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Let $p_1 = 1 - 2t - t^2$, $p_2 = t + t^2 + t^3$, $p_3 = 1 - t + t ^3$ and $p_4 = 3 + 4t + t^2 + 4t^3$.

Let $S$ be the set of these four functions. Find a subset of $S$ that is a basis for the span of $S$.

So I've started out by taking these functions, making them vectors and putting them into a matrix:

$$ \begin{bmatrix} 0 & -1 & -2 & 1\\ 1 & 1 & 1 & 0\\ 1 & 0 & -1 & 1\\ 4 & 1 & 4 & 3 \end{bmatrix} $$

Then I reduced it:

$$ \sim \begin{bmatrix} 1 & 0 & 0 & 1\\ 0 & 1 & 0 & -1\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix} $$

I guess my question is, is this the correct way to do this? Where do I go from here? What does a zero row mean for a basis?

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I did not check the details of whether or not your calculations are correct, but assuming what you have done is correct it means that your $4$ elements span a $3$ dimensional vector space. Hence you only need three vectors. In particular, if you are given $v_1,...,v_k$ to be linearly dependent, you can delete one, leaving the space spanned intact. After repeating this process finite many times you will get a linearly independent set that spans your space, (that is a basis). Now the question is which one to delete? Delete the one that can be written in terms of the others. –  Daniel Montealegre Apr 2 '12 at 22:06
    
There is a single zero row, so (i) we have linear dependence and (ii) the dimension is $3$, so your basis will have $3$ elements. You can check, using your row reduction, whether taking the first three vectors will do the job. –  André Nicolas Apr 2 '12 at 22:07

2 Answers 2

up vote 1 down vote accepted

A better way is to put them into as matrix as columns instead of as rows. The elementary row operations preserve linear dependence relations among the columns, so when you get to reduced form, if, say, columns 1, 2, and 4 are a basis for the column space, then $p_1,p_2,p_4$ are a basis for the original space.

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Yes, its the right way, assuming that you reduced your matrix without any mistakes. Now, the zero row mean, that this row are linear dependable on the first three rows, hence your basis for $S$ is $\text{Span}\{p_1,p_2,p_3\}$.

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Not so fast. Row reduction allows you to interchange rows, so you know there is one polynomial you can delete, but you don't know for sure that it's $p_4$. –  Gerry Myerson Apr 2 '12 at 23:16

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