Section 3.7 Sum and direct sum of vector subspaces
This section describes a couple of ways to obtain a new vector space from given vector spaces.
Definition 3.7.1. (Sum of vector subspaces).
Let \(W_1,W_2,W_3,\ldots, W_r\) be subspaces of a vector space \(V\) over a field \(F\text{.}\) The sum of \(W_i\) is denoted by \(W_1+W_2+\cdots+W_r\) and is defined by
\begin{equation*}
W_1+W_2+\cdots+W_r=\{w_1+w_2+\cdots+w_r:w_i\in W_i\}.
\end{equation*}
It is a subspace of \(V\text{.}\)We have the following result for the sum of two vector subspaces.
Theorem 3.7.3.
Let \(W_1,W_2\) be finite-dimensional subspaces of a vector space \(V\text{.}\) Then \(W_1+W_2\) is finite-dimensional and
\begin{equation*}
\dim_F(W_1+W_2)=\dim_FW_1+\dim_FW_2-\dim_F(W_1\cap W_2).
\end{equation*}
Proof.
Assume that \(\mathfrak{B}_i\) is an \(F\)-basis for \(W_i\text{.}\) By the definition of the sum of vector subspaces, we get that \(\mathfrak{B}_1\cup\mathfrak{B}_2\) generates \(W_1+W_2\text{.}\) Hence, \(W_1+W_2\) is finite-dimensional. Since \(W_1\cap W_2\leq W_i\text{,}\) the subspace \(W_1\cap W_2\) is also finite-dimensional. Consider \(\{v_1,v_2,\ldots,v_r\}\) to be a basis of \(W_1\cap W_2\text{.}\) By Remark 3.4.6, \(\{v_1,v_2,\ldots,v_r\}\) can be completed to a basis for \(W_i\text{.}\) Say \(\{v_1,v_2,\ldots,v_r,x_1,\ldots,x_n\}\) is a basis for \(W_1\) and \(\{v_1,v_2,\ldots,v_r,y_1,\ldots,y_m\}\) is a basis for \(W_2\text{.}\) The set \(\{v_1,\ldots,v_r,x_1,\ldots,x_n,y_1,\ldots,y_m\}\) spans \(W_1+W_2\text{.}\) Now suppose
\begin{gather*}
\sum\alpha_iv_i+\sum\beta_jx_j+\sum\gamma_ky_k=0\\
\text{i.e.,}\quad\sum\alpha_iv_i+\sum\beta_jx_j=-\sum\gamma_ky_k
\end{gather*}
We thus get \(\sum\gamma_ky_k\in W_1\cap W_2\text{,}\) and hence, \(\gamma_k=0\) for every \(k\text{.}\) Therefore, from the above equations, \(\sum\alpha_iv_i+\sum\beta_jx_j=0\text{.}\) As the set \(\{v_i,x_j\}\) is linearly independent, \(\alpha_i=0=\beta_j\) for every \(i,j\text{.}\) Thus, \(\{v_1,\ldots,v_r,x_1,\ldots,x_n,y_1,\ldots,y_m\}\) is linearly independent and hence a basis for \(W_1+W_2\text{.}\) The equality for dimension now follows.
Definition 3.7.4. (Direct sum of vector subspaces).
Let \(V\) be a vector space over a field \(F\text{,}\) and let \(W_i\) (\(1\leq i\leq n\)) be vector subspaces. A subspace \(W\leq V\) is said to be the direct sum of \(W_i\) if every vector \(w\in W\) can be uniquely represented in the form \(\sum_{i=1}^{n}w_i\text{,}\) where \(w_i\in W_i\text{.}\) In this case, we write \(W=\bigoplus_{i=1}^{n} W_i\text{.}\)
Theorem 3.7.5.
Let \(W\) be a finite-dimensional vector space over a field \(F\text{.}\) Let \(W_i\leq W\) be subspaces of a vector space \(W\text{.}\) Then, \(W=\bigoplus_{i=1}^{n}W_i\) if and only if any of the following conditions hold:
\(W=\sum_{i=1}^{n}W_i\) and \(W_j\cap\big(\sum_{i\neq j}W_i\big)=\{0\}\text{.}\) for all \(1\leq j\leq n\text{.}\)
\(W=\sum_{i=1}^{n}W_i\) and \(\dim_FW=\sum_{i=1}^{n}\dim_FW_i\text{.}\)
Proposition 3.7.6.
Let \(V\) be a finite-dimensional vector space over a field \(F\) and let \(W\) be a proper subspace of \(V\text{.}\) Then there exists a subspace \(W^\prime\) of \(V\) such that
\begin{equation*}
V=W\bigoplus W^\prime.
\end{equation*}
Proof.
Let \(\{w_1,\ldots,w_r\}\) be a basis for \(W\text{.}\) By Remark 3.4.6, there exists \(\{v_1,\ldots,v_s\}\subset V\) such that \(\{w_1,\ldots,w_r,v_1,\ldots,v_n\}\) is a basis for \(V\text{.}\) Consider \(W^\prime\) to be the subspace spanned by \(\{v_1,\ldots,v_s\}\text{.}\) We claim that \(V=W\bigoplus W^\prime\text{.}\) Since \(\{w_1,\ldots,w_r,v_1,\ldots,v_s\}\) is a basis of \(V\text{,}\) given any \(v\in V\) there exists unique scalars \(\alpha_i,\beta_j\in F\) such that \(v=\sum_i\alpha_iw_i+\sum_j\beta_jv_j\text{.}\) In particular, \(V=W+W^\prime\text{.}\)
If \(v\in W\cap W^\prime\) then, for some scalars \(\gamma_i,\delta_j\in F\text{,}\) we have the following.
\begin{equation*}
v=\sum_i\gamma_iw_i=\sum_j\delta_jv_j\quad\text{i.e.,}\quad 0=\sum_i\gamma_iw_i-\sum_j\delta_jv_j.
\end{equation*}
The linear independence of \(\{w_1,\ldots,w_r,v_1,\ldots,v_s\}\) implies that all \(\gamma_i=0\) and all \(\delta_j=0\text{.}\) Therefore, \(W\cap W^\prime=\{0\}\text{.}\)
By using the first condition in
Theorem 3.7.5, we get the required result.
