This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.
Dimension of Vector Spaces
From §Bases of Vector Spaces, ⁋Example 9 and §Bases of Vector Spaces, ⁋Example 11, we see that a basis of a vector space \(V\) need not be unique. However, looking at these examples, we can also verify that the number of elements in each basis remains the same. This is not a coincidence.
Theorem 1 For a \(\mathbb{K}\)-vector space \(V\), if two bases \(\mathcal{B}_1\), \(\mathcal{B}_2\) of \(V\) are given, then \(\lvert \mathcal{B}_1\rvert=\lvert \mathcal{B}_2\rvert\) holds.
This theorem also includes the case where \(\mathcal{B}_1\), \(\mathcal{B}_2\) are infinite. To show this, three steps are needed.
- First, if any basis of \(V\) is infinite, then all other bases must also be infinite and have the same cardinality.
- Therefore, if any basis of \(V\) is finite, then all other bases must also be finite.
- Finally, if two finite bases of \(V\) are given, then the number of elements in these two bases is the same.
Of course, there is nothing preventing us from proving this theorem now, but like §Bases of Vector Spaces, ⁋Theorem 10, proving it requires a bit of set-theoretic knowledge, so we separate it into another post. However, the last step can be proved without much background knowledge.
Lemma 2 For a \(\mathbb{K}\)-vector space \(V\), if \(\mathcal{B}_1\) and \(\mathcal{B}_2\) are both bases of \(V\) and finite, then \(\lvert \mathcal{B}_1\rvert=\lvert \mathcal{B}_2\rvert\) holds.
Proof
Let \(\mathcal{B}_1=\{x_1,x_2,\ldots, x_m\}\), and \(\mathcal{B}_2=\{y_1,y_2,\ldots, y_n\}\); we must show \(m=n\). Suppose for contradiction that \(m>n\).
First, since \(x_1\in V\), we can express \(x_1\) as a linear combination of \(y_1\), \(y_2\), \(\ldots\), \(y_n\). Thus, by §Bases of Vector Spaces, ⁋Proposition 6, the set \(\{x_1,y_1,y_2,\ldots, y_n\}\) is linearly dependent. That is, there exist scalars \(\beta_1\), \(\alpha_1\), \(\alpha_2\), \(\ldots\), \(\alpha_n\), not all zero, such that
\[\beta_1x_1+\alpha_1y_1+\alpha_2y_2+\cdots+\alpha_n y_n=0\tag{1}\]holds. It is obvious that \(\beta_1\) cannot be zero. If \(\beta_1=0\), then the above equation becomes
\[\alpha_1y_1+\alpha_2y_2+\cdots+\alpha_ny_n=0\]which contradicts the assumption that \(y_1\), \(y_2\), \(\ldots\), \(y_n\) are linearly independent. Also, if all \(\alpha_i\) are zero, then \(\beta_1x_1=0\), and since \(\beta_1\neq 0\), we have \(x_1=0\). In this case \(\{x_1, x_2, \ldots, x_m\}\) would be trivially linearly dependent, so we may assume that some \(\alpha_i\) is nonzero. Then we can rearrange equation (1) to obtain the equation
\[y_i=\frac{\beta_1}{\alpha_i}x_1-\frac{\alpha_1}{\alpha_i}y_1-\cdots-\frac{\alpha_{i-1}}{\alpha_i}y_{i-1}-\frac{\alpha_{i+1}}{\alpha_i}y_{i+1}-\cdots-\frac{\alpha_n}{\alpha_i}y_n\]Therefore, even if we remove \(y_i\) from the set \(\{x_1, y_1, y_2, \ldots, y_n\}\), this set still spans \(V\).
On the other hand, this set is linearly independent. For any scalars \(\beta_1'\), \(\alpha_1'\), \(\ldots\), \(\alpha_n'\), if
\[\beta_1'x_1+\alpha_1'y_1+\alpha_2'y_2+\cdots+\alpha_{i-1}'y_{i-1}+\alpha_{i+1}'y_{i+1}+\cdots+\alpha_n'y_n=0\]then by the same reasoning as above, \(\beta_1'\neq 0\), and therefore
\[x_1=-\frac{\alpha_1'}{\beta_1'}y_1-\frac{\alpha_2'}{\beta_1'}y_2-\cdots-\frac{\alpha_{i-1}'}{\beta_1'}y_{i-1}-\frac{\alpha_{i+1}'}{\beta_1'}y_{i+1}-\cdots-\frac{\alpha_n'}{\beta_1'}y_n\]substituting this into the previous equation gives
\[0=\left(\frac{\alpha_1'\beta_1}{\alpha_i\beta_1'}+\frac{\alpha_1}{\alpha_i}\right)y_1+\cdots+\left(\frac{\alpha_{i-1}'\beta_1}{\alpha_i\beta_{i-1}'}+\frac{\alpha_{i-1}}{\alpha_i}\right)y_{i+1}+y_i+\left(\frac{\alpha_{i+1}'\beta_{i+1}}{\alpha_i\beta_{i+1}'}+\frac{\alpha_{i+1}}{\alpha_i}\right)y_{i+1}+\cdots+\left(\frac{\alpha_n'\beta_n}{\alpha_i\beta_n'}+\frac{\alpha_n}{\alpha_i}\right)y_n\]Since the coefficient of \(y_i\) is not zero, this contradicts the assumption that \(\{y_1,y_2,\ldots,y_n\}\) is linearly independent.
Thus we have obtained a new basis \(\{x_1,y_1,y_2,\ldots,y_{i-1}, y_{i+1},\ldots, y_n\}\) of \(V\). Without loss of generality, if the vector we removed was \(y_n\), then this new basis is \(\{x_1, y_1, \ldots, y_{n-1}\}\). Now add \(x_2\) to this basis again and consider \(\{x_2, x_1, y_1, y_2, \ldots, y_n\}\).
If
\[\beta_2x_2+\beta_1x_1+\alpha_1y_1+\alpha_2y_2+\ldots+\alpha_{n-1}y_{n-1}=0\]then by the same logic as above, \(\beta_2\neq 0\), and excluding the case \(x_2=0\), there exists a nonzero coefficient among \(\beta_1\), \(\alpha_1\), \(\ldots\), \(\alpha_{n-1}\).
Here, if \(\beta_1\) is the unique nonzero coefficient, then the above equation becomes \(\beta_2x_2+\beta_1x_1=0\), so \(\{x_1,x_2\}\) is linearly dependent and the proof is complete.
Otherwise, if some \(\alpha_i\neq 0\) exists, we repeat the same process to obtain a new basis \(\{x_2,x_1,y_1,y_2,\ldots,y_{n-2}\}\).
Repeating this process, there are two possibilities.
-
If this process stops somewhere,
\[\beta_kx_k+\beta_{k-1}x_{k-1}+\cdots+\beta_1x_1=0\]then this becomes \(\beta_kx_k+\beta_{k-1}x_{k-1}+\cdots+\beta_1x_1=0\), so \(\{x_1,x_2,\ldots,x_m\}\) is linearly dependent.
-
Otherwise, after repeating \(n\) times, we will completely replace the original basis \(\{y_1,y_2,\ldots, y_n\}\) with a new basis \(\{x_1, x_2, \ldots, x_n\}\). In this case, \(x_{n+1}\in V\) can be expressed as a linear combination of \(\{x_1, x_2, \ldots, x_n\}\), so \(\{x_1,x_2,\ldots, x_{n+1}\}\) is linearly dependent, and hence so is \(\{x_1,x_2,\ldots, x_m\}\).
In either case, \(\{x_1,x_2,\ldots, x_m\}\) is linearly dependent and therefore cannot be a basis, which is a contradiction.
In fact, the proof of the above proposition actually proves the following slightly stronger proposition than the original:
Let a \(\mathbb{K}\)-vector space \(V\) have a finite basis \(\mathcal{B}\). Then any subset of \(V\) having more elements than \(\mathcal{B}\) must be linearly dependent.
Anyway, by Theorem 1, all bases of \(V\) have the same size, so the following definition makes sense.
Definition 3 For a \(\mathbb{K}\)-vector space \(V\), the cardinality of a basis of \(V\) is called the dimension of \(V\), denoted \(\dim V\), or \(\dim_\mathbb{K}V\) when we need to emphasize \(\mathbb{K}\). If \(\dim V\) is finite, then \(V\) is a finite-dimensional vector space; otherwise, \(V\) is an infinite-dimensional vector space.
Example 4
- The basis of the trivial vector space \(\{0\}\) is \(\emptyset\), so the dimension of this space is \(\lvert\emptyset\rvert=0\).
- For any field \(\mathbb{K}\), \(\mathbb{K}\) itself is a 1-dimensional \(\mathbb{K}\)-vector space.
- For any field \(\mathbb{K}\), the dimension of the Euclidean \(n\)-space \(\mathbb{K}^n\) is \(\dim \mathbb{K}^n=n\).
- \(\dim_\mathbb{R}\mathbb{C}=2\).
- \(\mathbb{K}[\x]\) is an infinite-dimensional vector space.
In what follows, we always assume that the vector spaces we treat are finite-dimensional.
Sometimes results for finite-dimensional vector spaces also hold in infinite dimensions, and sometimes they do not. For example, the following proposition can be extended to the infinite-dimensional case, but in this post we shall restrict ourselves to the finite-dimensional case.
Proposition 5 For a \(\mathbb{K}\)-vector space \(V\) and any linearly independent subset \(S\) of \(V\), there exists a basis \(\mathcal{B}\) of \(V\) containing \(S\).
Proof
If \(\langle S\rangle=V\), there is nothing more to prove. Otherwise, there exists \(v\in V\) with \(v\not\in\langle S\rangle\). Now set \(S_1=S\cup\{v\}\). Then \(S_1\) is linearly independent. Obviously \(v\neq 0\), and now for any linear combination of \(S_1\)
\[\sum_{x\in S_1} \alpha_xx=\sum_{x\in S}\alpha_xx+\alpha_vv=0\]if this equals zero, then in the case \(\alpha_v\neq 0\), moving \(\alpha_vv\) to the other side and multiplying by \(-\alpha_v^{-1}\) would allow us to express \(v\) as a linear combination of elements of \(S\), which contradicts the choice of \(v\). Therefore \(\alpha_v=0\), and then since the elements of \(S\) are linearly independent, \(\alpha_x=0\) holds for all \(x\in S\). Hence \(\alpha_x=0\) holds for all \(x\in S_1\).
Now if \(\langle S\rangle_1=V\), the proof is again complete; otherwise, we can define \(S_2=S_1\cup\{v'\}\) in the same way and repeat. Of course we must show that \(S_2\) is linearly independent, but since \(v'\) was chosen from \(V\setminus\langle S\rangle_1\), this is possible by exactly the same logic as shown above.
By the preceding Lemma 3, this process terminates in at most \(\dim V\) steps, and when this process ends we obtain the desired basis \(S_n\).
A basis of \(V\) is a set that is both linearly independent and spans \(V\). The above proposition says that we can add vectors appropriately to a linearly independent set so that it spans \(V\). Conversely, if there is a set that spans \(V\), we can remove some redundant elements appropriately so that the linear independence condition is also satisfied. The basic idea of the proof of this proposition is the same as that of Proposition 5, but since \(S\) may be an infinite set, the proof does not work by removing elements from \(S\) one by one.
Proposition 6 For a \(\mathbb{K}\)-vector space \(V\) and a subset \(S\) that spans \(V\), some subset of \(S\) is a basis of \(V\).
Proof
Let \(S_0=\emptyset\). Then \(\langle S\rangle_0=\{0\}\). Now choose an element \(x_1\) from \(S\setminus\langle S\rangle_0\) and set \(S_1=\{x_1\}=S_0\cup\{x_1\}\), and similarly choose an element \(x_2\) from \(S\setminus\langle S\rangle_1\) and repeat the process of making \(S_2=\{x_1,x_2\}=S_1\cup \{x_2\}\).
The sets \(S_i\) obtained in this way are linearly independent subsets by construction, and as long as \(\langle S\rangle_i\) is not equal to \(S\), the number of elements in \(S_{i+1}\) is always one more than in \(S_i\). Therefore it suffices to show that \(S\setminus\langle S\rangle_i\) is nonempty for all \(i < n = \dim V\).
Choose a natural number \(m\) such that \(S\setminus\langle S\rangle_m=\emptyset\). That is, \(S\subseteq\langle S\rangle_m\). Now from §Bases of Vector Spaces, ⁋Lemma 4, taking the \(\span\) preserves inclusion relations between sets, so
\[\langle S\rangle\subseteq\span\bigl(\langle S\rangle_m\bigr)\]holds, and since \(\langle S\rangle_m\) on the right-hand side is already a subspace of \(V\), from §Bases of Vector Spaces, ⁋Lemma 3 we know that \(\span\bigl(\langle S\rangle\bigr)=\langle S\rangle_m\). Therefore
\[V=\langle S\rangle\subseteq\span\bigl(\langle S\rangle_m\bigr)=\langle S\rangle_m\]from \(V=\langle S\rangle\subseteq\span\bigl(\langle S\rangle_m\bigr)=\langle S\rangle_m\) we know that \(\langle S\rangle_m=V\).
Finally, let us look at two slightly more general examples.
Example 7 Let two \(\mathbb{K}\)-vector spaces \(V\), \(W\) be given. Then their product \(V\times W\) is the vector space consisting of vectors of the form \((v,w)\) for arbitrary \(v\in V\), \(w\in W\). Their operations are given by
\[(v_1, w_1)+(v_2,w_2)=(v_1+v_2,w_1+w_2),\quad\alpha(v,w)=(\alpha v,\alpha w)\]respectively. It is not difficult to check that if \(\mathcal{B}_1\), \(\mathcal{B}_2\) are bases of \(V\), \(W\) respectively, then the subset
\[\mathcal{B}=\{(x, y)\mid x\in \mathcal{B}_1\text{ and }y\in \mathcal{B}_2\}\]of \(V\times W\) is a basis of \(V\times W\). In particular, if \(V\), \(W\) are both finite-dimensional, then so is \(V\times W\), and \(\dim(V\times W)\) equals \((\dim V)+(\dim W)\).
Example 8 Now let a \(\mathbb{K}\)-vector space \(V\) be given, and let \(W_1\), \(W_2\) be two subspaces of \(V\). Then the subspace \(W_1+W_2\) of \(V\) is defined as
the smallest subspace of \(V\) containing both subspaces \(W_1\), \(W_2\).
In symbols, we can write \(W_1+W_2=\span(W_1\cup W_2)\). Now assuming that \(W_1,W_2\) are both finite-dimensional,
\[\dim(W_1+W_2)=\dim W_1+\dim W_2-\dim(W_1\cap W_2)\]holds.
Proof
Let \(W_1,W_2\) have dimensions \(m\), \(n\) respectively, and let \(W_1\cap W_2\) have dimension \(k\). Then there exists a basis \(\mathcal{B}_0=\{x_1,\ldots, x_k\}\) of \(W_1\cap W_2\). Since this set is a linearly independent subset of both \(W_1\) and \(W_2\), there exist bases of \(W_1\) and \(W_2\) respectively containing this set. Let these be \(\mathcal{B}_1\) and \(\mathcal{B}_2\). Then
\[\mathcal{B}_1=\{y_1,\ldots, y_m\},\quad \mathcal{B}_2=\{z_1,\ldots, z_n\},\qquad y_1=z_1=x_1,\ldots, y_k=z_k=x_k\]we may write. Now the set
\[\mathcal{B}_1\cup\mathcal{B}_2=\{x_1=y_1,\ldots, x_k=y_k, \quad y_{k+1}, \ldots, y_m,\quad z_{k+1},\ldots, z_n\}\]spans \(W_1+W_2\). Moreover, this set is linearly independent. To show this, suppose
\[\alpha_1x_1+\cdots+\alpha_kx_k+\beta_{k+1}y_{k+1}+\cdots+\beta_{m}y_m+\gamma_{k+1}z_{k+1}+\cdots+\gamma_{n}z_n=0\tag{2}\]For any scalars \(\beta_i\), \(\gamma_i\) (\(i\leq k\)) satisfying \(\alpha_i=\beta_i+\gamma_i\),
\[\beta_1y_1+\cdots+\beta_ky_k+\beta_{k+1}y_{k+1}+\cdots+\beta_{m}y_m=-\gamma_1z_1-\cdots-\gamma_kz_k-\gamma_{k+1}z_{k+1}-\cdots-\gamma_{n}z_n\]writing this equation, the left-hand side is an element of \(W_1\) and the right-hand side is an element of \(W_2\), so this common vector is an element of \(W_1\cap W_2\). Since \(\mathcal{B}_0\) is a basis of \(W_1\cap W_2\), choosing suitable scalars \(\alpha_i'\),
\[\beta_1y_1+\cdots+\beta_my_m=\alpha_1'x_1+\cdots+\alpha_k'x_k=-\gamma_1z_1-\cdots-\gamma_nz_n\]we can write. In the first equation, moving the \(\alpha_i'x_i\) terms back to the left-hand side and combining with the \(\beta_iy_i\) terms,
\[(\beta_1-\alpha_1')y_1+\cdots+(\beta_k-\alpha_k')y_k+\beta_{k+1}y_{k+1}+\cdots+\beta_my_m=0\]we obtain zero, so by the linear independence of \(\mathcal{B}_1\) all coefficients are zero, and in particular \(\beta_{k+1}=\cdots=\beta_m=0\). Similarly, from the second equation \(\gamma_{k+1}=\cdots=\gamma_n=0\), then the only remaining equation from (2) is \(\alpha_1x_1+\cdots+\alpha_kx_k=0\), and since \(x_1,\ldots,x_k\) form a basis of \(W_1\cap W_2\), they are all zero again by linear independence. Thus, \(\mathcal{B}_1\cup\mathcal{B}_2\) is a linearly independent subset spanning \(W_1+W_2\), and therefore a basis of \(W_1+W_2\). Now
\[\dim(W_1+W_2)=\lvert\mathcal{B}_1\cup\mathcal{B}_2\rvert=\lvert\mathcal{B}_1\rvert+\lvert\mathcal{B}_2\rvert-\lvert\mathcal{B}_0\rvert=\dim W_1+\dim W_2-\dim(W_1\cap W_2).\]References
[Goc] M.S. Gockenbach, Finite-dimensional linear algebra, Discrete Mathematics and its applications, Taylor&Francis, 2011.
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