# subspaces...

• Jan 28th 2011, 01:41 AM
AkilMAI
subspaces...
For each of the following subsets U of the vector space V I have to decide whether or not U is a subspace of V . In each case when U is a subspace, I also must find a
basis for U and state dim U:
(i) V = R^4; U = {x = (x1; x2; x3; x4) : 3x1 - x2 -2x3 + x4 = 0}:
(ii) V = R^3; U = {x = (x1; x2; x3) : x1^2 = x2^2 + 4*x3^2}
(iii) V = P3; U = {p in P3 : p'(0) = p(1)}
I have some questions for each problem...

I need to know that my solutions are ok or not and if so, how to find the basis and dim of U.
(i) we take x,y in U=> x+y=0 and a(a scalar),a*x=a*0=0 ,so it is closed under addition and scalar multiplication,therefore it is a subspace.If my approach is correct how can I find basis U and dim U?
(ii)Here I did something similair,I wrote x1^2 = x2^2 + 4*x3^2 as x1^2 - x2^2 - 4*x3^2=0
=> x,y in U where x+y =0+0=0 and a*x=a*0=0.Again If my approach is correct how cand I find basis U and dim U?
(iii)take p and g two polyn from U=>(g+p)(1)=p(1)+g(1)=p'(0)+g'(0)=(g+p)(0) and (a*p)(1)=a*p(1)=a*p(0)=(a*p)(0).Same thing, If my approach is correct how cand I find basis U and dim U?
Thank you
• Jan 28th 2011, 06:46 AM
TheEmptySet
Quote:

Originally Posted by AkilMAI
1. The problem statement, all variables and given/known data
For each of the following subsets U of the vector space V I have to decide whether or not U is a subspace of V . In each case when U is a subspace, I also must find a
basis for U and state dim U:
(i) V = R^4; U = {x = (x1; x2; x3; x4) : 3x1 - x2 -2x3 + x4 = 0}:
(ii) V = R^3; U = {x = (x1; x2; x3) : x1^2 = x2^2 + 4*x3^2}
(iii) V = P3; U = {p in P3 : p'(0) = p(1)}
I have some questions for each problem...

2. The attempt at a solution
I need to know that my solutions are ok or not and if so, how to find the basis and dim of U.
(i) we take x,y in U=> x+y=0 and a(a scalar),a*x=a*0=0 ,so it is closed under addition and scalar multiplication,therefore it is a subspace.If my approach is correct how can I find basis U and dim U?
(ii)Here I did something similair,I wrote x1^2 = x2^2 + 4*x3^2 as x1^2 - x2^2 - 4*x3^2=0
=> x,y in U where x+y =0+0=0 and a*x=a*0=0.Again If my approach is correct how cand I find basis U and dim U?
(iii)take p and g two polyn from U=>(g+p)(1)=p(1)+g(1)=p'(0)+g'(0)=(g+p)(0) and (a*p)(1)=a*p(1)=a*p(0)=(a*p)(0).Same thing, If my approach is correct how cand I find basis U and dim U?
Thank you

To show that $\displaystyle U$ is a subspace of $\displaystyle V$.

You need to verify that it is
1: Show that $\displaystyle \vec{0} \in U$
2: that it is closed $\displaystyle \vec{x},\vec{y} \in U \implies \vec{x}+\vec{cy} \in U$

You are correct that i) is a subspace. The equation of the plane given shows you what properties the vector must satisfy.

$\displaystyle 3x1 - x2 -2x3 + x4 = 0$
This "system" of equations has 4 variables and 1 equation so it will have 3 free variables. Let $\displaystyle x_1,x_2,x_3$ be the free variables then
$\displaystyle x_4=-3x_1+x_2+2x_3$

Now writing the vector solution gives
$\displaystyle \begin{pmatrix}x_1 \\ x_2 \\ x_3 \\ -3x_1+x_2+2x_3\end{pmatrix} = \begin{pmatrix}1\\ 0 \\ 0 \\ -3 \end{pmatrix}x_1 + \begin{pmatrix}0\\ 1 \\ 0 \\ 1\end{pmatrix}x_2+ \begin{pmatrix}0\\ 0 \\ 1 \\ 2\end{pmatrix}x_3$

As you can check these vectors are the basis of you space. Each on by itself satisfy the equation above.

For ii) it is not a subspace as it is not linear. Note that

$\displaystyle \vec{v} \in U \implies \vec{v}=(\sqrt{x_2^2+4x_3^2},x_2,x_3)$

Now show that $\displaystyle (\sqrt{(cx_2)^2+4(cx_3)^2},(cx_2),(cx_3)) \ne c\vec{v}$

Remember that $\displaystyle \sqrt{c^2}=|c|$

For the last one here is a hint:
$\displaystyle p(x)=z+bx+cx^2+dx^3 \implies p'(x)=b+2cx+3dx^2$

$\displaystyle p'(0)=p(1) \implies b=a+b+c+d \iff d=-a-c$

This gives you a form for the polynomial in your space.
• Jan 28th 2011, 12:20 PM
HallsofIvy
Quote:

Originally Posted by AkilMAI
For each of the following subsets U of the vector space V I have to decide whether or not U is a subspace of V . In each case when U is a subspace, I also must find a
basis for U and state dim U:
(i) V = R^4; U = {x = (x1; x2; x3; x4) : 3x1 - x2 -2x3 + x4 = 0}:
(ii) V = R^3; U = {x = (x1; x2; x3) : x1^2 = x2^2 + 4*x3^2}
(iii) V = P3; U = {p in P3 : p'(0) = p(1)}
I have some questions for each problem...

I need to know that my solutions are ok or not and if so, how to find the basis and dim of U.
(i) we take x,y in U=> x+y=0 and a(a scalar),a*x=a*0=0 ,so it is closed under addition and scalar multiplication,therefore it is a subspace.If my approach is correct how can I find basis U and dim U?
(ii)Here I did something similair,I wrote x1^2 = x2^2 + 4*x3^2 as x1^2 - x2^2 - 4*x3^2=0
=> x,y in U where x+y =0+0=0 and a*x=a*0=0.Again If my approach is correct how cand I find basis U and dim U?
(iii)take p and g two polyn from U=>(g+p)(1)=p(1)+g(1)=p'(0)+g'(0)=(g+p)(0) and (a*p)(1)=a*p(1)=a*p(0)=(a*p)(0).Same thing, If my approach is correct how cand I find basis U and dim U?
Thank you

No, your approach is completely wrong. You have no reason at all to think that, for x and y in any of these subsets, x+y= 0 or that ax= 0. The fact that their components set some formula to 0 does NOT mean the vectors themselves are the 0 vector, which you appear to be assuming.

For example, the vector x= (-1, 0, 1, -1) is in U for (i) because -1+ 0+ 2(1)+ (-1)= 0. Also y= (1, -1, 0, 0) is in U because 1+ (-1)+ 2(0)+ 0= 0. Their sum is x+ y= (0,-1, 1, -1) which is NOT equal 0. But it is true that 0+ (-1)+ 2(1)+ (-1)= 0 so x+ y is also in U.