2
$\begingroup$

Let $V,W$ be two finite dimensional vector spaces and $f$ is linear mapping from $V$ to $W$. Show that $\dim \operatorname{im}(f) + \dim \ker(f) = \dim V.$


Here is what I did. Please kindly give me a recommendation or help me to show this. Thank in advance!

Let $(e_1,e_2, \dots, e_n)$ be a basis for $V$ and $(e_1,e_2, \dots, e_k)$ be a basis for $\ker(f),k<n$.

We can get $$\dim V=n,\quad \dim \ker(f)=k$$ Then we need to prove that $\dim \operatorname{im}(f)=n-k$

Since $(e_1,e_2, \dots,e_n)$ is basis for $V$. $(e_1,e_2, \dots,e_n)$ spans V.

Let $x\in V$.There exists $\lambda_1,\lambda_2,\dots,\lambda_n\in\mathbb{K}$ such that $$x=\lambda_1e_1+\lambda_2e_2+\cdots+\lambda_ne_n$$ $f$ is linear mapping $$f(x)=f(\lambda_1e_1+\lambda_2e_2+\cdots+\lambda_ne_n)$$ $$f(x)=\lambda_1f(e_1)+\lambda_2f(e_2)+\cdots+\lambda_nf(e_n)$$ Which that $\operatorname{im}(f)=\{f(v)\in W | v\in V\}$ $$\operatorname{im}(f)=\operatorname{span}\{f(e_1),f(e_2),\dots,f(e_n)\}$$ Since $(e_1,e_2,\dots,e_n)$ is linearly independent.There exists $\alpha_1,\alpha_2,\dots,\alpha_k,\dots\alpha_n\in\mathbb{K},\quad k<n$

Which $\alpha_1=\alpha_2=\dots=\alpha_n=0$ $$\alpha_1e_1+\alpha_2e_2+\cdots+\alpha_ke_k+\alpha_{k+1}e_{k+1}+\cdots+\alpha_ne_n=0$$ $$f(\alpha_1e_1+\alpha_2e_2+\cdots+\alpha_ke_k+\alpha_{k+1}e_{k+1}+\cdots+\alpha_ne_n)=f(0)$$ $$\alpha_1f(e_1)+\alpha_2f(e_2)+\cdots+\alpha_kf(e_k)+\alpha_{k+1}f(e_{k+1})+\cdots+\alpha_nf(e_n)=0$$ Since $\ker(f)=\operatorname{span}\{e_1,e_2,\dots,e_k\}\Longrightarrow f(e_1)=f(e_2)=\dots=f(e_k)=0$.We can get $$\alpha_{k+1}f(e_{k+1})+\cdots+\alpha_nf(e_n)=0$$ So that $\{f(e_{k+1}),\dots,f(e_n)\}$ is linearly independent.

We can get $\{f(e_{k+1}),\dots,f(e_n)\}$ is a basis for $\operatorname{im}(f)\Longrightarrow \dim \operatorname{im}(f)=n-k$.

$\endgroup$
6
  • $\begingroup$ $V/\ker(f)\cong\text{im}(f)$. $\endgroup$
    – Kenta S
    Apr 1 at 16:59
  • $\begingroup$ Should I add or do anything more on my proof? $\endgroup$
    – LaVendEr
    Apr 1 at 17:04
  • $\begingroup$ @LaVendEr I think you had a typo - I changed it. Compare to see that you agree? $\endgroup$
    – peter a g
    Apr 1 at 17:12
  • 2
    $\begingroup$ No this would not fly... You have to show 1) $f(e_{k+1}), \cdots f(e_{n})$ span $f(V)$ and 2) $f(e_{k+1}), \cdots f(e_n)$ are independent. Your argument should have these two parts addressed explicitly [you have part 1 somewhat implicitly]. For Part 2), (presumably after finishing with part 1 ) you should start off with something like "Suppose we have a relation $\alpha_{k+1} f(e_{k+1}) + \cdots +\alpha_n f(e_n) =0.$" and, then, after an argument, CONCLUDE that the $\alpha_{k+1}=\cdots = \alpha_n = 0$. Together, 1) and 2) imply that $f(e_{k+1}), \cdots, f(e_n)$ forms a basis for the image. $\endgroup$
    – peter a g
    Apr 1 at 17:45
  • 1
    $\begingroup$ Also, small note, just for the sake of writing, I think you should introduce the basis for $\ker f$ first, and then say you extend it to a basis of $V$, to make clear that these are intended to be the "same" $e_i$ (and so you don't have to sort of "retroactively" make the first $k$ of $(e_i)_{i=1}^n$ a basis for $\ker f$ after you've already introduced them.) $\endgroup$
    – thorimur
    Apr 1 at 18:23
0
$\begingroup$

Your proof is better than mine because yours deals with linear transformations. I'm just giving you another perspective. Let $A$ be the $m\times n$ matrix for $f$ (wrt any choice of bases). Let $r$ be the rank of $A.$ $$\dim \operatorname{im}(f)=\dim(\text{column space of }A)=r$$ $\dim \ker(f) = n-r= $ the number of parameters when solving $AX=0.$

$\endgroup$
0
$\begingroup$

Unsure what you're allowed to assume, but I think about it like this. $\mathrm{Ker}(f) = K$ is a linear subspace and together with its orthogonal complement, the direct product spans $V$: $$ V = K \oplus K^\perp $$

Since $f$ restricted to $K^\perp$ is a linear function with trivial kernel, it is necessaritly true that $\left. f \right|_{K^{\perp}}$ is injective. This implies that $\mathrm{dim} \, K^\perp = \mathrm{dim} \, f(K^\perp)$.

The desired result follows from the fact that the dimension of $V$ is equal to the sum of the dimensions of the spaces in any equivalent direct product.

So:

$$ \begin{aligned} \mathrm{dim} \, V & = \mathrm{dim} \, K \oplus K^\perp \\ & = \mathrm{dim} \, K + \mathrm{dim} \, K^{\perp} \\ & = \mathrm{dim} \, K + \mathrm{dim} \, f(K^{\perp}) \\ & = \mathrm{dim} \, K + \mathrm{dim} \, f(V) \\ & = \mathrm{dim} \, \mathrm{Ker}(f) + \mathrm{dim} \, \mathrm{Im}(f) \end{aligned} $$

$\endgroup$
5
  • 1
    $\begingroup$ Not every vector space is an inner product space. $\endgroup$ Apr 2 at 13:04
  • $\begingroup$ Fair enough. But they're finite dimensional, so they should be isomorphic to $\mathbb{R}^n$ and $\mathbb{R}^m$, right? $\endgroup$
    – user20672
    Apr 2 at 13:27
  • 1
    $\begingroup$ Not if the ground field is not $\mathbb{R}$. Do you see anything in the problem stating that this is a real or complex vector space? I don’t. The result holds over arbitrary fields, and in fact holds in infinite dimension as well. $\endgroup$ Apr 2 at 13:28
  • $\begingroup$ Also a good point $\endgroup$
    – user20672
    Apr 2 at 13:31
  • $\begingroup$ It doesn't need to be an orthogonal complement. Any complement of $\ker f$ will suffice. $\endgroup$ Apr 2 at 13:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.