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From Ex 6.43 of P474 asks
... we check that $\{1 + x, 1 - x, x^2\}$ is linearly independent. Can you see a quick way to tell this?

I thought to consider a harder question instead:

P476 6.2.50. Find a basis for $\operatorname{span}\{p_1 = 1 -2x,\; p_2 = 2x - x^2, \; p_3 = 1 - x^2,\; p_4 = 1 + x^2\}$.

I grasp the traditional way, given by http://www.math.odu.edu/~bogacki/cgi-bin/lat.cgi:
In summary, $c_1p_1 + c_2p_2 + c_3p_3 + c_4p_4 = 0 \Longrightarrow RREF \Longrightarrow \begin{bmatrix} \color{#B8860B}{1} & ~ & 1 & ~ \\ ~ & \color{#B8860B}{1} & 1 & ~ \\ ~ & ~ & ~ & \color{#B8860B}{1} \\ \end{bmatrix}\begin{bmatrix} c_1 \\ c_2 \\ c_3 \\ c_4 \\ \end{bmatrix} \mathbf{=0}$.
The (bronze) pivots are in columns $1, 2, 4$, so $\operatorname{basis}\operatorname{span}\{p_1, p_2, p_3, p_4\} = \operatorname{span}\{p_1, p_2, \require{cancel} \cancel{p_3}, p_4\}.$
Moreover, this RREF of the homogenous system reveals : $c_1 = c_2 = -c_3 \in \mathbb{C}$ and $c_4 = 0$, so $p_3 \in \operatorname{span}\{p_1,p_2\}.$

$\Large{{1.}}$ What are the quick/devious ways to do the above?

$\Large{2.}$ If two "vectors"/functions are of different degrees, is the set formed by these $2$ automatically linearly independent? I'm thinking that since $1 - 2x$ is of degree $1$ and all the others "vectors"/functions are of degree $2$, $\operatorname{span}\{1 - 2x, \text{any single degree 2 polynomial}\}$ must be linearly independent. This agrees with the answer.

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More generally, if $n$ polynomials are of different degrees, then they are linearly independent. If the polynomials are $p_1,p_2,\dots,p_n$ of degrees $d_1\gt d_2\gt\dots\gt d_n$, and $a_1p_1+a_2p_2\cdots+a_np_n=0$, then we must have $a_1=0$, since there's nothing else to cancel out the term of degree $d_1$; then we must have $a_2=0$, since there's nothing else to cancel out the term of degree $d_2$; and so on.

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  • $\begingroup$ I just finished teaching a linear algebra course and called (essentially) this the eyeball criterion for linear independence: if you have a finite set of vectors in $\mathbb{R}^n$ such that reading from left to right each is nonzero in some component for which the previous vectors are all zero, then the set is necessarily linearly independent. Then one can think about Gaussian reduction as a process that massages a set of vectors until the eyeball criterion applies. $\endgroup$ – Pete L. Clark Dec 13 '13 at 18:40
  • $\begingroup$ @PeteL.Clark: Many thanks. Unfortunately, the book doesn't publicise this. How would I apply this to the last 3 vectors to the span above? $\endgroup$ – NNOX Apps Dec 16 '13 at 6:12
  • $\begingroup$ @LePressentiment: I am having a lot of trouble following what you've written. It looks to me like you have several misconceptions, perhaps beyond the scope of a focused Q&A site to fully address. To pick out a few: linear independence is a property of a set of vectors, rather than of a single vector or a pair of vectors. Or more precisely, linear independence makes sense for a single vector but it just means that that vector is nonzero, so "spotting that $1-2x$ is linearly dependent" is a strange thing to say (even assuming that you meant "linearly independent, which you may have)... $\endgroup$ – Pete L. Clark Dec 16 '13 at 8:31
  • $\begingroup$ Since $1-2x$ is not the zero polynomial, it is linearly independent. Your question about two different polynomials having different degrees being linearly independent has a more trivial answer than you seem to think: two nonzero vectors are linearly dependent exactly when one is a scalar multiple of the other. Clearly this cannot happen for two nonzero polynomials of different degrees. Gerry's answer is adjusting your question so as to give an answer which could be useful in practice. However his criterion does not apply to your examples: I think you may be confused about that too. $\endgroup$ – Pete L. Clark Dec 16 '13 at 8:35
  • $\begingroup$ Your sentence "On the condition that this is true, since 1−2x is of degree 1 and all the others "vectors"/functions are of degree 2, thus 1−2x must be linearly independent" has several problems with it: first, you can't be asking whether $1-3x$ is linearly independent: it is nonzero, so it is linearly independent: I think you mean to ask whether it is in the span of the other three polynomials (although I'm not sure why you're asking that). But this has nothing to do with Gerry's criterion (nor did he claim that it did)... $\endgroup$ – Pete L. Clark Dec 16 '13 at 8:39

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