Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $W$ is a set of functions with degree at most 1, and so $$W_1=\{a_0 +a_1 \cdot x | a_0,a_1 \in \mathbb{R} \}$$

I know that the basis of this is the set $B_1 = \{1,x\}$. But I am having trouble showing linearly independance. I know we choose scalers $c_1, c_2$ such that $c_1(1) + c_2(x) = 0$ but I dont know where to go from here. Also I know the dimension is 2 but why is this? Is this because the number of elements in the basis is 2 or because dim(W) = n+1 for polynomials.

share|improve this question
2  
"A set of functions", or "a set of polynomials with real coefficients"? (Not the same thing!) –  Arturo Magidin Sep 29 '11 at 19:43
    
The dimension of a vector space is the number of vectors in a basis for that space. The reason that $\dim W_n=n+1$ is that $\{1,x,x^2,\dots,x^n\}$ is a basis for $W_n$ that has $n+1$ members. –  Brian M. Scott Sep 29 '11 at 19:52

2 Answers 2

up vote 1 down vote accepted

The expression $c_1(1) + c_2(x) = 0$ is an equality of two polynomials. But two polynomials are equal if and only if they are identical; that is, $$a_0 + a_1x + \cdots + a_nx^n = b_0+b_1x + \cdots b_nx^n$$ if and only if $a_0=b_0$, $a_1=b_1,\ldots,a_n=b_n$.

Here you have $c_1(1) + c_2(x) = 0 = 0(1) + 0(x)$, so $c_1=c_2=0$.

To prove they span $W$, let $a+bx$ be an arbitrary element of $W$. Then $a+bx = a(1)+b(x) \in \mathrm{span}(1,x)$, so $W\subseteq \mathrm{span}(1,x)$. Since each of $1$ and $x$ lie in $W$, $\mathrm{span}(1,x)\subseteq W$, giving equality.

share|improve this answer

$c_1(1) + c_2(x) = 0$ is an equality of functions. So, evaluate both sides at $x=0$ to get $c_1=0$ and then at $x=1$ to get $c_2=0$.

share|improve this answer
    
So that shows in linearly independant. How would I show span(1, x) = W? –  Tyler Hilton Sep 29 '11 at 19:42
1  
@Tyler: They span $W$ because $W$ is defined as the span of $1$ and $x$ (your definition). –  Fredrik Meyer Sep 29 '11 at 19:46
    
Hi, can you show how evaluating them is the right way to do it. I thought we are supposed to show linear independance for ALL x. so why would concrete values for x –  Tyler Hilton Sep 30 '11 at 1:57
    
@Tyler, there are only 2 unknowns here: $c_1$ and $c_2$. So you only need 2 equations to find them. Actually, any two different values if $x$ will do. –  lhf Sep 30 '11 at 2:51

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.