Row Space and Column Space

References are to Anton–Rorres

PROBLEM: Compute everything about the 4×5 matrix

A = [ \begin{matrix}1 & −2 & 0 & 0 & 3 \\ 2 & −5 & −3 & −2 & 6 \\ 0 & 5 & 15 & 10 & 0 \\ 2 & 6 & 18 & 8 & 6 \end{matrix} ]  (1)

(This includes Example 8 (p. 267) in §5.5.)

The nullspace of A Find the dimension (= nullity(A)) and a basis. In effect, solve the linear system Ax = 0. Therefore we use elementary row operations to reduce A to row echelon form (not uniquely, so your answer may vary)

R = [ \begin{matrix}1 & −2 & 0 & 0 & 3 \\ 0 & 1 & 3 & 2 & 0 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{matrix} ]  (2)

with the leading ones boxed. Or we can go all the way to reduced row echelon form

R′ = [ \begin{matrix}1 & 0 & 0 & −2 & 3 \\ 0 & 1 & 0 & −1 & 0 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{matrix} ]  (3)

which is unique. From (2) or (3), it is clear that x_{4} and x_{5} are free variables and may be chosen arbitrarily; we put x_{4} = r and x_{5} = s. Then from (2), by back substitution, or directly from (3), the general solution of Ax = 0 can be written

x_{1} = 2r − 3s, x_{2} = r, x_{3} = −r, x_{4} = r, x_{5} = s

Thus the nullspace has dimension 2, as it needs two coordinates, and has the basis

{(2, 1, −1, 1, 0), (−3, 0, 0, 0, 1)}

(Here, the first vector is obtained by setting r = 1 and s = 0 and the second by r = 0 and s = 1; equivalently, we read off the coefficients of r and s in each x_{j}.)

The row space of A Find the dimension (= rank(A)) and a basis. By Theorem 5.5.4, the row space of A is the same as the row space of R (or R′). But by Theorem 5.5.6, we see from (2) that the first three rows of R form a basis. (None of these rows is a linear combination of later rows, and the zero row has no effect on the row space.) Thus the row space of A has dimension rank(A) = 3 and has the basis

{(1, −2, 0, 0, 3), (0, 1, 3, 2, 0), (0, 0, 1, 1, 0)}

The column space of A Find the dimension (= rank(A)) and a basis. Write u_{j} for column j of R′. It is clear that u_{1} = e_{1}, u_{2} = e_{2}, and u_{3} = e_{3}, and that these form a basis of the column space of R′. Explicitly, we read off that u_{4} = −2u_{1} − u_{2} + u_{3} and u_{5} = 3u_{1}. The column space of R′ is not the same as the column space of A; however, Theorem 5.5.5 allows us to conclude that the corresponding columns c_{j} of A do the same job for A. Namely, the column space of A has dimension rank(A) = 3 and has the basis

{c_{1} = (1, 2, 0, 2), c_{2} = (−2, −5, 5, 6), c_{3} = (0, −3, 15, 18)}

Further, the remaining columns of A are expressed in terms of these as

c_{4} = (0, −2, 10, 8) = −2c_{1} − c_{2} + c_{3}, c_{5} = (3, 6, 0, 6) = 3c_{1}

as is easily checked from (1).

The row space of A, revisited Better: Select a basis from the rows of A, {r_{1}, r_{2}, r_{3}, r_{4}}. Theorem 5.4.6 says that this is always possible, and sometimes, this is what we need. We just did it for the columns of A. [It is perfectly possible to develop a theory of column operations, but we and Anton–Rorres choose not to go this route.] IDEA: Consider the transpose of A, the 5×4 matrix

A^{T} = [ \begin{matrix}1 & 2 & 0 & 2 \\ −2 & −5 & 5 & 6 \\ 0 & −3 & 15 & 18 \\ 0 & −2 & 10 & 8 \\ 3 & 6 & 0 & 6 \end{matrix} ]

Elementary row operations reduce this to the row echelon form (see p. 268)

[ \begin{matrix}1 & 2 & 0 & 2 \\ 0 & 1 & −5 & −10 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix} ]  (4)

and to the reduced row echelon form

[ \begin{matrix}1 & 0 & 10 & 0 \\ 0 & 1 & −5 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix} ]  (5)

Write v_{j} for column j of this matrix. This time, we have the basis {v_{1}, v_{2}, v_{4}}, and v_{3} = 10v_{1} − 5v_{2}. Again by Theorem 5.5.5 (applied to A^{T}), we deduce the basis

{r_{1} = (1, −2, 0, 0, 3), r_{2} = (2, −5, −3, −2, 6), r_{4} = (2, 6, 18, 8, 6)} of the row space of A, and the relation (easily verified from (1))

r_{3} = (0, 5, 15, 10, 0) = 10r_{1} − 5r_{2}

The nullspace of A^{T} Find the dimension and a basis. From (5), we see that this time there is only one free variable, x_{3}. The dimension is 1 and the basis consists of the single vector (−10, 5, 1, 0). Note that 1 = 4 − 3, as in Theorem 5.6.3 (for A^{T}).