W) then U=W
A Little Bit of More Theory Regarding Forming a Basis
“Any spanning set of vectors can be cut down to form a basis”
Suppose a set S spans V. If they are linearly independent, then they’re already a basis.
Otherwise, they are linearly dependent and one vector can be expressed as a linear
combination of the others. Thus it can be deleted without changing the span of the set.
Continue until all vectors in S are linearly independent.
“Any set of linearly independent vectors can be extended to form a basis”
Suppose a set S is linearly independent but does not span V. Then there is a vector in V
that is not in the span of S, and this vector is linearly independent w.r.t. S. So we can add
this vector to S and it will still be linearly independent. Now the span of S has increased
and includes the vector we just added. Continue until S spans V.
These results apply to finite dimensional vector spaces.
Infinite dimensional space: Rn: (x1, …, xn, …)