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Methods for Finding Bases
1 Bases for the subspaces of a matrix
Row-reduction methods can be used to find bases. Let us now look at an example illustrating how to obtain bases for the row space, null space, and column space of a matrix A. To begin, we look at an example, the matrix A on the left below. If we row reduce A, the result is U on the right.
A =
⇐⇒ U =
Let the rows of A be denoted by rj , j = 1, 2 , 3, and the columns of A by ak, k = 1, 2 , 3 , 4. Similarly, ρj denotes the rows of U. (We will not need the columns of U .)
1.1 Row space
The row spaces of A and U are identical. This is because elementary row operations preserve the span of the rows and are themselves reversible op- erations. Let’s see in detail how this works for A. The row operations we used to row reduce A are these.
step 1 : R 2 = R 2 − 2 R 1 = r 2 − 2 r 1 R 3 = R 3 − 2 R 1 = r 3 − 2 r 1 step 2 : R 3 = R 3 + 12 R 2 = 0 step 3 : R 2 = 12 R 2 = 12 r 2 − r 1 step 4 : R 1 = R 1 − R 2 = 2 r 1 − 12 r 2
Inspecting these row operations shows that the rows of U satisfy
ρ 1 = 2r 1 −
r 2 ρ 2 =
r 2 − r 1 ρ 3 = 0.
It’s not hard to run the row operations backwards to get the rows of A in terms of those of U.
r 1 = ρ 1 + ρ 2 r 2 = 2ρ 1 + 4ρ 2 r 3 = 2ρ 1 + ρ 2.
Thus we see that the nonzero rows of U span the row space of A.
They are also linearly independent. To test this, we begin with the equation c 1 ρ 1 + c 2 ρ 2 = ( 0 0 0 0 )
Inserting the rows in the last equation we get
( c 1 c 2 3 c 1 − c 2 − 2 c 1 + 2c 2 ) = ( 0 0 0 0 ).
This gives us c 1 = c 2 = 0, so the rows are linearly independent. Since they also span the row space of A, they form a basis for the row space of A. This is a general fact:
Theorem 1.1 The nonzero rows in U , the reduced row echelon form of a matrix A, comprise a basis for the row space of A.
1.1.1 Rank
The rank of a matrix A is defined to be the dimension of the row space. Since the dimension of a space is the number of vectors in a basis, the rank of a matrix is just the number of nonzero rows in the reduced row echelon form U. That number also equals the number of leading entries in the U , which in turn agrees with the number of leading variables in the corresponding homogeneous system.
Corollary 1.2 Let U be the reduced row echelon form of a matrix A. Then, the number of nonzero zero rows in U , the number of leading entries in U , and the number of leading variables in the corresponding homogeneous sustem Ax = 0 all equal rank(A).
As an example, consider the matrices A and U in (1). U has two nonzero rows, so rank(A) = 2. This agrees with the number of leading entries, which are U 1 , 1 and U 2 , 1. (These are in boldface in (1)). Finally, the leading variables for the homogeneous system Ax = 0 are x 1 and x 2 , again there are two.
1.2 Null space
We recall that the null spaces of A and U are identical, because row oper- ations don’t change the solutions to the homogeneous equations involved. Let’s look at an example where A and U are the matrices in (1). The
1.3 Column space
We now turn to finding a basis for the column space of the a matrix A. To begin, consider A and U in (1). Equation (2) above gives vectors n 1 and n 2 that form a basis for N (A); they satisfy An 1 = 0 and An 2 = 0. Writing these two vector equations using the “basic matrix trick” gives us:
− 3 a 1 + a 2 + a 3 = 0 and 2 a 1 − 2 a 2 + a 4 = 0.
We can use these to solve for the free columns in terms of the leading columns, a 3 = 3a 1 − a 2 and a 4 = − 2 a 1 + 2a 2.
Thus the column space is spanned by the set {a 1 , a 2 }. (a 1 and a 2 are in boldface in our matrix A above in (1).) This set is also linearly independent because the equation
0 = x 1 a 1 + x 2 a 2 = x 1 a 1 + x 2 a 2 + 0a 3 + 0a 4 = A
x 1 x 2 0 0
implies that ( x 1 x 2 0 0 )T^ is in the null space of A. Matching this vector with the general form of a vector in the null space shows that the corresponding t 1 and t 2 are 0, and therefore so are x 1 and x 2. It follows that {a 1 , a 2 } is linearly independent. Since it spans the columns as well, it is a basis for the column space of A. Note that these columns correspond to the leading variables in the problems, x 1 and x 2. This is no accident. The argument that we used can be employed to show that this is true in general:
Theorem 1.4 Let A ∈ Rm×n. The columns of A that correspond to the leading variables in the associated homogeneous problem, U x = 0 , form a basis for the column space of A. In addition, the dimension of the column space of A is rank(A).
2 Another matrix example
Let’s do another example. Consider the matrix A and the matrix U , its row reduced form, shown below.
A =
⇐⇒^ U^ =
From U , we can read off a basis for the row space, {( 1 0 − 1 − 1 − (^65)
Again, from U we see that the leading variables are x 1 and x 2 , so the leading columns in A are a 1 and a 2. Thus, a basis for the column space is the set
To get a basis for the null space, note that the free variables are x 3 through x 5. Let t 1 = x 3 , etc. The system corresponding to U x = 0 then has the form
x 1 − t 1 − t 2 −
t 3 = 0
x 2 + t 2 +
t 3 = 0.
To get n 1 , set t 1 = 1, t 2 = t 3 = 0 and solve for x 1 and x 2. This gives us n 1 =
)T
. For n 2 , set t 1 = 0, t 2 = 1, t 3 = 0, in the system
above; the result is n 2 =
)T
. Last, set t 1 = 0, t 2 = 0,
t 3 = 1 to get n 3 =
5 −^
7 5 0 0 1
)T
. The basis for the null space is thus (^)
n 1 =
, n 2 =
, n 3 =
6 5 − (^75) 0 0 1
We want to make a few remarks on this example, concerning the dimensions of the spaces involved. The common dimension of both the row space and the column space is rank(A) = 2, which is also the number of leading variables. The dimension of the null space is the nullity of A. Here, nullity(A) = 3. Thus, in this case we have verified that
rank(A) + nullity(A) = 5,
the number of columns of A.
and find its reduced row echelon form, U :
[A|b] ⇐⇒ U =
Because of the way the row reduction process is done, the first two columns of U are the reduced row echelon form of A. that is,
A ⇐⇒
From the equations above, we see that rank(A) = 2 < rank[A|b] = 3. By part 1 of Theorem 3.1, the system is inconsistent.
Example 3.3 Let A =
and b = (3 − 1)T^. Determine
whether the system Ax = b is consistent.
The augmented form of the system and its reduced row echelon form are given below.
[A|b] =
⇐⇒ U =
As before, the first four columns of U comprise the reduced row echelon form of A; that is,
A =
Inspecting the matrices, we see that rank(A) = rank[A|b] = 2 < 4. By part 3 of Theorem 3.1, the system is consistent and has infinitely many solutions. We close by pointing out that for any system [A|b] ∈ Rm×(n+1), the first n columns of the reduced echelon form of [A|b] always comprise the reduced echelon form of A. As above, we can use this to easily find rank(A) and rank[A|b], without any additional work.
