I am not sure how to distinguish between a multivariate function and a vector-valued function when they are defined. For example, if I define a function as $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$, how do I know whether it's defining something like $f(x, y) = x + 2y$ or $\vec{v}(x, y) = (x + y)\hat{i} + (x - y)\hat{j}$?

  • $\begingroup$ $f(x,y)=x+2y$ is not a function $\mathbb R^2\rightarrow\mathbb R^2$. If you instead had, for instance, $f(x,y)=(x+y,x-y)$, then this is actually the same function as the one you wrote in vector form. $\endgroup$
    – Jason
    Sep 30, 2015 at 0:40

1 Answer 1


The difference is in if the domain or the target space has multiple dimensions. We say that a function is multivariate if it has multiple variables as an input (another way to view this is that it has a vector as an input.) A function that is mapped into a vector is vector-valued. It is possible for the function to be both, as the following examples show:

(i)Multivariate: $f(x,y) = x^2 + y^2$

(ii)Vector-Valued: $g(x) = x^3\hat{i} + 4x\hat{j}$

(iii)Multivariate and Vector-Valued: $h(x,y) = (x^2 + y^2)\hat{i} + (x^2 - y^2)\hat{j}$

So as far as the function you defined it would be like (iii) (so closest to $\vec{v}$ in your example)

  • $\begingroup$ So (i) would be $\mathbb{R}^2 \rightarrow \mathbb{R}$, (ii) would be $\mathbb{R} \rightarrow \mathbb{R}^2$, and (iii) would be $\mathbb{R}^2 \rightarrow \mathbb{R}^2$? $\endgroup$
    – Wesley
    Sep 30, 2015 at 2:14
  • $\begingroup$ @MilesDavis That is exactly right. $\endgroup$
    – JDThinking
    Oct 1, 2015 at 3:11

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