# Dimension of span(Z)

• Jan 17th 2010, 09:05 AM
MMAfanatic
Dimension of span(Z)
thanks for the help!
• Jan 17th 2010, 10:36 AM
Dinkydoe
First: You can write these polynomials $\displaystyle f_1,\cdots f_5$ with respect to the basis $\displaystyle \left\{1,x,x^2,x^3,x^4\right\}$

Thus we can write:

$\displaystyle f_1 = (1,0,1,1,0)$
$\displaystyle f_2= (2,1,2,0,1)$
$\displaystyle f_3 = (-4,-3,-4,2,-3)$
$\displaystyle f_4=(1,-3,0,1,0)$
$\displaystyle f_5= (13,-1,11,3,5)$

Put these vectors in a matrix and by performing elementary row-operations we get:

\displaystyle \left[ \begin {array}{ccccc} 1&0&1&1&0\\\noalign{\medskip}2&1&2&0&1 \\\noalign{\medskip}-4&-3&-4&2&-3\\\noalign{\medskip}1&-3&0&1&0 \\\noalign{\medskip}13&-1&11&3&5\end {array} \right] \to \left[ \begin {array}{ccccc} 1&0&1&1&0\\\noalign{\medskip}0&1&0&-2&1 \\\noalign{\medskip}0&0&-1&-6&3\\\noalign{\medskip}0&0&0&0&0 \\\noalign{\medskip}0&0&0&0&0\end {array} \right]

We see that span$\displaystyle (Z)\subset \mathcal{P}_4(\mathbb{R})$ is a 3-dimensional subset and:

Now, letting:

$\displaystyle g_1 = 1+x^2+x^3$
$\displaystyle g_2 = x-2x^3+x^4$
$\displaystyle g_3=-x^2-6x^3+3x^4$:

$\displaystyle Y: = \left\{g_1,g_2,g_3 \right\}$ is such that span($\displaystyle Y$)= span($\displaystyle Z$)
• Jan 17th 2010, 11:24 AM
MMAfanatic
hey thank you for your help.

could u please send the operations you used to row reduce the matrix please.

when i did it i just did row 2 - 2 x row 1, row 3 - 4 x row 1, row 4 - row 1, row 5 - 13 x row 1, row 3 + 3x row 2

so im left with

1 0 1 1 0
0 1 0 -2 1
0 -3 -1 0 0
0 -1 -2 -10 5

i know that i am wrong because im not sure how to row reduce properly i dont think :-(

thanks for you help again.
• Jan 17th 2010, 12:18 PM
Dinkydoe
Actually I let Maple calculate the reduced form. I admit it can be quite tedious work, but it's still probably the simplest method.
• Jan 17th 2010, 12:40 PM
MMAfanatic
oh right...fair enough lol.

umm i know row 2 is row 2 - 2 x row 1, dont know how the other rows came about?

can anyone help on the row reducing to get the final product.

sorry guys for this.

major thanks to the original helper and anyone else who may be able to help me out :-)
• Jan 17th 2010, 12:53 PM
MMAfanatic
i figured it

:-)

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• Jan 17th 2010, 01:00 PM
Dinkydoe
Very good. Now find even 2 more linear independant vectors: $\displaystyle v_1,v_2\in\mathbb{R}^5$ , write them on the basis $\displaystyle \left\{1,x,x^2,x^3,x^4\right\}$ to find polynomials $\displaystyle p_1,p_2$

Then together with $\displaystyle Y$ you have a basis for $\displaystyle \mathcal{P}_4(\mathbb{R})$
• Jan 17th 2010, 02:53 PM
MMAfanatic
Quote:

Originally Posted by Dinkydoe
Very good. Now find even 2 more linear independant vectors: $\displaystyle v_1,v_2\in\mathbb{R}^5$ , write them on the basis $\displaystyle \left\{1,x,x^2,x^3,x^4\right\}$ to find polynomials $\displaystyle p_1,p_2$

Then together with $\displaystyle Y$ you have a basis for $\displaystyle \mathcal{P}_4(\mathbb{R})$

• Jan 17th 2010, 03:11 PM
Dinkydoe
No, it's the part where they ask to extend $\displaystyle Y$ to find a basis for $\displaystyle \mathcal{P}_4(\mathbb{R})$.

But they can be used to extend Y.

Namely $\displaystyle p_0(x) = 1$ is independant of Y and $\displaystyle p_1(x)= x$ is independant of Y

We can even show that: $\displaystyle Y\cup \left\{p_0(x),p_1(x)\right\}$ is a basis for $\displaystyle \mathcal{P}_4(\mathbb{R})$

You can do this by showing that:

\displaystyle \left[ \begin {array}{ccccc} 1&0&1&1&0\\\noalign{\medskip}0&1&0&-2&1 \\\noalign{\medskip}0&0&-1&-6&3\\\noalign{\medskip}0&1&0&0&0 \\\noalign{\medskip}1&0&0&0&0\end {array} \right]

has rank 5.
• Jan 17th 2010, 03:36 PM
MMAfanatic
Quote:

Originally Posted by Dinkydoe
No, it's the part where they ask to extend $\displaystyle Y$ to find a basis for $\displaystyle \mathcal{P}_4(\mathbb{R})$.

But they can be used to extend Y.

Namely $\displaystyle p_0(x) = 1$ is independant of Y and $\displaystyle p_1(x)= x$ is independant of Y

We can even show that: $\displaystyle Y\cup \left\{p_0(x),p_1(x)\right\}$ is a basis for $\displaystyle \mathcal{P}_4(\mathbb{R})$

You can do this by showing that:

\displaystyle \left[ \begin {array}{ccccc} 1&0&1&1&0\\\noalign{\medskip}0&1&0&-2&1 \\\noalign{\medskip}0&0&-1&-6&3\\\noalign{\medskip}0&1&0&0&0 \\\noalign{\medskip}1&0&0&0&0\end {array} \right]

has rank 5.

Ok now im totally lost. I agree that matrix has rank 5 but where did you get that matrix from? Its different to our row operationed matrix we had before.
• Jan 17th 2010, 04:23 PM
Dinkydoe
It's the same matrix with the 3 upper rows:

with $\displaystyle p_0 = (1,0,0,0,0)$ and $\displaystyle p_1 = (0,1,0,0,0)$ instead of the zero-rows.

Where $\displaystyle p_0,p_1$ are the vector representations of $\displaystyle p_1(x), p_2(x)$w.r.t. the basis $\displaystyle \left\{1,x,x^2,x^3,x^4,\right\}$

You want to show that $\displaystyle Y$ is linear independant of $\displaystyle p_1(x),p_2(x)$. You can do that by showing that $\displaystyle g_1,g_2,g_3,p_1,p_2$ are linear independant, that is, the given matrix has rank 5. (meaning, having 5 linear independant rows).
• Jan 17th 2010, 05:49 PM
MMAfanatic
oh i see.

i know why i was confused. i disagree with what your final row operated product looks like

i have

1 0 1 1 0

0 1 0 -2 1

0 0 1 6 -3

0 0 0 0 0

0 0 0 0 0
• Jan 17th 2010, 05:55 PM
Dinkydoe
You're allways free to disagree with a computer output

I didn't do the calculations, do whatever suits you.

edit: Actually they're the same matrices.