Show that $\dim \operatorname{im}(f) + \dim \ker(f) = \dim V$

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.

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

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 $$\operatorname{im}(f)\Longrightarrow \dim \operatorname{im}(f)=n-k$$.

• $V/\ker(f)\cong\text{im}(f)$. Apr 1 at 16:59
• Should I add or do anything more on my proof? Apr 1 at 17:04
• @LaVendEr I think you had a typo - I changed it. Compare to see that you agree? Apr 1 at 17:12
• 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. Apr 1 at 17:45
• 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.) Apr 1 at 18:23

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 $$AX=0.$$

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}

• Not every vector space is an inner product space. Apr 2 at 13:04
• Fair enough. But they're finite dimensional, so they should be isomorphic to $\mathbb{R}^n$ and $\mathbb{R}^m$, right? Apr 2 at 13:27
• 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. Apr 2 at 13:28
• Also a good point Apr 2 at 13:31
• It doesn't need to be an orthogonal complement. Any complement of $\ker f$ will suffice. Apr 2 at 13:38