Linear Dependent Span

Question

$\{x \cos x, x, \cos x \}$ is a subspace of $V$.

I need to find if it's a linear dependent or linear independent.

So I thought that its dependent since $x \cos x$ is multiplication of $x$ and $\cos x$.

am I right? and if i'm wrong then why?

Answer 1

What is $V$? Sounds like a space of functions, but you'll have to be more specific. In any case, if you have three objects in a vector space, you don't just need one to be a product of the other two, for them to be dependent, you need it to be a linear combination $a_1v_1+a_2v_2$. So in the space of continuous functions $\Bbb R\to\Bbb R$ these three are independent. This is a special space where multiplication of vectors is defined, but you cannot use that operation to interrogate the vector space properties.

Answer 2

The set $\{x\cos x, x, \cos x\}$ is linearly independent as shown below,

Let $a,b,c \in \mathbb{R}$ such that

$ax\cos x+bx+c\cos x=0 . . . . . . (1)$

putting $x=0$ in (1) we get $c=0$

putting $x=\frac{\pi}{2}$ in (1) we get $b=0$

putting $x=2 \pi$ in (1) we get $a=0$

Answer 3

The set $V$ contains all functions in the form:

$$v(x) = c_1x \cos x + c_2 x +c_3 \cos x,$$

for any real valued coefficients $c_1, c_2$ and $c_3$.

Maybe your question is:

Is the base of $V$ (let's call it $B = \{x\cos x, x, \cos x\}$) minimal?

Or equivalently:

Are all the elements of $B$ linearly independent?

For this, you need to prove that

$$v(x) = 0 \iff c_1 = c_2 = c_3 = 0.$$

Case "$\Leftarrow$"

It's obvious. Just put $c_1 = c_2 = c_3 = 0$ in the definition of $v(x)$ and obtain that $v(x) = 0$.

Case "$\Rightarrow$"

Suppose that $v(x) = c_1x \cos x + c_2 x +c_3 \cos x = 0$. This must be true for all $x$. Let's consider $x= 0$. Then:

$$v(0) = c_3 = 0 \Rightarrow c_3 = 0$$

Now, with $c_3 = 0$, consider $x=\pi$. Then:

$$v(\pi) = -c_1 \pi + c_2 \pi = 0 \Rightarrow c_1=c_2$$

At this point, we need to work on $v(x) = c_1 x (\cos x - 1)$. This function is identically null iff $c_1 = 0$. Then we proved that $v(x) = 0 \Rightarrow c_1 = c_2 = c_3 = 0$. We conclude saying that $B$ is minimal or all its elements are linearly independent.