basis of span with functions

Question

I'm not sure how to solve the following problem:

Determine a basis and the dimension of $\langle e^{2t}, t^2, t \rangle$ (in the space of functions from $\mathbb{R}$ to $\mathbb{R}$)

When finding the basis of a span of vectors, I create a matrix and reduce the matrix to row-echolon form. If there is a row without a pivot, I know that the corresponding vector is not part of the basis. Since this is a span of functions instead of vectors, I'm not sure how to solve this.

From what I have found on the exchange here, I know that, in order for this span to be a basis, it should satisfy the following:
$$a\cdot e^{2t}+b\cdot t^2+c\cdot t = 0 \quad\text{ if and only if }\quad a = b = c = 0$$

1. Choosing $t = 0$, shows that $a$ has to be $0$
2. Choosing $t = 1$, shows that $b = -c$ or $-b = c$

But step $2$ means that $b$ and $c$ can basically be any number right? However, according to the answer sheet the given span should be a basis. What am I missing?

And is there a way I can solve questions like these with a matrix as well?

Answer 1

In order for the linear combination of functions to be zero, it must be true that the linear combination $ae^t + bt^2 + ct = 0$ for all values of $t$. You showed that $a = 0$ by noting this for the specific value $t = 0$.

Similarly, by choosing $t = 1$ you conclude that $b = -c$. However, this doesn't tell you that any such values will in fact work (just that they will work for $t = 1$, but remember you need the values to work for any $t$). If you choose some other $t$ value, you will have more information.

For instance, choosing $t = 2$ gives you $4b = -2c$. This along with the fact that $b = -c$ tells you that $4b = 2b \implies b = c = 0$. We have now shown that in order for $ae^t + bt^2 + t = 0$ at each of the values $t = 0,1,2$ it must be that $a = b = c = 0$. This is enough to prove that the functions are linearly independent, because certainly if $ae^t + bt^2 + t = 0$ for all values of $t$, then it first must be zero for $t = 1,2,3$.

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Edit: As for solving this with a matrix as you have done in other cases, I don't see an obvious way to do it. In order to associate each of the functions with a column vector to place into a matrix, you first need to have a basis, but a priori, you don't know a basis for this span.

Answer 2

Look at $t=2$.

(FFR, even if you use $n$ values of $t$ and don't prove that the functions are independent, it doesn't necessarily mean that they aren't. To prove that they're dependent, you could e.g. algebraically manipulate a linear combination of some of the functions to show that it's equal to zero or a different function in the list.)

Answer 3

You can solve it with matrices, sort of.

To check linear independence of a set $\{f_1, f_2, f_3\}$ of twice differentiable functions on $\mathbb{R}$, you can calculate the Wronskian, the determinant defined as:$$W(f_1, f_2, f_3)(t) = \begin{vmatrix} f_1(t) & f_2(t) & f_3(t) \\ f_1'(t) & f_2'(t) & f_3'(t)\\ f_1''(t) & f_2''(t) & f_3''(t)\end{vmatrix}$$

Notice that if your functions satisfy $\alpha f_1 + \beta f_2 + \gamma f_3 = 0$ on $\mathbb{R}$, then linearity of the derivative also implies $$\alpha f_1' + \beta f_2' + \gamma f_3' = 0$$ $$\alpha f_1'' + \beta f_2'' + \gamma f_3'' = 0$$

so $W(f_1, f_2, f_3)(t)$ will be zero on $\mathbb{R}$, since the matrix columns will be dependent.

Thus, if $W(f_1, f_2, f_3)$ is not identically zero, we can conclude that $\{f_1, f_2, f_3\}$ are linearly independent. However, if $W(f_1, f_2, f_3) = 0$ you cannot outright conclude that $\{f_1, f_2, f_3\}$ is linearly dependent.

In your case, we have:

$$W(e^{2t}, t^2, t)(t) = \begin{vmatrix} e^{2t} & t^2 & t \\ 2e^{2t} & 2t & 1\\ 4e^{2t} & 2 & 0\end{vmatrix} = -2 e^{2 t} (2t^2 - 2 t + 1) \ne 0$$

for all $t \in \mathbb{R}$ so $\{e^{2t}, t^2, t\}$ is certainly linearly independent, and hence a basis for your space.

Answer 4

You can use just one point if you go to derivatives: if $ae^t+bt^2+ct=0$ for all $t$, then also \begin{gather} ae^t+2bt+c=0\\ ae^t+2b=0 \end{gather} for all $t$. Evaluating the three identities at $t=0$ yields \begin{cases} a=0\\ a+c=0\\ a+2b=0 \end{cases} so $a=b=c=0$.

Not really different from the Wronskian method, but doesn't require it.

For working with matrices, you should have at hand a (finite) basis of a subspace containing the span of the three functions, which is unlikely to be found.