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Let $W= \text{span}(p_1,p_2,p_3), W \subseteq R_2[x]$

$p_1(x)=1+2x+x^2$

$p_2(x)=3-9x^2$

$p_3(x)=1+4x+5x^2$

From $p_1,p_2,p_3$ choose a basis B for $W$.

I got a problem because they are asking for a basis but my solution says that they are linearly dependent... How can I get a basis anyway?

Form these polynomials to vectors, the form is $a+bx+cx^2$, so we have:

$$\vec{p_1}=\begin{pmatrix} 1\\ 2\\ 1 \end{pmatrix}, \vec{p_2}=\begin{pmatrix} 3\\ 0\\ -9 \end{pmatrix}, \vec{p_3}=\begin{pmatrix} 1\\ 4\\ 5 \end{pmatrix}$$

Now we need to check if they are linearly independent. I used determinant trick:

$$\begin{vmatrix} 1 & 3 & 1\\ 2 & 0 & 4\\ 1 & -9 & 5 \end{vmatrix}\begin{matrix} 1 & 3\\ 2 & 0\\ 1 & -9 \end{matrix}$$

If we use Saruss, indeed, the determinant is zero and thus the vectors are linearly dependent. So we cannot choose any of them as a basis..

Is there a way to get a basis anyway? Maybe I can choose one of the polynomials $p_1,p_2,p_3$ as a basis because a single one of them should be linearly independent.

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  • $\begingroup$ What is $R_2[x]$? $\endgroup$ Mar 14, 2017 at 11:59
  • $\begingroup$ @RobertoRastapopoulos It means that the degree of polynomials is $\leq 2$ $\endgroup$
    – cnmesr
    Mar 14, 2017 at 12:00
  • $\begingroup$ What is $M$? It looks like it should be $W$. $\endgroup$ Mar 14, 2017 at 12:16
  • $\begingroup$ @CatalinZara Thank you corrected! $\endgroup$
    – cnmesr
    Mar 14, 2017 at 12:17

1 Answer 1

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Use row-reduction to find the redundant vector/column/polynomial(s); the remaining one(s) form a basis of $W$.

More details: $$\begin{bmatrix} 1 & 3 & 1 \\ 2 & 0 & 4 \\ 1 & -9 & 5 \end{bmatrix} \Longrightarrow \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & -1/3 \\ 0 & 0 & 0 \end{bmatrix} $$ tells you that $p_3(x) = 2p_1(x) - 1/3 p_2(x)$, hence $p_3$ is redundant of $p_1 $ and $p_2$.

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  • $\begingroup$ Awesome, this helped! I did that and I have the matrix: $\begin{pmatrix} 3 & 0 & 0\\ 6 & 6 & 0\\ 3 & 12 & 0 \end{pmatrix}$. Assuming my calculation to this matrix is correct, would you say I can conclude the basis is $B = \left\{ \begin{pmatrix} 3\\ 6\\ 3 \end{pmatrix}, \begin{pmatrix} 0\\ 6\\ 12 \end{pmatrix}\right\}$? $\endgroup$
    – cnmesr
    Mar 14, 2017 at 12:26
  • $\begingroup$ Yes, $B$ is a basis for the column space of the matrix (i.e. subspace spanned by the columns of the matrix). Not surprisingly, since the last column is the zero vector and the first two are not proportional - hence those two columns are linearly independent. $\endgroup$ Mar 14, 2017 at 12:30
  • $\begingroup$ Great thanks! What if I would still end up with linearly dependent vectors after that process, or if I wouldn't get zero columns? $\endgroup$
    – cnmesr
    Mar 14, 2017 at 12:31
  • $\begingroup$ After you row-reduce, look for the columns with pivots. The corresponding columns of the initial matrix are linearly independent and form a basis for the column space. See my answer to math.stackexchange.com/questions/1697260/… $\endgroup$ Mar 14, 2017 at 12:36

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