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Linear Algebra, 2020a

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Section III. Basis and Dimension 139<br />

Thus, the first point for this subsection is that the tools of this chapter give<br />

us a more conceptual understanding of Gaussian reduction.<br />

For the second point observe that row operations on a matrix can change its<br />

column space. ( )<br />

(<br />

1 2 −2ρ 1 +ρ 2 1 2<br />

−→<br />

2 4<br />

0 0<br />

)<br />

The column space of the left-hand matrix contains vectors with a second component<br />

that is nonzero but the column space of the right-hand matrix contains<br />

only vectors whose second component is zero, so the two spaces are different.<br />

This observation makes next result surprising.<br />

3.10 Lemma Row operations do not change the column rank.<br />

Proof Restated, if A reduces to B then the column rank of B equals the column<br />

rank of A.<br />

This proof will be finished if we show that row operations do not affect linear<br />

relationships among columns, because the column rank is the size of the largest<br />

set of unrelated columns. That is, we will show that a relationship exists among<br />

columns (such as that the fifth column is twice the second plus the fourth) if and<br />

only if that relationship exists after the row operation. But this is exactly the<br />

first theorem of this book, Theorem One.I.1.5: in a relationship among columns,<br />

⎛ ⎞<br />

⎛ ⎞ ⎛ ⎞<br />

a 1,1<br />

a 1,n 0<br />

a 2,1<br />

c 1 ·<br />

⎜<br />

⎝<br />

⎟<br />

. ⎠ + ···+ c a 2,n<br />

n ·<br />

⎜ . ⎝<br />

⎟<br />

. ⎠ = 0<br />

⎜<br />

⎝<br />

⎟<br />

. ⎠<br />

a m,1 a m,n 0<br />

row operations leave unchanged the set of solutions (c 1 ,...,c n ).<br />

QED<br />

Another way to make the point that Gauss’s Method has something to say<br />

about the column space as well as about the row space is with Gauss-Jordan<br />

reduction. It ends with the reduced echelon form of a matrix, as here.<br />

⎛<br />

⎜<br />

1 3 1 6<br />

⎞<br />

⎛<br />

⎟<br />

⎜<br />

1 3 0 2<br />

⎞<br />

⎟<br />

⎝2 6 3 16⎠ −→ ··· −→ ⎝0 0 1 4⎠<br />

1 3 1 6<br />

0 0 0 0<br />

Consider the row space and the column space of this result.<br />

The first point made earlier in this subsection says that to get a basis for the<br />

row space we can just collect the rows with leading entries. However, because<br />

this is in reduced echelon form, a basis for the column space is just as easy: collect<br />

the columns containing the leading entries, 〈⃗e 1 ,⃗e 2 〉. Thus, for a reduced echelon

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