Linear Algebra, 2020a

138 Chapter Two. Vector Spaces
meaning 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 4
to get a basis for the column space, temporarily turn the columns into rows and
reduce.
⎛
⎜
1 2 0 4
⎞
⎛
⎟
⎝3 3 1 0⎠ −3ρ 1+ρ 2 −2ρ 2 +ρ 3 ⎜
1 2 0 4
⎞
⎟
−→ −→ ⎝0 −3 1 −12⎠
−7ρ 1 +ρ 3
7 8 2 4
0 0 0 0
Now turn the rows back to columns.
⎛ ⎞ ⎛ ⎞
1 0
2
〈 ⎜ ⎟
⎝0⎠ , −3
⎜ ⎟
⎝ 1 ⎠ 〉
4 −12
The result is a basis for the column space of the given matrix.
3.8 Definition The transpose of a matrix is the result of interchanging its rows
and columns, so that column j of the matrix A is row j of A T and vice versa.
So we can summarize the prior example as “transpose, reduce, and transpose
back.”
We can even, at the price of tolerating the as-yet-vague idea of vector spaces
being “the same,” use Gauss’s Method to find bases for spans in other types of
vector spaces.
3.9 Example To get a basis for the span of {x 2 + x 4 ,2x 2 + 3x 4 , −x 2 − 3x 4 } in
the space P 4 , think of these three polynomials as “the same” as the row vectors
(0 0 1 0 1), (0 0 2 0 3), and (0 0 −1 0 −3), apply Gauss’s Method
⎛
⎞
⎛
0 0 1 0 1
⎜
⎟
⎝0 0 2 0 3 ⎠ −2ρ 1+ρ 2 2ρ 2 +ρ 3 ⎜
0 0 1 0 1
⎞
⎟
−→ −→ ⎝0 0 0 0 1⎠
ρ 1 +ρ 3
0 0 −1 0 −3
0 0 0 0 0
and translate back to get the basis 〈x 2 + x 4 ,x 4 〉. (As mentioned earlier, we will
make the phrase “the same” precise at the start of the next chapter.)