I'm presented with the question:
Show that the given vectors are linearly independent in $\mathcal{C}[0,1]$:
$x^{3/2}, x^{5/2}$
I'm having a terrible time understanding linearly algebra in general. I think my part of my problem with this question is understanding what the $\mathcal{C}[0,1]$ notation means. Beyond that, I'm still not exactly sure how to show this. Any help would be greatly appreciated.
$C[0,1]$ is the vector space of continuous functions from $[0,1]$ into $\mathbb{R}$. You can add functions, subtract them, there is a zero function, multiply by elements in $\mathbb{R}$, etc. Thus, it makes sense to ask about linearly independent elements of this vector space. Take the elements $x^{3/2}$ and $x^{5/2}$. We want to show that these are linearly independent, so that one is not the multiple of another by an element in $\mathbb{R}$. Can you show this?
$\mathcal{C}[0,1]$ is the set of continuous functions from the interval $[0,1]$ to $\mathbb R$. So you're asked to show that the functions $f$ and $g$ on $[0,1]$ defined by $f(x) = x^{3/2}$ and $g(x) = x^{5/2}$ are linearly independent.
To show this, assume that $\lambda, \mu \in {\mathbb R}$ satisfy $\lambda f + \mu g = 0$, i.e., for all $x \in [0,1]$, $\lambda x^{3/2} + \mu x^{5/2} = 0$. Now plug in some values for $x$ and derive that $\lambda = \mu = 0$.
$\alpha x^{3/2}+\beta x^{5/2} = 0$ $\Longrightarrow $ $$ x^{3/2}(\alpha + \beta x)=0 $$ $$ \beta x = -\alpha $$ or $$ x^{3/2}=0 $$ Can you finish?
$C[0,1]$ usually denotes the collection of continuous functions $f: [0,1]\to \mathbb{R}$. This is a vector space over $\mathbb{R}$, with multiplication defined pointwise: $(af)(x) = af(x)$.
What would it mean if two functions, $f,g\in C[0,1]$, were linearly dependent? It would meant that there were two scalars, $a,b\in\mathbb{R}$, not both zero, such that $af + bg = 0$ as functions. In other words, for all $x$, $af(x) + bg(x) = 0$.
To show that $f=x^{3/2}$ and $g=x^{5/2}$ are linearly independent, let's assume that such $a$ and $b$ exist, and try to show that they're both zero.. If such $a,b\in\mathbb{R}$ exist, then $ax^{3/2} + bx^{5/2} = 0$ for all $x$. Let's try plugging in $x=1$—we get $a+b=0$. Plugging in $x=1/4$, we get $a/8 + b/32=0$ (note: you could also plug in different values of $x$, but I chose these to keep the math as simple as possible).
Now, we can solve the two equations $a+b=0$ and $a/8+b/32=0$, to get $a=b=0$. Ah, but this is exactly what we wanted to show: $x^{3/2}$ and $x^{5/2}$ cannot have a linear sum which equal zero unless it is the trivial sum, i.e. $0\cdot f + 0\cdot g = 0$.
Hint: You can use the Wronskian.
$$ W(f_1, \ldots, f_n) (x)= \begin{vmatrix} f_1(x) & f_2(x) & \cdots & f_n(x) \\ f_1'(x) & f_2'(x) & \cdots & f_n' (x)\\ \vdots & \vdots & \ddots & \vdots \\ f_1^{(n-1)}(x)& f_2^{(n-1)}(x) & \cdots & f_n^{(n-1)}(x) \end{vmatrix},\qquad x\in I. $$
In order for the functions to be linearly independent, then the Wronskian does not equal zero.
Added:
In your case, you want to prove
$$ W(x^{3/2}, x^{5/2}) (x)= \begin{vmatrix} x^{3/2} & x^{5/2} \\ \frac{3}{2}x^{1/2} & \frac{5}{2}x^{3/2} \end{vmatrix}$$
Now, you need to calculate the above determinant and prove it does not equal to zero.
(Essentially, put anything you want to format between two $ signs.)
– Tom Oldfield Oct 03 '13 at 21:14