# Dimensions of span

1. Feb 6, 2009

### gtfitzpatrick

1. The problem statement, all variables and given/known data

if x$$_{1}$$=$$\begin{pmatrix}2 \\ 1 \\ 3\end{pmatrix}$$

x$$_{2}$$=$$\begin{pmatrix}3 \\ -1 \\ 4\end{pmatrix}$$

x$$_{3}$$=$$\begin{pmatrix}2 \\ 6 \\ 4\end{pmatrix}$$

(i) show that x1,x2 and x3 are linearly dependent
(ii) show that x1 and x2 are linearly independent
(iii)what is the dimension of span (x1,x2,x3)
(iv)give a geometric discription of span (x1,x2,x3)

for (i) i solved the matrices
$$\begin{pmatrix}2 & 3 & 2 \\ 1 & -1 & 6 \\ 3 & 4 & 4\end{pmatrix}$$$$\begin{pmatrix}c1 \\ c2 \\ c3\end{pmatrix}$$ = $$\begin{pmatrix}0\\ 0 \\ 0\end{pmatrix}$$
and i got a not trivial solution which when i let c3=1 i get -4(x1)+2(x2)+(x3)=0
firstly i would like to know if i formed the matrix correctly,should i have put the values accross instead of down?
Thanks

2. Feb 6, 2009

### Staff: Mentor

The matrix was how it should be, and your values for c1, c2, and c3 check, so all is good.

3. Feb 6, 2009

### gtfitzpatrick

thanks Mark,
for part (ii)

i did the same as (i) and got c1=c2=0 which proves they are linearly independent.

but from part (i) if i let c3 = 1 it gives
$$\begin{pmatrix}-8 \\ -4 \\ -12\end{pmatrix}$$+ $$\begin{pmatrix}6 \\ -2 \\ 8\end{pmatrix}$$+ $$\begin{pmatrix}2 \\ 6 \\ 4\end{pmatrix}$$=0

i think that if you can write one vector as a combination of the other vectors it will be linear dependant but if you cant write a vector as combination of the other vectors it is linear independent but this contradicts what i got when i worked it out as i seem to be able to write each vector as a combination of the other vectors!confussed!

4. Feb 6, 2009

### Staff: Mentor

Any time you have two vectors, it's very simple to see if the set is linearly dependent: each vector will be a some multiple of the other. By nothing more complicated that observation I can tell the {x1, x2} is a linearly independent set, as is {x2, x3}, but {x1, x3} is a linearly dependent set, since x3 is a multiple of x1 (and x1 is a different multiple of x3).

The terms linearly dependent and linearly independent are always used when talking about a set of vectors, and never in the context of a single vector. For example, and considering your vectors x1 and x3, it doesn't make any sense to describe x1 as linearly independent or linearly dependent, nor does it to describe x3 this way. Again, we're always talking about a set of vectors.

5. Feb 6, 2009

### gtfitzpatrick

Thanks for all the replies Mark, i get the linear (in)dependance now but parts (iii) and (iv) are driving my head round and round, i'll have to do more reading and then try them a bit later...

6. Feb 6, 2009

### HallsofIvy

Staff Emeritus
Well, now that you've done (i) and (ii), (iii) is trivial isn't it? You are told that the set is spanned by $x^1$, $x^2$ and $x^3$ and have shown that $x^3$ can be written in terms of $x^1$ and $x^2$ while $x^1$ and $x^2$ are independent- that means that $\{x^1, x^2\}$ is a basis for the space.

Of course, geometrically, this is the plane containing (0,0,0) and the lines in the directions of $x^1$ and $x^2$.

7. Feb 6, 2009

### gtfitzpatrick

so it has a dim of 2 i think i finally see, thanks a mill, onward...

8. Feb 7, 2009

### gtfitzpatrick

for (iv)
the subspace spaned by these 3 vectors is a plane through the origin in R$$^{3}$$. The origin is in every subspace since the 0 vector is in every subspace?
i'm not sure i follow this though...

9. Feb 7, 2009

### HallsofIvy

Staff Emeritus
Yes. All subspaces of R3 are of theform
1) R3 itself (dimension 3)
2) planes containing the origin (dimension 2)
3) lines containing the origin (dimension 1)
4) the origin alone (dimension 0)