Null space of a vector

Assuming I have a vector $v_1 \in \mathbb{R}^n$ of dimension $n$. (edit what I mean is e.g. $v_1 \in \mathbb{R}^n$)

I want to describe the set of "all the vectors that are orthogonal to $v_1$".

• Would it be correct to call this set the "null space of $v_1$"?
• Would "orthogonal complement of $V=\{v_1\}$ in the whole vector space of dimension n $\mathbb{R}^n$" be a better / more correct definition?

Suppose I have the set of "all the vectors that are orthogonal to $v_1$" at hand and I also found a basis $B$ for this set: $$B = \{b_1, b_2,..., b_{n-1}\}$$

And suppose further that another vector $n$-dimensional vector $v_2 \in \mathbb{R}^n$, $v_2 \neq v_1$ is expressed in this basis $B$: $$[v_2]_B = \sum_{k=0}^{n-1}b_k<b_k,v_2>$$

• What would be an intuitive explanation for the relationship between $v_1$ and $[v_2]_B$? Did this process of obtaining $[v_2]_B$ somehow "remove traces" of $v_1$ in $v_2$? Is there any intuition for this?

EDIT

Sorry some further question:

• If $v_1 = v_2$, then $[v_2]_B = 0$. So can $|[v_2]_B|$ be seen as some sort of similarity measurement?
• Dimension is a property of vector spaces, not of vectors. Null space is something a matrix has, not something a vector has. It's important to get these things right, or you wind up talking nonsense. – Gerry Myerson Dec 8 '12 at 11:55
• @GerryMyerson Yes, this is exactly why I am asking. So a "null space of a vector" does not exist? The "orthogonal complement" definition would work however? I am not sure about your comment with respect to using dimension as a property of vectors. I believe this definition is alright. – Tobold Dec 8 '12 at 12:03
• If you view the vector $v_1$ as a $1 \times n$ matrix, then it looks OK to say the solutions are the null space of that matrix. – coffeemath Dec 8 '12 at 12:15
• @coffeemath: thank you for your input! Any advice on what would be the "better" definition? I tend towards the "orthogonal complement" one but in either case the vector needs to be represented in a different way (as the only element of a subspace or a 1×n matrix). – Tobold Dec 8 '12 at 12:27

An $m \times n$ matrix $M$ represents a linear map from $R^n$ to $R^m$ in the standard bases of these spaces. As such, if a vector $v=(v_1,...,v_n)$ is turned into the $1 \times n$ matrix $M=[v_1,v_2,...,v_n]$ then $M$ represents a linear map from $R^n$ to $R^1$, and the null space of $M$ is the same as the set of vectors $w$ for which $v \cdot w=0$.
Another term used for this is the orthogonal complement of the vector $v$ in $R^n$.
1. Perhaps the cleaner if we are talking about the linear functional induced by the given vector $v_1$ (and by the inner product!): $$\Bbb R^n\to \Bbb R:\quad x\mapsto \langle v_1,x\rangle$$ Then, the subspace ${v_1}^\perp=\{x\mid \langle v_1,x\rangle=0\}$ is just the nullspace of this mapping.
2. Note that only the orthonormal bases satisfy that for any vector $v_2$, its coordinates can be calculated by the inner product: $$v_2=\sum_{k<n}b_k\cdot \langle b_k,v_2 \rangle .$$ The general definition for $[v_2]_{B}=(\beta_k)_k$ is not less and not more than this one: $$v_2=\sum_k \beta_k\cdot b_k\ .$$
3. And note that in this case, if $(b_k)_k$ is indeed a basis for ${v_1}^\perp$, this expressibility of $v_2$ says exactly that $v_2$ is in there, i.e. $v_2\in{v_1}^\perp$, i.e. $v_2\perp v_1$.
4. In this case, if $v_1=v_2$, then indeed $v_2=0$ follows.
5. If you want to measure something like how orthogonal $v_2$ is to $v_1$, then the inner product itself would just do it fine: $$\langle v_1,v_2\rangle$$ or maybe if lengths are ignored (and neither of them is $0$), you might want to use $$\frac{\langle v_1,v_2\rangle}{|v_1|\cdot |v_2|}$$ where $|v|=\sqrt{\langle v,v\rangle}$ according to the Pythagorean. Anyway, it is just the cosine of the angle between $v_1$ and $v_2$...
• Thank you very much this helps a lot! You are right, I forgot to mention the orthonormality. It think the brackets $[]_B$ around $v_2$ are missing in in '4.'. And shouldn't it be $n-1$ in '2.'? I am not sure if I understood '3.': Any vector $v_2$ can be expressed as described in '2.' but still you are mentioning 'expressibility'. Is $v_2$ somehow special if it is expressable in $B$? – Tobold Dec 8 '12 at 13:24
• BTW No I am not trying to measure orthogonality. I saw this method of expressing $v_2$ in the orthonormal basis $B$ spanning the "null space of $v_1$" (sorry I am using this term again but that's how it was described) somewhere and was wondering what's the reasoning behind doing so. Think of $v_1$ and $v_2 \in \mathbb{R}^n$ as time series of grey values. Still I am lacking this intuition behind what's going on. – Tobold Dec 8 '12 at 13:31