Question

Consider the polynomials Bold p 1 (t )p1(t) equals = 3 3 plus + 8 8​t, Bold p 2 (t )p2(t) equals = 3 3 minus − 8 8​t, and Bold p 3 (t )p3(t) equals = 6 6 ​(for all​ t). By​ inspection, write a linear dependence relation among Bold p 1 p1​, Bold p 2 p2​, and Bold p 3 p3. Then find a basis for Span StartSet Bold p 1 comma Bold p 2 comma Bold p 3 EndSet Spanp1, p2, p3.

Answer 1

Apparently
`\mathbf{p}_1(t) + \mathbf{p}_2(t) - \mathbf{p}_3(t) = 0`.

One possible basis for this span is
`\left\{3 + 8\, t,\, 3 - 8\, t\right\}`.

Explanation:

Linear Dependence

A set of vectors is linearly-dependent if one of the vectors is a linear combination of the others.

Alternatively, to show linear dependence, show that the equation
`a\, \mathbf{p}_1 + b\, \mathbf{p}_2 + c\, \mathbf{p}_3 = 0` has a non-trivial solution (where at least one of the three scalars
`a`,
`b`, and
`c` is non-zero.)

For this set of polynomials, it can be shown that:

`\mathbf{p}_3(t) = \mathbf{p}_1(t) + \mathbf{p}_2(t)`, or equivalently,

`\mathbf{p}_1(t) + \mathbf{p}_2(t) - \mathbf{p}_3(t) = 0`.

Either way, the set
`\left\{\mathbf{p}_1,\, \mathbf{p}_2,\, \mathbf{p}_3\right\}` would be linearly-dependent.

Basis of the Span

`\mathbf{p}_3(t) = \mathbf{p}_1(t) + \mathbf{p}_2(t)` implies that
`\mathbf{p}_3` is a linear combination of
`\mathbf{p}_1` and
`\mathbf{p}_2`. Therefore,
`\mathbf{p}_3` is in the span of
`\mathbf{p}_1` and
`\mathbf{p}_2` (in other words,
`\mathbf{p}_3 \in \mathrm{Span}\left\{\mathbf{p}_1,\, \mathbf{p}_2\right\}`.) Hence:

`\mathrm{Span}\left\{\mathbf{p}_1,\, \mathbf{p}_2,\, \mathbf{p}_3\right\} = \mathrm{Span}\left\{\mathbf{p}_1,\, \mathbf{p}_2\right\}`.

Therefore, any basis of the set
`\mathrm{Span}\left\{\mathbf{p}_1,\, \mathbf{p}_2\right\}` would also be a basis of the set
`\mathrm{Span}\left\{\mathbf{p}_1,\, \mathbf{p}_2,\, \mathbf{p}_3\right\}`.

On the other hand, it can be shown that
`\mathbf{p}_1` and
`\mathbf{p}_2` are linearly-independent. Therefore,
`\mathrm{Span}\left\{\mathbf{p}_1,\, \mathbf{p}_2\right\}` should have a dimension of
`2`. As a result, there should be exactly
`2` linearly-independent vectors in a basis of
`\mathrm{Span}\left\{\mathbf{p}_1,\, \mathbf{p}_2\right\}`.

There are many different choices for the basis of
`\mathrm{Span}\left\{\mathbf{p}_1,\, \mathbf{p}_2\right\}`. One possible choice is the set
`\left\{\mathbf{p}_1,\, \mathbf{p}_2\right\}`, which is equal to
`\left\{3 + 8\, t,\, 3 - 8\, t\right\}`. Make sure that this set is indeed linearly-independent and contains two vectors.

Because it has already been shown that
`\mathrm{Span}\left\{\mathbf{p}_1,\, \mathbf{p}_2,\, \mathbf{p}_3\right\} = \mathrm{Span}\left\{\mathbf{p}_1,\, \mathbf{p}_2\right\}`,
`\left\{3 + 8\, t,\, 3 - 8\, t\right\}` should be a basis of the set
`\mathrm{Span}\left\{\mathbf{p}_1,\, \mathbf{p}_2,\, \mathbf{p}_3\right\}`, as well.