I have a homework question that is seriously stumping me. It was a true/false statement. The statement is

"If $v$ is any vector in an inner product space $V$, then $v/\|v\|$ is a unit vector"

and according to my professor, the statement is false. Isn't dividing a vector by its length the very definition of normalizing a vector?

Edit: Thanks everyone! I didn't even consider the null vector.

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    $\begingroup$ The statement as it is does not exclude $v=0$ . $\endgroup$
    – Jimmy R.
    Apr 23, 2016 at 17:02
  • 1
    $\begingroup$ maybe the thing you have to note is that this doesn't make sense if $v=0$ $\endgroup$ Apr 23, 2016 at 17:03

4 Answers 4


It's not true for any vector. There is at least one vector for which it's undefined : the null vector.

Now, it's true that for every non null vector, it's an unit vector


$\vec{v} = 0$, then: $$\frac{1}{\|v\|}\vec{v} = ?$$

The norm, $\| \|$, is normally $>0$ except only if $v = 0$.

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    $\begingroup$ "If you write $||v||$ instead of $\|v\|$, then you'll have "the norm $||||$", with four equally spaced vertical lines, rather than "the norm $\|\|$". I changed it to the latter, which is standard usage. Notice the conspicuous diffence between $||a||||b||$ and $\|a\|\|b\|$. $\qquad$ $\endgroup$ Apr 23, 2016 at 17:32

I consider your professor's question to be badly stated.

The expression

If $v$ is any vector in an inner product space $V$, then $v/\|v\|$ is a unit vector.

can be translated into symbols as

$$\forall v \in V: \left\| \frac{v}{\|v\|} \right\| =1. \tag{1}\label{first}$$

If your professor has true/false as possible answers then either \eqref{first} is true or its negation, which is

$$ \exists v \in V: \left\| \frac{v}{\|v\|} \right\| \ne 1. \tag{2}\label{second}$$

Now note that \eqref{second} isn't true either since it "breaks" if you try to plug in $0\in V$ (for one can't divide by zero).

So while one can see what was meant by the question and that your professor meant false to be the correct answer, I, being a student, am always somewhat peeved when I see something like this on an exam.

  • $\begingroup$ I agree. The statement isn't false; its ill-typed. $\endgroup$ Apr 25, 2016 at 8:50

Great question. Yes, the zero vector is the counter-example.

I want to also add that the zero matrix is good to remember when needing counter-examples -- it has been a good one, especially for exams, from my own experience.


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