Both Linearly Independent and Dependent?

Question

Is it possible for two vector functions of, for the moment's simplicity, one variable be both independent and dependent?

The reason I'm asking this is because on a problem from a book of mine (not homework), they put the following exercise:

Let $x^{(1)}(t)=\left (\begin{array}{cc} e^t \\ te^t\end{array} \right)$ and $x^{(2)}(t) = \left ( \begin{array}{cc} 1 \\ t \end{array} \right )$ . Show that $x^{(1)} (t)$ and $x^{(2)}(t)$ are linearly dependent at each point in the interval $0 ≤ t ≤ 1$. Nevertheless, show that $x^{(1)}(t)$ and $x^{(2)}(t)$ are linearly independent on $0≤t≤1$.

I would think that they're linearly dependent because $x^{(1)}(t)$ can simply be divided by the scalar $\frac{1}{e^t}$ (this is allowed because it is a never-zero exponential) to be equal to $x^{(2)}(t)$ $\forall t$, but because of the question I'm not too sure.

Could you give me some insight and/or guidance?

Answer 1

Bad notation is bad....

Show that $x^{(1)}(t)$ and $x^{(2)}(t)$ are linearly dependent at each point in the interval $0 ≤ t ≤ 1$.

What you're being asked to prove here is that given $t\in [0,1]$, the vectors $x^{(1)}(t)$ and $x^{(2)}(t)$ are linearly dependent. There is nothing wrong with this.

Nevertheless, show that $x^{(1)}(t)$ and $x^{(2)}(t)$ are linearly independent on $0≤t≤1$.

What you're being asked to prove here, is that the vectors $x^{(1)}, x^{(2)}\colon [0,1]\to \mathbb R^{2\times 1}$ (vectors as in elements of a certain vector space - for example that of functions from $[0,1]$ to $\mathbb R^{2\times 1}$ - this vectors happen to be functions) are linearly independent.

That is, you're being asked to prove that $$\forall \alpha, \beta \in \mathbb R\left[\forall t\in [0,1]\left(\alpha x^{(1)}(t)+\beta x^{(2)}(t)=\begin{pmatrix}0\\0\end{pmatrix}\right)\implies \alpha =0=\beta\right].$$

Here the author is looking at $x^{(1)}(t)$ and $x^{(2)}(t)$ as if they were functions, which they are not. The notation is wrong. Correct would be:

Nevertheless, show that $x^{(1)}$ and $x^{(2)}$ are linearly independent on $0≤t≤1$.

Answer 2

The question mixes two different vector spaces. You have two functions from $[0,1]\to\mathbb{R}^2$, $$ x^{(1)}\colon t\mapsto\begin{bmatrix}e^t\\te^t\end{bmatrix},\qquad x^{(2)}\colon t\mapsto\begin{bmatrix}1\\t\end{bmatrix} $$ These functions belong to the vector space $V$ of continuous maps from $[0,1]$ to $\mathbb{R}^2$ and form a linearly independent set in $V$. Indeed, if $$ \alpha x^{(1)}+\beta x^{(2)} $$ is the zero function, then it's easy to prove that $\alpha=\beta=0$ (just evaluate the functions at two suitable values of $t$). However, for each $t\in[0,1]$, you have $x^{(1)}(t)\in\mathbb{R}^2$ and $x^{(2)}(t)\in\mathbb{R}^2$; it can happen that, for some value of $t$ or even for all values of $t\in[0,1]$, these vectors form a linearly dependent set in the space $\mathbb{R}^2$.