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Linear Algebra, 2020a

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102 Chapter Two. Vector Spaces<br />

unrestricted linear combinations of the two {c 1 (3x − x 2 )+c 2 (2x) | c 1 ,c 2 ∈ R}.<br />

Clearly polynomials in this span must have a constant term of zero. Is that<br />

necessary condition also sufficient?<br />

We are asking: for which members a 2 x 2 + a 1 x + a 0 of P 2 are there c 1 and c 2<br />

such that a 2 x 2 + a 1 x + a 0 = c 1 (3x − x 2 )+c 2 (2x)? Polynomials are equal when<br />

their coefficients are equal so we want conditions on a 2 , a 1 , and a 0 making that<br />

triple a solution of this system.<br />

−c 1 = a 2<br />

3c 1 + 2c 2 = a 1<br />

0 = a 0<br />

Gauss’s Method and back-substitution gives c 1 =−a 2 , and c 2 =(3/2)a 2 +<br />

(1/2)a 1 , and 0 = a 0 . Thus as long as there is no constant term a 0 = 0 we can<br />

give coefficients c 1 and c 2 to describe that polynomial as an element of the<br />

span. For instance, for the polynomial 0 − 4x + 3x 2 , the coefficients c 1 =−3 and<br />

c 2 = 5/2 will do. So the span of the given set is [S] ={a 1 x + a 2 x 2 | a 1 ,a 2 ∈ R}.<br />

Incidentally, this shows that the set {x, x 2 } spans the same subspace. A space<br />

can have more than one spanning set. Two other sets spanning this subspace<br />

are {x, x 2 , −x + 2x 2 } and {x, x + x 2 ,x+ 2x 2 ,...}.<br />

2.19 Example The picture below shows the subspaces of R 3 that we now know<br />

of: the trivial subspace, lines through the origin, planes through the origin, and<br />

the whole space. (Of course, the picture shows only a few of the infinitely many<br />

cases. Line segments connect subsets with their supersets.) In the next section<br />

we will prove that R 3 has no other kind of subspace, so in fact this lists them all.<br />

This describes each subspace as the span of a set with a minimal number<br />

of members. With this, the subspaces fall naturally into levels — planes on one<br />

level, lines on another, etc.<br />

⎛<br />

⎞<br />

⎛<br />

⎞<br />

{x ⎝ 1 0⎠ + y ⎝ 0 1⎠}<br />

0 0<br />

⎛<br />

✄<br />

✄<br />

⎞<br />

{x ⎝ 1 0⎠}<br />

0<br />

✘✘✘✘✘✘✘<br />

✘✘⎛<br />

✘ ⎞ ⎛ ⎞ ✏ ✏✏✏✏ ⎛ ⎞<br />

{x ⎝ 1 0⎠ + z ⎝ 0 0⎠}<br />

0 1<br />

⎛<br />

{x ⎝ 1 ⎞ ⎛<br />

0⎠ + y ⎝ 0 ⎞ ⎛<br />

1⎠ + z ⎝ 0 ⎞<br />

0⎠}<br />

0 0 1<br />

<br />

⎛<br />

{x ⎝ 1 1⎠ + z ⎝ 0 0<br />

0 1<br />

❆<br />

❍<br />

✏<br />

✏ ✏✏✏ ❍❍❍ ❆<br />

<br />

⎛<br />

{y ⎝ 0 ⎞ ⎛<br />

1⎠}<br />

{y ⎝ 2 ⎞ ⎛<br />

1⎠}<br />

{y ⎝ 1 ⎞<br />

1⎠} ···<br />

❳ 0<br />

0<br />

1<br />

❳ ❳<br />

❳❳ ❍<br />

❳ ❳❳ ❍❍❍<br />

❳<br />

❆ ⎛ ⎞<br />

❳❳<br />

❳<br />

{ ⎝ 0 0⎠}<br />

0<br />

⎞<br />

⎠} ···

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