Determine the dimension of the image and kernel and a basis of the kernel of a linear transformation

Question

Given are the following transformations $$f_1:\mathbb{R}^n\to \mathbb{R},\; (x_1,\dots,x_n)\mapsto x_1+\dots +x_n$$ and

$$f_2:\mathbb{R}^2\to \mathbb{R},\; (x,y)\mapsto x\cdot y$$

First, I need to determine whether they're linear or not.

$f_1$ should be linear, because $\forall u,v \in \mathbb{R}^n:f(u+v)=f(u)+f(v)$ and $\forall \lambda \in \mathbb{R},\; \forall u \in \mathbb{R}^n:f(\lambda u) = \lambda f(u) $

$f_2$ is not linear, because $\forall u,v \in \mathbb{R}^2:f(u+v)\neq f(u)+f(v)$ and that's why we ignore $f_2$ for the rest of the task.

Now I should determine $\dim\operatorname{im}f$ and $\dim\operatorname{ker}f$ and one basis of $\operatorname{ker} f$, but I'm not sure how to do it in this case. If I had a transformation matrix from $f_1$, it would be simple.

Answer 1

Observe that

$$x_1+\ldots +x_n=0\iff x_n=-(x_1+\ldots+x_{n-1})\implies \dim\ker f=n-1$$

as we have $\;n-1\;$ free variables in the above equation. This is called sometimes hyperplane = maximal subspace of a linear space, in this case $\;\Bbb R^n\;$ . Any hyperplane is of this form: the kernel of a non-zero linear functional.

For a basis now is easy: for example, $\;(1,-1,0,...,0) , (1,0,-1,0,...,0),\ldots\;$ ...finish the argument and prove you get a l.i. set which is a then a basis of $\;\ker f\;$

Answer 2

$\mathbb{R}$ is a vector space of dimension $1$ over $\mathbb{R}$, so since the image is a subspace its dimension can be either $0$ or $1$. Of course it is not $0$, so it has to be $1$. From here we conclude that the dimension of the kernel must be $n-1$. As for a basis, this is the easiest choice:

$\{(1,0,...,0,-1),(0,1,0,...,0,-1),...,(0,0,...,0,1,-1)\}$.

Answer 3

I'll call the map $f$ instead of $f_1$ for convenience. Note that $\text{image}(f)$ is a subspace of $\Bbb{R}$, which means it is either the zero vector space $\{0\}$, or it is the whole space $\Bbb{R}$ (these are the only subspaces of $\Bbb{R}$).

However, note that $f(1, 0,\dots, 0) = 1 \neq 0$, which means $\text{image}(f) \neq \{0\}$. This shows that $\text{image}(f) = \Bbb{R}$; hence its dimension is $1$. By the rank-nullity theorem, it follows that \begin{align} \text{dim}\ker(f) = n-1 \end{align} A possible choice for a basis is $\{e_i-e_n\}_{i=1}^{n-1}$.

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By the way, in my argument above I tried to give a "dimension based argument" for why $\text{image}(f) = \Bbb{R}$, but in this case it is easy to argue directly as well, because for any $c \in \Bbb{R}$, we have $f(c,0,\dots, 0) = c$, proving that $\text{image}(f) = \Bbb{R}$.