Prove that the mapping $f:\mathbb{R}^n\rightarrow \mathbb{R}^m$ is linear if and only if its coordinate functions $f_1,...,f_m$ are linear.

Question

• Prove that the mapping $f:\mathbb{R}^n\rightarrow \mathbb{R}^m$ is linear if and only if its coordinate functions $f_1,...,f_m$ are linear.

  To prove the question, assume that the following theorem is proven: The mapping $f:\mathbb{R}^n\rightarrow \mathbb{R}^m$ is linear if and only if there exists a matrix $A$ such that $f(x)=Ax$ for all $x\in\mathbb{R}^n$.

My attempt. $(\Rightarrow)$. Assume the mapping $f:\mathbb{R}^n\rightarrow \mathbb{R}^m$ is linear. By the theorem there exists a matrix $A$ such that $f(x)=Ax$ for all $x\in\mathbb{R}^n$. So

$$ f_1(x+y)=a_{11}(x_1+y_1)+...+a_{1n}(x_n+y_n)=(a_{11}x_1+a_{12}x_2+...+a_{1n}x_n)+(a_{11}y_1+...+a_{1n}y_n) =f_1(x)+f_1(y), \\ \vdots \\ f_m(x+y)=a_{1m}(x_1+y_1)+...+a_{mn}(x_n+y_n) =f_m(x)+f_m(y) $$

Similarly, $cf(x)=f(cx)$. So, its coordinate functions $f_1,...,f_m$ are linear

$(\Leftarrow)$. Assume the coordinate functions $f_1,...,f_m$ are linear.So how can I show the mapping $f:\mathbb{R}^n\rightarrow \mathbb{R}^m$ is linear?

Can you check my attempt?

Answer 1

Your approach to the first direction is missing the proof of the homogeneity condition (that $f_i(\lambda x)=\lambda f_i(x)$), but the addition is ok.

However, you're using a handgun at a knife fight. Why not simply use the definition of linear function?

For shorter notation, let us write $f(x)=(f_i(x))_i$. I'll also use a shorter condition for linearity (e.g. see https://en.wikipedia.org/wiki/Linear_map#Definition_and_first_consequences)

Then \begin{align*} f\text{ is linear}&\iff f(x+\lambda y)=f(x)+\lambda f(y)&\text{for all }x,y,\lambda\\ &\iff (f_i(x+\lambda y))_i=(f_i(x))_i+\lambda(f_i(y))&\text{for all }x,y,\lambda\tag{1}\\ &\iff (f_i(x+\lambda y))_i=(f_i(x)+\lambda f_i(y))_i&\text{for all }x,y,\lambda\tag{2}\\ &\iff f_i(x+\lambda y)=f_i(x)+\lambda f_i(y)&\text{for all }x,y,\lambda\text{ and }i\tag{3}\\ &\iff f_i\text{ is linear}&\text{ for all }i\tag{4} \end{align*} where $x$ and $y$ denote vectors and $\lambda$ denotes an scalar. (1) follows from the definition of $f$; (2) follows from the definition of the operations on $\mathbb{R}^m$; (e) follows because two $m$-tuples are equal iff respective coordinates are equal; (4) follows from the definition of linearity.

This more abstract approach, which is easier in my opinion, works in general for linear maps between products of vector spaces.

Answer 2

Here is another approach that uses the result you were given:

Suppose the $f_k$ are linear. Then the theorem says there is a matrix $a_k$ such that $f_k(x) = a_k x $ for all $x$. Now let $A = \begin{bmatrix} a_1 \\ \vdots \\ a_n \end{bmatrix}$. Then $\phi(x) = Ax$ is linear and $[\phi(x)]_k = a_k x = f_k(x)$, so we have $\phi=f$ and hence $f$ is linear.