12.07.2015 Views

Linear Algebra

Linear Algebra

Linear Algebra

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

124 Chapter Two. Vector SpacesThe column space of the left-hand matrix contains vectors with a second componentthat is nonzero but the column space of the right-hand matrix is differentbecause it contains only vectors whose second component is zero. It is thisobservation that makes next result surprising.3.10 Lemma Row operations do not change the column rank.Proof Restated, if A reduces to B then the column rank of B equals the columnrank of A.We will be done if we can show that row operations do not affect linearrelationships among columns because the column rank is just the size of thelargest set of unrelated columns. That is, we will show that a relationship existsamong columns (such as that the fifth column is twice the second plus thefourth) if and only if that relationship exists after the row operation. But thisis exactly the first theorem of this book, Theorem One.I.1.5: in a relationshipamong columns,⎛ ⎞⎛ ⎞ ⎛ ⎞a 1,1a 1,n 0a 2,1c 1 ·⎜ ⎟⎝ . ⎠ + · · · + c a 2,nn ·⎜ ⎟⎝ . ⎠ = 0⎜ ⎟⎝.⎠a m,1 a m,n 0row operations leave unchanged the set of solutions (c 1 , . . . , c n ).QEDAnother way, besides the prior result, to state that Gauss’s Method hassomething to say about the column space as well as about the row space is withGauss-Jordan reduction. Recall that it ends with the reduced echelon form of amatrix, as here.⎛⎜1 3 1 6⎞⎛⎟⎜1 3 0 2⎞⎟⎝2 6 3 16⎠−→ · · · −→ ⎝0 0 1 4⎠1 3 1 60 0 0 0Consider the row space and the column space of this result. Our first point madeabove says that a basis for the row space is easy to get: simply collect together allof the rows with leading entries. However, because this is a reduced echelon formmatrix, a basis for the column space is just as easy: take the columns containingthe leading entries, that is, 〈⃗e 1 ,⃗e 2 〉. (<strong>Linear</strong> independence is obvious. The othercolumns are in the span of this set, since they all have a third component ofzero.) Thus, for a reduced echelon form matrix, we can find bases for the rowand column spaces in essentially the same way: by taking the parts of the matrix,the rows or columns, containing the leading entries.3.11 Theorem The row rank and column rank of a matrix are equal.Proof Bring the matrix to reduced echelon form. At that point, the row rankequals the number of leading entries since that equals the number of nonzero

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!