Can basis vectors have fractions? So I was diagonalizing a matrix in a book, and one of the basis vectors was [3/2, 1], after doing the problem, the answer in the book was different than mine. It came with an explanation, and in it the basis vector was [3,2]. They are the same thing, just multiples of each other, so I was curious, is it mandatory to take fractions out of a basis vector?
 A: It is not. If $v$ is a member of some basis, then we can substitute $cv$ for any non-zero constant $c$ of the reals (or the underlying field in the general case).
A: In short, the answer is no, it is not mandatory - but there is some terminological confusion both in the question and other answers, which I will try to say something about.
You shouldn't really say that $(3,2)$ and $(\frac{3}{2},1)$ are the same thing - they aren't - but they are scalar multiples of one another. This means they are both bases for the same $1$-dimensional space (although not the same basis). Since you mention a matrix, you are probably looking for eigenvectors (i.e. a basis of each eigenspace), so it might be less confusing to say eigenvector than basis vector (or not, depending on context).
One further point - more things are vector spaces than $\mathbb{Q}^n$, so it doesn't even always make sense to "take out fractions". (Even in $\mathbb{R}^n$, no scalar multiple of $(1,\pi)$ has integer entries, and there are vector spaces like the set of continuous functions $\mathbb{R}\to\mathbb{R}$, whose elements aren't "lists of numbers").
A: You are right and they are also right. This is because multiplication by non-zero constants does not affect the span of the basis nor the linear independence of the vectors in that basis. You should be able to see this quite obviously once you think about it.
