Linear Algebra

Section III. Basis and Dimension 1253.5 Example From any matrix, we can produce a basis for the row space byperforming Gauss’ method and taking the nonzero rows of the resulting echelonform matrix. For instance,⎛⎝ 1 3 1 ⎞⎛1 4 1⎠ −ρ 1+ρ 2 6ρ 2 +ρ 3−→ −→ ⎝ 1 3 1 ⎞0 1 0⎠−2ρ2 0 51+ρ 30 0 3produces the basis 〈 ( 1 3 1 ) , ( 0 1 0 ) , ( 0 0 3 ) 〉 for the row space. Thisis a basis for the row space of both the starting and ending matrices, since thetwo row spaces are equal.Using this technique, we can also find bases for spans not directly involvingrow vectors.3.6 Definition The column space of a matrix is the span of the set of itscolumns. The column rank is the dimension of the column space, the numberof linearly independent columns.Our interest in column spaces stems from our study of linear systems. Anexample is that this systemc 1 + 3c 2 + 7c 3 = d 12c 1 + 3c 2 + 8c 3 = d 2c 2 + 2c 3 = d 34c 1 + 4c 3 = d 4has a solution if and only if the vector of d’s is a linear combination of the othercolumn vectors, ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞1 3 7 d 1c 1⎜2⎟⎝0⎠ + c ⎜3⎟2 ⎝1⎠ + c ⎜8⎟3 ⎝2⎠ = ⎜d 2⎟⎝d 3⎠4 0 4 d 4meaning that the vector of d’s is in the column space of the matrix of coefficients.3.7 Example Given this matrix,⎛ ⎞1 3 7⎜2 3 8⎟⎝0 1 2⎠4 0 4to get a basis for the column space, temporarily turn the columns into rows andreduce.⎛⎝ 1 2 0 4⎞⎛3 3 1 0⎠ −3ρ 1+ρ 2 −2ρ 2 +ρ 3−→ −→ ⎝ 1 2 0 4⎞0 −3 1 −12⎠−7ρ7 8 2 41+ρ 30 0 0 0