Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

Thanks in advance!


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?
share|cite|improve this question
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
up vote 1 down vote accepted

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 $R^n$.

share|cite|improve this answer
I accepted this answer since I feel it is closer to what is meant by the concept of a "null space of a vector". @Berci's answer is also really useful though. About the intuition behind this whole process I am still unsure and I think I will need to ask a more specific question for that. – Tobold Dec 12 '12 at 7:09
  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 $v_2$...
share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.