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

Answer:

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

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

Step-by-step 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.