# Prove $\dim V / U$ equals $\dim V - \dim U$ without rank-nullity

From Algebra by Artin: Based on this, we see that the map from $$\varphi(G) \longrightarrow G/K$$ defined by $$\varphi(g) \mapsto gK$$ is a group isomorphism from the image (of $$\varphi$$) to the cosets of the kernel, so $$\varphi(G) \cong G/K$$.

If $$V$$ is a vector space and $$K$$ is a subspace of $$V$$, call $$V/K = \{ v + K : v \in V\}$$ a quotient space. Under the intuitive operations $$(v + K) + (u + K) = (v + u) + K$$ and $$\lambda(v + K) = (\lambda v) + K$$, any quotient space is a vector space.

The group isomorphism above naturally extends to a vector space isomorphism $$T:V \longrightarrow V'$$, proving $$\text{im}T \cong V/K$$, where $$K = \ker T$$.

Now if we prove $$\dim V / K = \dim V - \dim K$$, the rank-nullity theorem falls out as a corollary.

Let $$\pi$$ be the canonical map from $$V$$ to $$V/K$$, i.e. $$\pi(v) = v + K$$, which is surjective with kernel $$K$$. The rank-nullity theorem completes the proof that $$\dim V / K = \dim V - \dim K$$.

But how can we prove, when $$K$$ is a subspace of finite dimensional $$V$$, that $$\dim V / K = \dim V - \dim K$$? WITHOUT using the rank-nullity theorem.

EDIT: to clarify, the rank-nullity theorem states that if $$T:V \longrightarrow W$$ and $$V$$ is finite dimensional, then the rank (dimension of $$\text{im}T$$) plus the nullity (dimension of $$\ker T$$) equals $$\dim V$$.

• What is your statement of Rank-Nullity Theorem? I thought the statement you want to prove was the Rank-Nullity. Aug 21, 2020 at 18:05
• You could begin with a basis $u_{1}, \ldots, u_{m}$ for $K$, extend that basis to a basis $u_{1}, \ldots, u_{m}, v_{1}, \ldots, v_{n}$ for $V$ and check that $v_{1} + K, \ldots, v_{n} + K$ is a basis for $V / K$. Aug 21, 2020 at 18:08
• There is only one rank-nullity theorem: that the sum of rank and nulity of any linear map is the dimension of the domain. It immediately implies the fact in the OP. Aug 21, 2020 at 18:13

Take a basis of $$K$$ it has $$m=dim K$$ elements. It is linearly indendent in $$V$$, so it can be extended to a basis in $$V$$ by adding $$r=dim V-m$$ elements $$v_1,...v_r$$ Then $$v_1+K,...,v_r+K$$ are linearly independent in $$V/K$$ and span it. Hence the dim of the factor space is $$r$$ as claimed.

• Let $T:V \to W$ be a linear transformation. Take a basis of $K$ and extend to a basis of (finite dimensional) $V$ with $v_1, \dots, v_r$ additions. Then $T(v_1), \dots, T(v_r)$ is a basis for the image $T(V)$, which proves the rank nullity theorem. This is a nice proof, but I'm hoping to see an alternative approach. Aug 21, 2020 at 19:16
• The rank-nullity theorem is usually proved using RREF. Aug 21, 2020 at 19:18

What about using the following result:

Proposition. If $$K$$ is a subspace of a vector space $$V$$ and $$V/K$$ is finite-dimensional, then $$V \cong K \times (V/K) .$$

Proof. Let $$v_{1} + K, \ldots, v_{n} + K$$ be a basis for $$V/K$$. Then, for any $$v \in V$$ there exist scalars $$\alpha_{1}, \ldots, \alpha_{n}$$ such that $$v + K = \alpha_{1}(v_{1} + K) + \ldots + \alpha_{n}(v_{n} + K).$$ Now consider the linear map $$\varphi: V \rightarrow K \times (V/K)$$ defined by mapping $$v \in V$$ to
$$\left( v - \sum_{i=1}^{n}\alpha_{i} v_{i} \hspace{0.2cm}, \hspace{0.2cm} v + K \right) .$$ This linear map is an isomorphism. $$\square$$

Edit 1. Now as a corollary, suppose $$V$$ is finite-dimensional. Then $$K$$ is finite-dimensional and $$V/K$$ must be finite-dimensional as well, because for any basis $$v_{1}, \ldots, v_{n}$$ of $$V$$, the list $$v_{1} + K, \ldots, v_{n} + K$$ generates $$V/K$$. Using our previous result:

$$\dim V = \dim \left( K \times (V/K) \right) = \dim K + \dim V/K.$$

Edit 2. Let's prove that $$\varphi$$ is bijective. First suppose $$v \in V$$ is such that $$\varphi(v) = ( 0_{V}, K )$$ . Notice that $$0_{V}$$ is the additive identity of $$K$$ and $$K$$ is the additive identity of $$V/K$$, so $$(0_{V}, K)$$ is the additive identity of $$K \times (V/K)$$. By the definition of $$\varphi$$, it follows that $$v + K = K = 0 \cdot (v_{1} + K) + \ldots + 0 \cdot (v_{n} + K) ,$$ so $$v - \sum_{i=1}^{n} 0 \cdot v_{i} = 0_{V}$$ and $$v = 0_{V}$$. Hence $$\ker \varphi = \{ 0_{V} \}$$ and $$\varphi$$ is injective.

To prove surjectivity, consider an arbitrary element $$(u, v + K)$$ of $$K \times (V/K)$$. Since $$V/K$$ is finite-dimensional, we can write $$v + K = \alpha_{1}(v_{1} + K) + \ldots + \alpha_{n}(v_{n} + K).$$ Let's now take a look at the vector $$u + \sum_{i=1}^{n} \alpha_{i} v_{i}$$ in $$V$$. The equivalence class of this vector is precisely $$\alpha_{1}(v_{1} + K) + \ldots + \alpha_{1}(v_{n} + K) = v + K,$$ so $$\varphi \left( u + \sum_{i=1}^{n} \alpha_{i} v_{i} \right) = \left( u + \sum_{i=1}^{n} \alpha_{i} v_{i} - \sum_{i=1}^{n} \alpha_{i} v_{i} \hspace{0.2cm}, \hspace{0.2cm} v + K \right) = (u, v + K).$$

• This is how I think about it, and it generalizes nicely to projective objects in other categories. Aug 22, 2020 at 1:38
• Don't we need $K$ to be finite dimensional for the last line? Does $\dim V/K < \infty \implies \dim K < \infty$? I'd like to use the assumption that $V$ is finite dimensional instead of $V/K$. Does $\dim V < \infty \implies \dim V/K < \infty$? Aug 22, 2020 at 3:53
• This proof is exactly the sort of thing I was hoping for, but I'm having trouble verifying that the linear transformation $\varphi$ is bijective/invertible. Am I missing a simple way to prove this? Aug 22, 2020 at 4:04
• I added details on how $V$ being finite-dimensional implies that $V/K$ is finite-dimensional and why $\varphi$ is bijective. Let me know if something remains unclear. Aug 22, 2020 at 6:22

If $$K = \left\{ \mathbf{0}_V \right\}$$, where $$\mathbf{0}_V$$ denotes the zero vector of $$V$$, then $$\dim K = 0$$, and also $$V/K = \big\{ \, \{ v \} \colon v \in V \, \big\},$$ and so $$\dim V/K = \dim V = \dim V - \dim K.$$

So let us suppose that the subspace $$K$$ has non-zero vectors as well.

Let us suppose that $$\dim K = m$$, and let $$\left( e_1, \ldots, e_m \right)$$ be a basis (in fact an ordered basis) for $$K$$.

Let us suppose that $$\dim V = n$$.

If $$K = V$$, then of course $$V/K = \big\{ K \big\}$$ so that $$\dim V/K = 0 = \dim V - \dim K.$$

So let us suppose that $$K$$ is a proper subspace of $$V$$. Then of course $$n > m$$, and the ordered basis $$\left( e_1, \ldots, e_m \right)$$ of the subspace $$K$$ can be extended to an ordered basis $$\left( e_1, \ldots, e_m, e_{m+1}, \ldots, e_n \right)$$ for the whole space $$V$$, for some vectors $$e_{m+1}, \ldots, e_n \in V \setminus K$$.

We now show that the (ordered) set $$\left( e_{m+1} + K, \ldots, e_n + K \right)$$ forms a basis (i.e. an ordered basis) for the quotient space $$V/K$$.

Let $$v+K$$ be an arbitrary element of $$V/K$$, where $$v \in V$$.

As $$v \in V$$ and as $$\left( e_1, \ldots, e_m, e_{m+1}, \ldots, e_n \right)$$ is an ordered basis for $$V$$, so this $$v$$ can be expressed uniquely as a linear combination of the vectors $$e_1, \ldots, e_m, e_{m+1}, \ldots, e_n$$; that is, there exists a unique $$n$$-tuple $$\left( \alpha_1, \ldots, \alpha_m, \alpha_{m+1}, \ldots, \alpha_n \right)$$ of scalars such that $$v = \alpha_1 e_1 + \cdots + \alpha_m e_m + \alpha_{m+1} e_{m+1} + \cdots + \alpha_n e_n.$$ And, as $$e_1, \ldots, e_m \in K$$ and as $$K$$ is a (vector subspace) of $$V$$, so we obtain \begin{align} v+K &= \left( \alpha_1 e_1 + \cdots + \alpha_m e_m + \alpha_{m+1} e_{m+1} + \cdots + \alpha_n e_n \right) + K \\ &= \left( \alpha_1 e_1 + K \right) + \cdots \left( \alpha_m e_m + K \right) + \left( \alpha_{m+1} e_{m+1} + K \right) + \cdots + \left( \alpha_n e_n + K \right) \\ &= \alpha_1 \left( e_1 + K \right) + \cdots \alpha_m \left( e_m + K \right) + \alpha_{m+1} \left( e_{m+1} + K \right) + \cdots + \alpha_n \left( e_n + K \right) \\ &= \alpha_1 K + \cdots + \alpha_m K + \alpha_{m+1} \left( e_{m+1} + K \right) + \cdots + \alpha_n \left( e_n + K \right) \\ &= \underbrace{K + \cdots + K}_{\mbox{m terms}} + \alpha_{m+1} \left( e_{m+1} + K \right) + \cdots + \alpha_n \left( e_n + K \right) \\ &= K + \alpha_{m+1} \left( e_{m+1} + K \right) + \cdots + \alpha_n \left( e_n + K \right) \\ &= \alpha_{m+1} \left( e_{m+1} + K \right) + \cdots + \alpha_n \left( e_n + K \right). \end{align} Note that $$K$$ is the so-called zero vector of the quotient (vector) space $$V/K$$. Thus the ordered set $$\left( e_{m+1} + K, \ldots, e_n + K \right)$$ spans $$V/K$$.

We now show that $$\left( e_{m+1} + K, \ldots, e_n + K \right)$$ is linearly independent. For this suppose that, for some scalars $$\beta_{m+1}, \ldots, \beta_n$$, we have $$\beta_{m+1} \left( e_{m+1} + K \right) + \cdots \beta_n \left( e_n + K \right) = K.$$ Note once again that $$K$$ is the so-called zero vector of the quotient (vector) space $$V/K$$. The preceding equation can be rewritten as $$\left( \beta_{m+1} e_{m+1} + \cdots + \beta_n e_n \right) + K = K,$$ which implies that $$\beta_{m+1} e_{m+1} + \cdots + \beta_n e_n \in K,$$ and as $$\left( e_1, \ldots, e_m \right)$$ is an ordered basis for $$K$$, so there exists a unique $$m$$-tuple $$\beta_1, \ldots, \beta_m$$ of scalars such that $$\beta_{m+1} e_{m+1} + \cdots + \beta_n e_n = \beta_1 e_1 + \cdots + \beta_m e_m,$$ which implies that $$\beta_1 e_1 + \cdots + \beta_m e_m - \beta_{m+1} e_{m+1} - \cdots - \beta_n e_n = \mathbf{0}_V,$$ where $$\mathbf{0}_V$$ denotes the zero vector in $$V$$, and since the vectors $$e_1, \ldots, e_m, e_{m+1}, \ldots, e_n$$ being basis vectors are linearly independent, therefore we can conclude that $$\beta_1 = \cdots = \beta_m = \beta_{m+1} = \cdots = \beta_n = 0,$$ and thus in particular we obtain $$\beta_{m+1} = \cdots = \beta_n = 0,$$ thus showing the linear independence of $$\left( e_{m+1} + K, \ldots, e_n + K \right)$$.

Hence $$\left( e_{m+1} + K, \ldots, e_n + K \right)$$ is a (an ordered) basis for $$V/K$$, which shows that $$\dim V/K = n - m = \dim V - \dim K,$$ as required.

• Not sure why this got downvoted. You included the first part so that you wouldn't have to deal with the span of an empty sequence, correct? Aug 21, 2020 at 19:39
• This seems to be an elaboration on JCAA's answer. You can use essentially the same argument to prove the rank nullity theorem. $T(v) = \sum_{i=1}^n c_i T(e_i) = \sum_{i=m+1}^n c_i T(e_i)$ since $e_1, \dots, e_n \in K$, and $\sum_{i=m+1}^n c_i T(e_i) = 0$ implies $\sum_{i=m+1}^n c_i e_i \in K$ so all the scalars $c_{m + 1}, \dots, c_n$ are zero, so $T(e_{m + 1}), \dots, T(e_n)$ is a basis for the image of $T$. Aug 21, 2020 at 19:45
• @jskattt797 I'm sorry but I hadn't seen the earlier answer by JCAA. Moreover, I think my answer is easier to follow for a beginner. Aug 21, 2020 at 19:49
• @SaaqibMahmood,it is a nice answer
– MAS
Aug 22, 2020 at 6:17

All of these arguments are consequences of the following basic result about vector spaces:

Lemma: If $$V$$ is a vector space over a field $$\mathsf k$$ and $$S\subseteq V$$ is a linearly independent set, then there exists a basis of $$V$$ containing $$S$$.

Proof: The is true for any vector space, at least if you are happy to use the axiom of choice in the form of Zorn's Lemma: consider the set $$\mathcal T = \{T\subseteq V: S\subseteq T, T \text{ linearly independent}\}.$$ Any nested collection of elements of $$\mathcal T$$ has a least upper bound -- namely their union -- hence by Zorn's Lemma $$\mathcal T$$ has a maximal element $$B$$ say. But then if $$v \in V$$ we must have $$v\in \text{span}(B)$$, since otherwise $$B \cup \{v\}$$ would lie in $$\mathcal T$$ contradicting the maximality of $$B$$. $$\phantom{asdaAaaaaaa}\fbox{\phantom{a}}$$

The argument above is often given to show that any vector space has a basis, i.e. with $$S=\emptyset$$. The formulation above is more flexible and, for example, makes it easy to prove the following:

Claim: If $$K$$ is a subspace of $$V$$ then $$K$$ has a basis and any such basis of $$K$$ can be extended to a basis of $$V$$.

Proof: Applying the Lemma with $$V=K$$ and $$S=\emptyset$$ shows that $$K$$ has a basis $$B_K$$ say. Applying it again to $$V$$ with $$S=B_K$$ shows that there is a basis $$B_V$$ of $$V$$ with $$B_V\supseteq B_K$$ as required. $$\fbox{\phantom{a}}$$

Using the above, you can also show that the following proposition holds without any assumption on the dimension of $$V/K$$:

Proposition If $$K$$ is a subspace of $$V$$ then $$V\cong K\oplus V/K$$.

Proof: Let $$q\colon V \to V/K$$ be the quotient map, and $$C$$ a basis of $$V/K$$. Then since $$q$$ is surjective, we may pick $$D \subset V$$ such that $$q$$ restricts to a bijection $$q_C\colon D \to C$$. Now since $$C$$ is a basis, the map $$q_C^{-1}$$ extends uniquely to a linear map $$s\colon V/K\to V$$.

Now by definition $$q\circ s(c)=c$$ for all $$c \in C$$, hence by linearity, we must have $$q\circ s = \text{id}_{V/K}$$. Let $$W = \text{span}(D)$$ so that $$\text{im}(s)=W$$ and $$q_{|W}\colon W \to V/K$$ and $$s\colon V/K \to W$$ are inverses of each other.

This shows that in particular, $$ker(q) \cap W = \{0\}$$. Since for any $$v \in V$$ we have $$v= (v-s(q(v))) + s(q(v))$$ where clearly $$s(q(v)) \in W$$ and $$q(v-s(q(v))=q(v)-q(s(q(v))) = q(v)-q(v)=0$$, so that $$(v-s(q(v))) \in \ker(q)=K$$, it follows that $$V= K \oplus W$$. But then the map $$(\text{id}_K,q_{|W})$$ gives an isomorphism from $$V= K\oplus W$$ to $$K\oplus V/K$$ as required. $$\phantom{aaaaaaaaaaaaaaaa}\fbox{\phantom{a}}$$