The Dimension of a Sum of Subspaces

# The Dimension of a Sum of Subspaces

We will now look at a very important theorem which relates the dimension of a sum of subspaces of a finite-dimensional vector space to the dimension of each of the individual subspaces and their set intersection.

Theorem 1: Let $V$ be a finite-dimensional vector space, and let $U_1$ and $U_2$ be subspaces of $V$. Then the dimension of the subspace sum $U_1 + U_2$ can be obtained with the formula $\mathrm{dim} (U_1 + U_2) = \mathrm{dim} (U_1) + \mathrm{dim} (U_2) - \mathrm{dim} (U_1 \cap U_2)$. |

**Proof:**Let $V$ be a finite-dimensional vector space and let $U_1$ and $U_2$ be subspaces of $V$. We have already seen that since $V$ is finite-dimensional then any subspace of $V$ will also be finite-dimensional. In particular, $U_1$ and $U_2$ will be finite-dimensional.

- Let $\{ u_1, u_2, ..., u_m \}$ be a basis of $U_1 \cap U_2$ so that $\mathrm{dim} (U_1 \cap U_2) = m$. We note that $\{ u_1, u_2, ..., u_m \}$ is linearly independent as a basis in $U_1 \cap U_2$, and so this set of vectors is also linearly independent in $U_1$, and so we can extend this set of vectors to a basis $\{ u_1, u_2, ..., u_m, v_1, v_2, ..., v_j \}$ of $U_1$. Therefore $\mathrm{dim} (U_1) = m + j$. We can also take the set of vectors $\{ u_1, u_2, ..., u_m\}$ and extend this set of vectors to a basis $\{ u_1, u_2, ..., u_m, w_1, w_2, ..., w_k \}$ of $U_2$ by the same logic. Therefore $\mathrm{dim} (U_2) = m + k$.

- Note that $\mathrm{dim} (U_1) + \mathrm{dim} (U_2) - \mathrm{dim} (U_1 \cap U_2) = (m + j) + (m + k) - m = m + j + k$. We thus want to show that $\mathrm{dim} (U_1 + U_2) = m + j + k$. We will do this by showing that $\{ u_1, u_2, ..., u_m, v_1, v_2, ..., v_j, w_1, w_2, ..., w_k \}$ is a basis of $U_1 + U_2$.

- To show that $\{ u_1, u_2, ..., u_m, v_1, v_2, ..., v_j, w_1, w_2, ..., w_k \}$ is a basis of $U_1 + U_2$ we must show that this set of vectors spans $U_1 + U_2$ and that it is linearly independent in $U_1 + U_2$.

- Let $u \in U_1 + U_2$. Then $u = v + w$ where $v \in U_1$ and $w \in U_2$. We already have bases for both of these subspaces though, and for some set of scalars $a_1, a_2, ..., a_m, b_1, b_2, ..., b_j \in \mathbb{F}$ and $c_1, c_2, ..., c_m, d_1, d_2, ..., d_k \in \mathbb{F}$ we have that:

\begin{align} \quad u = v + w \\ \quad u = \underbrace{a_1u_1 + a_2u_2 + ... + a_mu_m + b_1v_1 + b_2v_2 + ... + b_jv_j}_{= v \in U_1}+ \underbrace{c_1u_1 + c_2u_2 + ... + c_mu_m + d_1w_1 + d_2w_2 + ... + d_kw_k}_{=w \in U_2} \\ \quad u = (a_1 + c_1)u_1 + (a_2 + c_2)u_2 + ... + (a_m + c_m)u_m + b_1v_1 + b_2v_2 + ... + b_jv_j + d_1w_1 + d_2w_2 + ... + d_kw_k \end{align}

- As we can see, $u \in \mathrm{span} (u_1, u_2, ..., u_m, v_1, v_2, ..., v_j, w_1, w_2, ..., w_k)$. We now only need to show that this set of vectors is linearly independent. Consider the following vector equation:

\begin{align} \quad a_1u_1 + a_2u_2 + ... + a_mu_m + b_1v_1 + b_2v_2 + ... + b_jv_j + c_1w_1 + c_2w_2 + ... + c_kw_k = 0 \\ \quad c_1w_1 + c_2w_2 + ... + c_kw_k = -(a_1u_1 + a_2u_2 + ... + a_mu_m + b_1v_1 + b_2v_2 + ... + b_jv_j) \end{align}

- Now notice that the equation above implies that $c_1w_1 + c_2w_2 + ... + c_kw_k$ is a linear combination of the vectors $\{ u_1, u_2, ..., u_m, v_1, v_2, ..., v_j \}$ which is a basis of the subspace $U_1$. Therefore $( c_1w_1 + c_2w_2 + ... + c_kw_k ) \in U_1$. However, $(c_1w_1 + c_2w_2 + ... + c_kw_k) \in U_2$, and so $( c_1w_1 + c_2w_2 + ... + c_kw_k ) \in U_1 \cap U_2$. Now we note that $\{ u_1, u_2, ..., u_m \}$ is a basis of $U_1 \cap U_2$ and so for some set of scalars $d_1, d_2, ..., d_m \in \mathbb{F}$ we have that:

\begin{align} \quad c_1w_1 + c_2w_2 + ... + c_kw_k = d_1u_1 + d_2u_2 + ... + d_mu_m \\ \quad d_1u_1 + d_2u_2 + ... + d_mu_m - (c_1w_1 + c_2w_2 + ... + c_kw_k) = 0 \end{align}

- However, the list of vectors $\{ u_1, u_2, ..., u_m, w_1, w_2, ..., w_k \}$ is linearly independent which implies that $d_1 = d_2 = ... = d_m = 0$ and $c_1 = c_2 = ... = c_k = 0$. Therefore we have that:

\begin{align} \quad a_1u_1 + a_2u_2 + ... + a_mu_m + b_1v_1 + b_2v_2 + ... + b_jv_j = 0 \end{align}

- Now since $\{ u_1, u_2, ..., u_m, v_1, v_2, ..., v_j \}$ is also linearly independent, this implies that $a_1 = a_2 = ... = a_m = 0$ and $b_1 = b_2 = ... = b_j = 0$. Therefore all of the scalars are equal to zero and so $\{ u_1, u_2, ..., u_m, v_1, v_2, ..., v_j, w_1, w_2, ..., w_k \}$ is linearly independent.

- Since $\{ u_1, u_2, ..., u_m, v_1, v_2, ..., v_j, w_1, w_2, ..., w_k \}$ is spans $U_1 + U_2$ and is linearly independent in $U_1 + U_2$ we must have that $\{ u_1, u_2, ..., u_m, v_1, v_2, ..., v_j, w_1, w_2, ..., w_k \}$ is a basis of $U_1 + U_2$ and so $\mathrm{dim} (U_1 + U_2) = m + j + k$, and so:

\begin{align} \quad \mathrm{dim} (U_1 + U_2) = \mathrm{dim} (U_1) + \mathrm{dim} (U_2) - \mathrm{dim} (U_1 \cap U_2) \quad \blacksquare \end{align}