From Algebra by Artin:

Artin algebra group homomorphism kernel cosets

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 $\dim V$.

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    $\begingroup$ What is your statement of Rank-Nullity Theorem? I thought the statement you want to prove was the Rank-Nullity. $\endgroup$
    – Brian Shin
    Aug 21, 2020 at 18:05
  • $\begingroup$ 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 $V / K$. $\endgroup$ Aug 21, 2020 at 18:08
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    $\begingroup$ 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. $\endgroup$
    – markvs
    Aug 21, 2020 at 18:13

4 Answers 4


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.

  • $\begingroup$ 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. $\endgroup$
    – jskattt797
    Aug 21, 2020 at 19:16
  • $\begingroup$ The rank-nullity theorem is usually proved using RREF. $\endgroup$
    – markvs
    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 $$ \varphi \left( u + \sum_{i=1}^{n} \alpha_{i} v_{i} \right) = \left( u + \sum_{i=1}^{n} \alpha_{i} v_{i} - \sum_{i=1}^{n} \alpha_{i} v_{i} \hspace{0.2cm}, \hspace{0.2cm} v + K \right) = (u, v + K). $$

  • 1
    $\begingroup$ This is how I think about it, and it generalizes nicely to projective objects in other categories. $\endgroup$ Aug 22, 2020 at 1:38
  • 1
    $\begingroup$ 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 $\dim V < \infty \implies \dim V/K < \infty$? $\endgroup$
    – jskattt797
    Aug 22, 2020 at 3:53
  • 1
    $\begingroup$ 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? $\endgroup$
    – jskattt797
    Aug 22, 2020 at 4:04
  • $\begingroup$ 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. $\endgroup$ 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.

  • $\begingroup$ 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? $\endgroup$
    – jskattt797
    Aug 21, 2020 at 19:39
  • $\begingroup$ 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 $T$. $\endgroup$
    – jskattt797
    Aug 21, 2020 at 19:45
  • 1
    $\begingroup$ @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. $\endgroup$ Aug 21, 2020 at 19:49
  • $\begingroup$ @SaaqibMahmood,it is a nice answer $\endgroup$
    – 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}$}$


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