# Thread: [x,y]=0 is a subspace

1. ## [x,y]=0 is a subspace

Hi,

problem:

Prove that if y is a linear functional on an n-dimensional vector space V, then the set of all those vectors x for which [x,y]=0 is a subspace of V; what is the dimension of that subspace?

attempt:

Let X be the set of vectors x in V for which [x,y]=0 where y is a linear functional. Let $x_1\;and\;x_2$ be any two vectors in X and let $\alpha\;and\;\beta$ be arbitrary scalars. Then,
$[\alpha x_1+\beta x_2,y]=\alpha[x_1,y]+\beta[x_2,y]=0$ and so X is a subspace of V.

Don't know the dimension, maybe n-1

Thanks!

2. Originally Posted by Mollier
Hi,

problem:

Prove that if y is a linear functional on an n-dimensional vector space V, then the set of all those vectors x for which [x,y]=0 is a subspace of V; what is the dimension of that subspace?

attempt:

Let X be the set of vectors x in V for which [x,y]=0 where y is a linear functional. Let $x_1\;and\;x_2$ be any two vectors in X and let $\alpha\;and\;\beta$ be arbitrary scalars. Then,
$[\alpha x_1+\beta x_2,y]=\alpha[x_1,y]+\beta[x_2,y]=0$ and so X is a subspace of V.

Don't know the dimension, maybe n-1

Thanks!

Apparently you guys use the notation $y(x)=[x,y]$ ...weird. Anyway, use the dimensions theorem since, after all, alinear function is just a linear transformation from

a vector space $V_\mathbb{F}\rightarrow \mathbb{F}$ , $\mathbb{F}=$ a field.

About the dimension: if the functional is the zero func. then___, but if it is now the zero func. then it automatically is ONTO, so___

Dimensions theorem: if $T:V\rightarrow W$ is a linear transfromation between vectors spaces over the same field, and if $\dim V=n<\infty$ , we get

$\dim(ker\, T)+\dim(Im\, T)=n$

Tonio

3. Originally Posted by tonio
Apparently you guys use the notation $y(x)=[x,y]$ ...weird.
Halmos uses $y(x)=[x,y]$ in his books. I was just giving it a try

Originally Posted by tonio
About the dimension: if the functional is the zero func. then___, but if it is now the zero func. then it automatically is ONTO, so___
I let $\{x_1,\cdots,x_n\}$ be a basis of $V$. Any vector in the subspace, say $N$, can be written as
$x=\xi_1x_1+\cdots+\xi_nx_n$.
$y(x)=0 \rightarrow \xi_1y(x_1)+\cdots+\xi_ny(x_n)=0$

I know for a earlier theorem that if $V$ is an n-dimensional vector space with a basis $\{x_1,\cdots,x_n\}$, then there is a uniquely determined basis $\{y_1,\cdots,y_n\}$ (dual basis) with the property that $y_j(x_i)=\delta_{ij}$

So in my $y(x)$, there is some $y(x_k)$ that is equal to 1 and if $y(x)=0$, then $\xi_k=0$.

So I end up with $dim(N)=n-1$..

Originally Posted by tonio
Dimensions theorem: if $T:V\rightarrow W$ is a linear transfromation between vectors spaces over the same field, and if $\dim V=n<\infty$ , we get

$\dim(ker\, T)+\dim(Im\, T)=n$
I will have to read up on this a bit.

Thanks.