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How do we show that the set $B=\{1,x+1,x(x+1),x(x+1)(x-1)\}$ is a basis for a linear polynomial space?

I know that we need to do two things:

  1. Show that our set is a spanning set, meaning that every $w \in W$ can be written as $w=a_1\vec{w_1}+a_2\vec{w_2}+\ldots+a_n\vec{w_n}$
  2. Show that the polynomials in our set are linearly independent, meaning that the only way $a_1\vec{w_1}+a_2\vec{w_2}+\ldots+a_n\vec{w_n}=\vec{0}$ can be true is when $a_0=a_1=\ldots=a_n=0$ is true.

But how do we show that?

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1  
you mean for polynomials of degree at most $3$? –  Julian Kuelshammer Nov 11 '12 at 21:33
    
Ah, yes. I forgot to mention that. –  user1132363 Nov 11 '12 at 21:33

3 Answers 3

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For both parts you can simply work from the definitions.

The polynomial space in question is $P_3$, the space of polynomials of degree at most $3$. You show that $B$ spans $P_3$ by showing how to write an arbitrary $p(x)\in P_3$ as a linear combination of members of $B$, so let $p(x)=a_0+a_1x+a_2x^2+a_3x^3\in P_3$. You want to find coefficients $c_0,c_1,c_2,c_3$ such that

$$p(x)=c_0\cdot1+c_1(x+1)+c_2x(x+1)+c_3x(x+1)(x-1)\;.$$

Multiply out the righthand side: you want

$$p(x)=(c_0+c_1)+(c_1+c_2-c_3)x+c_2x^2+c_3x^3\;.$$

Now equate coefficients:

$$\left\{\begin{align*} &c_0+c_1=a_0\\ &c_1+c_2-c_3=a_1\\ &c_2=a_2\\ &c_3=a_3 \end{align*}\right.\tag{1}$$

Finally, show that the resulting system can always be solved for $c_0,c_1,c_2$ and $c_3$ in terms of $a_0,a_1,a_2$, and $a_3$; that assures you that every $p(x)\in P_3$ is a linear combination of elements of $B$.

Similarly, to show that $B$ is a linearly independent set, try to do what the definition of linear independence says you must do to show that a set is linearly independent: assume that $c_0,c_1,c_2$, and $c_3$ are scalars such that

$$c_0\cdot1+c_1(x+1)+c_2x(x+1)+c_3x(x+1)(x-1)=0\;,$$

and show that this implies that $c_0=c_1=c_2=c_3=0$. Most of the work has already been done in showing that $B$ spans $P_3$: you’re really just showing that when $a_0=a_1=a_2=a_3=0$, the system $(1)$ has the unique solution $c_0=c_1=c_2=c_3=0$, and if you’ve already done the first part, you already know that $(1)$ has a unique solution for every choice of $a_0,a_1,a_2$, and $a_3$.

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Thank you! Very understandable. PS: I think you made a small syntax error in the last coefficient equation (should probably be $c_3=a_3$). –  user1132363 Nov 12 '12 at 0:11
    
@user1132363: You’re welcome. And thanks for catching the typo: my little finger strayed one key to the right and hit + instead of _. –  Brian M. Scott Nov 12 '12 at 16:34

For $(2)$, to show that the polynomials in the set $$B=\{1,x+1,x(x+1),x(x+1)(x-1)\}=\{1, x+1, x^2+x, x^3 -x\}$$ are linearly independent:

We can use the Wronskian $W$ of a set of functions, in this case $B$, to establish linear independence.

Each function, in this case polynomials, is listed in the first row of a matrix; in this case, we want to facilitate the computation of the determinant, and so in the first row we go from the constant 1 in the first entry to the largest degree polynomial in the last entry of the row. The corresponding entries of each subsequent row is the derivative of the entries in the row immediately above it. The Wronskian is the determinant of the resulting matrix.

Applying this to the polynomials in $B$ gives us a "nice" upper triangular matrix from which we can easily compute the determinant $W$ as the product of the diagonal entries:

$$W[1, x+1, x^2 +x, x^3 - x] = \left|\begin{bmatrix} 1 & x+1 & x^2+x & x^3-x\\ 0 & 1 & 2x+1 & 3x^2-1\\ 0 & 0 & 2 & 6x\\ 0 & 0 & 0 & 6\\ \end{bmatrix} \right|= 1\cdot 1 \cdot 2 \cdot 6 = 12 \neq 0$$

Therefore, since the Wronskian (determinant) is not equal to $0$, the polynomials in $B$ must be linearly independent.


In general, the Wronskian $W$ is given by:

enter image description here

$$W[f_1(x), f_2(x), \dots f_n(x)] \neq 0 \implies f_1(x), f_2(x), \dots f_n(x) \text{ are linearly independent}.$$


One word of caution, if the Wronskian $W$ of such a matrix is identically $0$ (in other words, if $W \not\equiv 0$), we cannot conclude anything about whether or not the corresponding functions are linearly dependent.

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I was not familiar with Wronskian W. It seems extremely useful. Thank you for that! –  user1132363 Nov 11 '12 at 22:11
    
Yes, it is useful, especially with how cleanly they work with polynomials! –  amWhy Nov 11 '12 at 22:14

Hint1: To show that this is a spanning set, you can express a spanning set/basis with your new elements

Hint2: To show it is linearly independent, just use a dimension argument.

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