# Subspaces & spans

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• Dec 3rd 2009, 11:28 AM
SirOJ
Subspaces & spans
not the following subsets of
R3 are subspaces of R3:
(i)
{(x, y, z) | xy = 0}

(ii) {(x, y, z) | x = 3z}

for (i) I got no, the set is not a subset of R3
for (ii) Yes, the set is a subset of R3

Could someone please tell me if I'm right or not...?

What is the
span of a set of vectors in Rn? Does the set of vectors

{(1, 2, 4), (2, 1, 3), (4,1, 1)} span R3? Give reasons for your answer.

Would someone be able to step through this problem for me please...?
• Dec 3rd 2009, 11:41 AM
Defunkt

A set of three vectors spans $\mathbb{R}^3$ if and only if they are linearly independent over $\mathbb{R}$. Do you know how to find out if a set of vectors is linearly independent?
• Dec 3rd 2009, 12:52 PM
SirOJ
yes i think so... I just want to see if the set spans R3 not if it's linearly independent. I thought I would only need to check if it's linearly independent if I was lookin for a basis of R3

i Know that you can write any element in R3 as a combonation of S (where S is equal to the set)

e.g. (x,y,z) = a1(1,2,4) + a2(2,1,3) + a3(4,-1,1)

=> x = a1 + 2a2 + 4a3
y = 2a1 + a2 - a3
z = 4a1 + 3a2 + a3

It's at this stage I get stuck... I think I am ment to write a1, a2, a3 in terms of x, y, z...?

I'm not sure if im able to do this or if this is even the right method

Any help would be much appreciated
• Dec 3rd 2009, 01:56 PM
Raoh
Quote:

Originally Posted by Defunkt

A set of three vectors spans $\mathbb{R}^3$ if and only if they are linearly independent over $\mathbb{R}$. Do you know how to find out if a set of vectors is linearly independent?

for every $\vec{x}\in \mathbb{R}^{3}$ there's $\left (\alpha_{1},\alpha _{2},\alpha _{3} \right )\in \mathbb{R}^{3}$ such that
$\vec{x}=\alpha_{1}(1,2,4) + \alpha _{2}(2,1,3) + \alpha _{3}(4,-1,1)$
no ?
• Dec 3rd 2009, 02:01 PM
Raoh
Quote:

Originally Posted by SirOJ
yes i think so... I just want to see if the set spans R3 not if it's linearly independent. I thought I would only need to check if it's linearly independent if I was lookin for a basis of R3

i Know that you can write any element in R3 as a combonation of S (where S is equal to the set)

e.g. (x,y,z) = a1(1,2,4) + a2(2,1,3) + a3(4,-1,1)

=> x = a1 + 2a2 + 4a3
y = 2a1 + a2 - a3
z = 4a1 + 3a2 + a3

It's at this stage I get stuck... I think I am ment to write a1, a2, a3 in terms of x, y, z...?

I'm not sure if im able to do this or if this is even the right method

Any help would be much appreciated

(Happy)
Use Gauss elemination method.
• Dec 3rd 2009, 02:38 PM
Defunkt
Quote:

Originally Posted by Raoh
for every $\vec{x}\in \mathbb{R}^{3}$ there's $\left (\alpha_{1},\alpha _{2},\alpha _{3} \right )\in \mathbb{R}^{3}$ such that
$\vec{x}=\alpha_{1}(1,2,4) + \alpha _{2}(2,1,3) + \alpha _{3}(4,-1,1)$
no ?

That is incorrect.

Definiton: Let V be a vector space over a field $\mathbb{F}$ and $x_1,x_2,...,x_n \in V$. We say that $x_1,x_2,...,x_n$ are linearly dependent if there exist $\alpha_1,\alpha_2,...,\alpha_n \in \mathbb{F}$, not all zero, such that:

$\sum_{i=1}^{n}\alpha_ix_i = 0$.

We will say that $y_1,...,y_n$ are linearly independent if $\sum_{j=1}^{n}b_jy_j = 0 \Leftrightarrow ~ b_1 = b_2 = ... = b_n = 0$
• Dec 3rd 2009, 02:48 PM
Raoh
Quote:

Originally Posted by Defunkt
That is incorrect.

Definiton: Let V be a vector space over a field $\mathbb{F}$ and $x_1,x_2,...,x_n \in V$. We say that $x_1,x_2,...,x_n$ are linearly dependent if there exist $\alpha_1,\alpha_2,...,\alpha_n \in \mathbb{F}$, not all zero, such that:

$\sum_{i=1}^{n}\alpha_ix_i = 0$.

We will say that $y_1,...,y_n$ are linearly independent if $\sum_{j=1}^{n}b_jy_j = 0 \Leftrightarrow ~ b_1 = b_2 = ... = b_n = 0$

i know the definition (Rofl),i meant that a set spans $R^3$ if every vector in $R^3$ can be written as a linear combination of the vectors in that set.
• Dec 3rd 2009, 03:03 PM
Defunkt
Quote:

Originally Posted by Raoh
i know the definition (Rofl),i meant that a set spans $R^3$ if every vector in $R^3$ can be written as a linear combination of the vectors in that set.

For some reason I confused the spanning set with a base. My bad. The correct definition for a spanning set is what Raoh had written in his previous post.
• Dec 4th 2009, 02:05 AM
SirOJ
Thanks a lot guys, i'll give gaussian elimination a try...
• Dec 4th 2009, 02:43 AM
SirOJ
Quote:

Originally Posted by Raoh
(Happy)
Use Gauss elemination method.

x = a1 + 2a2 + 4a3
y = 2a1 + a2 - a3
z = 4a1 + 3a2 + a3

x = a1 + 2a2 + 4a3
y-2x = 0 - 3a2 - 9a3 R2-> R2 - 2R1
z-2y = 0 + a2 + 3a3 R3 -> R3 - 2R2

x = a1 + 2a2 + 4a3
y-2x = 0 - 3a2 - 9a3
(z-2y) + 1/3(y-2x) = 0 R3-> R3 + 1/3R2

Any ideas were to go from here...? I'm sorry for all tge questions, it's just it's really frustrating getting stuck mid-problem
• Dec 4th 2009, 03:23 AM
Raoh
Quote:

Originally Posted by SirOJ
x = a1 + 2a2 + 4a3
y = 2a1 + a2 - a3
z = 4a1 + 3a2 + a3

x = a1 + 2a2 + 4a3
y-2x = 0 - 3a2 - 9a3 R2-> R2 - 2R1
z-2y = 0 + a2 + 3a3 R3 -> R3 - 2R2

x = a1 + 2a2 + 4a3
y-2x = 0 - 3a2 - 9a3
(z-2y) + 1/3(y-2x) = 0 R3-> R3 + 1/3R2

Any ideas were to go from here...? I'm sorry for all tge questions, it's just it's really frustrating getting stuck mid-problem

you should cancel the last line with the first one.
$l_{3}=l'_{3}-4l_{1}$
• Dec 4th 2009, 03:44 AM
HallsofIvy
It is simpler to throw out the "a"s and just do the Gaussian elimination on the matrix of coefficients:
$\begin{bmatrix} 1 & 2 & 4 \\ 2 & 1 & 3 \\ 4 & -1 & 1\end{bmatrix}$
Subtract twice the first row from the second row and subtract 4 times the first row from the third row:
$\begin{bmatrix}1 & 2 & 4 \\ 0 & -3 & -5 \\ 0 & -9 & -15\end{bmatrix}$
Subtract 3 times the second row from the third
$\begin{bmatrix}1 & 2 & 4 \\ 0 & -3 & -6 \\ 0 & 0 & 0\end{bmatrix}$.

There you have it: Since the last row is all 0s those vectors are not independent. Since no two are multiples of each other, take any two of the three matrices as basis.
• Dec 4th 2009, 03:46 AM
Raoh
Quote:

Originally Posted by HallsofIvy
It is simpler to throw out the "a"s and just do the Gaussian elimination on the matrix of coefficients:
$\begin{bmatrix} 1 & 2 & 4 \\ 2 & 1 & 3 \\ 4 & -1 & 1\end{bmatrix}$
Subtract twice the first row from the second row and subtract 4 times the first row from the third row:
$\begin{bmatrix}1 & 2 & 4 \\ 0 & -3 & -5 \\ 0 & -9 & -15\end{bmatrix}$
Subtract 3 times the second row from the third
$\begin{bmatrix}1 & 2 & 4 \\ 0 & -3 & -6 \\ 0 & 0 & 0\end{bmatrix}$.

There you have it: Since the last row is all 0s those vectors are not independent. Since no two are multiples of each other, take any two of the three matrices as basis.

i think he's trying to prove if that set of vectors spans $\mathbb{R}^3$.
• Dec 4th 2009, 03:49 AM
Raoh
start like this
$\left\{\begin{matrix}
a_{1} + 2a_{2} + 4a_{3}=x \\
2a_{1} + a_{2} - a_{3}=y \\
4a_{1}+ 3a_{2} + a_{3}=z
\end{matrix}\right.$

use Gauss method well and you should end up with this equation(if i'm not mistaking)(Itwasntme),
$\frac{2}{3}x+\frac{5}{3}y-z=0$
• Dec 4th 2009, 04:27 AM
SirOJ
Quote:

Originally Posted by Raoh
start like this
$\left\{\begin{matrix}
a_{1} + 2a_{2} + 4a_{3}=x \\
2a_{1} + a_{2} - a_{3}=y \\
4a_{1}+ 3a_{2} + a_{3}=z
\end{matrix}\right.$

use Gauss method well and you should end up with this equation(if i'm not mistaking)(Itwasntme),
$\frac{2}{3}x+\frac{5}{3}y-z=0$

Yes I am only interested to see if the set spans R3

By the getting the above equation what does that prove or what is the next step...?

p.s. i'm not 100% sure how you got that equation (Worried) (and I do know how to use gaussian elimination)
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