Subspace of vectors orthogonal to an arbitrary vector.

[TheoEndre]

Homework Statement

Let $V$ be a vector space with dim$V=n$. Show that the set consisting of vectors orthogonal to a vector $\vec v \in V, \vec v \neq \vec 0$ forms a subspace with dimension $n-1$.

Relevant Equations

No relevant equations

The proof that the set is a subspace is easy. What I don't get about this exercise is the dimension of the subspace. Why is the dimension of the subspace $n-1$? I really don't have a clue on how to go through this.

 

[Infrared]

Think about the linear map $V\to\mathbb{R}$ given by $w\mapsto\langle w,v\rangle$ and apply rank-nullity.

 

[TheoEndre]

Think about the linear map $V\to\mathbb{R}$ given by $w\mapsto\langle w,v\rangle$ and apply rank-nullity.

I see, everything makes sense now. To be honest, I didn't even know such theorem exist (This exercise is from a quantum mechanics textbook, they didn't mention this theorem). But from what I've seen, with the help of the linear map you gave me, the subspace the exercise mentioned is just the kernel of the linear map. Since the dimension of the image of the linear map is $1$, by the rank-nullity theorem (I will denote the transformation $T$):
$$dim (Image T)+dim (kernel T)=dim V$$
$$1+dim(kernel T)=n$$
$$dim(kernel T)=n-1$$
correct?

 

[Infrared]

Yes, that's fine (if you want to 100% complete, you should say why $\text{im}(T)=\mathbb{R}$). If you don't want to appeal to rank-nullity, you could extend $v$ to an orthogonal basis $\{v,v_2,\ldots,v_n\}$ for $V$. Then you should check that $\{v_2,\ldots,v_n\}$ is a basis for the subspace of vectors orthogonal to $v$.

 

[TheoEndre]

Yes, that's fine (if you want to 100% complete, you should say why $\text{im}(T)=\mathbb{R}$). If you don't want to appeal to rank-nullity, you could extend $v$ to an orthogonal basis $\{v,v_2,\ldots,v_n\}$ for $V$. Then you should check that $\{v_2,\ldots,v_n\}$ is a basis for the subspace of vectors orthogonal to $v$.

I actually don't know why $\text{im}(T)=\mathbb{R}$, but I assumed the inner product to be completely defined (if that is the right word). I hope you tell me the reason. Also, the other solution is what I've been looking for (since the textbook didn't include the rank-nullity theorem), so thank you for that (It took me a while to actually understand it though).

 

[Infrared]

The image of $T$ is a subspace of $\mathbb{R}$, so it must be either $\mathbb{R}$ or $\{0\}$. But the second case is impossible because $T(v)=\langle v,v\rangle>0$ since $v\neq 0$. Recall that one of the axioms for an inner product is that $\langle v,v\rangle\geq 0$ for all $v\in V$ with equality only when $v=0$ (which we're assuming is not the case here).

Since you're unfamiliar with rank-nullity and your book doesn't mention it, I'd recommend finding a linear algebra textbook to read in parallel. It's likely that you have other significant gaps.

 

[TheoEndre]

The image of $T$ is a subspace of $\mathbb{R}$, so it must be either $\mathbb{R}$ or $\{0\}$. But the second case is impossible because $T(v)=\langle v,v\rangle>0$ since $v\neq 0$. Recall that one of the axioms for an inner product is that $\langle v,v\rangle\geq 0$ for all $v\in V$ with equality only when $v=0$ (which we're assuming is not the case here).

Since you're unfamiliar with rank-nullity and your book doesn't mention it, I'd recommend finding a linear algebra textbook to read in parallel. It's likely that you have other significant gaps.

Thank you very much for your time, everything makes sense to me now.