study inside a finite-dimensional inner product space.

Question

I am studying for my test and have completed most of the problems assigned to me, but am having trouble with a few, and this one in particular. Thank you in advance for any help.

Let $V$ be a finite-dimensional inner product space. Suppose that $U$ and $W$ are subspaces of $V$, such that there is an element $0\neq u ∈ U$ with $u\in W^\perp$. Prove that there exists an element $0\neq w \in W$ such that $w \in U^\perp$.

Is ${0}^\perp$ the whole space? – GEdgar Mar 18 '13 at 20:56 — GEdgar, Mar 18 '13 at 20:56
I do not know as a matter of fact. – Omar Mar 18 '13 at 21:06 — Omar, Mar 18 '13 at 21:06
Then you need to check the definitions and find out. – GEdgar Mar 18 '13 at 21:06 — GEdgar, Mar 18 '13 at 21:06

Alfonso Fernandez · Accepted Answer · 2013-03-18T21:31:53.453

This is false:

Let $V=\mathbb{R}^3$, $U=\mbox{span}\left(u_1=(0,0,1),u_2=(0,1,0)\right)$, and $W=\mbox{span}\left((0,0,1)\right)$.

$(0,1,0)\in U$ and $(0,1,0)\in W^\perp$, but every $w\in W$ satisfies $w\in U$, so $w \notin U^\perp$.

Following the comments, if you assume $W\not\subseteq U$:

Let $V=\mathbb{R}^3$, $U=\mbox{span}\left(u_1=(0,0,1),u_2=(0,1,0)\right)$, and $W=\mbox{span}\left((1,0,1)\right)$.

$(0,1,0)\in U$ and $(0,1,0)\in W^\perp$, but every $w\in W$ is of the form $(c,0,c)$, so $\left<u_1,w\right>=c \ne 0$, and $w\notin U^\perp$.

If you assume $W\cap U = \{0\}$:

Let $V=\mathbb{R}^3$, $U=\mbox{span}\left(u_1=(1,1,0),u_2=(0,1,1)\right)$, and $W=\mbox{span}\left((0,0,1)\right)$.

$u_1\in U$ and $u_1\in W^\perp$, but every $w\in W$ is of the form $(0,0,c)$, so $\left<u_2,w\right>=c \ne 0$, and $w\notin U^\perp$.

I thought this but then I just assumed he meant to say the subspaces intersected trivially. — muzzlator, Mar 18 '13 at 21:12
Could it be true if we only assumed that neither subspace was a subset of the other? — ferson2020, Mar 18 '13 at 21:12

study inside a finite-dimensional inner product space.

1 Answers1