The Dimension of a Direct Sum of Subspaces

# The Dimension of a Direct Sum of Subspaces

Recall from The Dimension of a Sum of Subspaces page that if $V$ is a finite-dimensional vector space and $U_1$ and $U_2$ are subspaces of $V$, then we have that:

(1)
\begin{align} \quad \mathrm{dim} (U_1 + U_2) = \mathrm{dim} (U_1) + \mathrm{dim} (U_2) - \mathrm{dim} (U_1 \cap U_2) \end{align}

Now suppose that $U_1$, $U_2$, …, $U_m$ are all subspaces to the finite-dimensional vector space $V$ and such that $V = U_1 \oplus U_2 \oplus ... \oplus U_m$. The following theorem gives us a formula for the dimension of $V$ in terms of the subspaces $U_1$, $U_2$, …, $U_m$.

 Theorem 1: Let $V$ be a finite-dimensional vector space such that $U_1$, $U_2$, …, $U_m$ are subspaces of $V$ and $V = U_1 \oplus U_2 \oplus U_m$. Then $\mathrm{dim} (V) = \mathrm{dim} (U_1) + \mathrm{dim} (U_2) + ... + \mathrm{dim} (U_m)$.
• Proof: Since $V$ is finite-dimensional then any subspace of $V$ is also finite-dimensional. Let $B_1$, $B_2$, …, $B_m$ be bases of $U_1$, $U_2$, …, $U_m$ respectively. Let $B = U_1 \cup U_2 \cup ... \cup U_m$. We note that every vector $v \in V$ is such that:
(2)
\begin{align} \quad v = u_1 + u_2 + ... + u_m \end{align}
• We note that $u_1 \in U_1$, $u_2 \in U_2$, …, $u_m \in U_m$. Therefore for each $u_i$ for $i = 1, 2, ..., m$ we have that $u_i$ is a linear combination of the vectors in $B_i$. Thus, $B$ is a spanning set of $V$.
• To show that $B$ is linearly independent, suppose that a linear combination of $B$ is zero. Then group terms coming from each basis $B_1$, $B_2$, …, $B_m$. Since $V = U_1 \oplus U_2 \oplus ... \oplus U_m$, this implies that each group in the equation equals zero, and since each group equations zero and $B_1$, $B_2$, …, $B_m$ are bases of each of these subspaces, then the coefficients all equal zero. Therefore $B$ is linearly independent.
• Since $B$ is linearly independent and has spans $V$ we have that $B$ is a basis of $V$. Note that $B$ also has $\sum_{i=1}^{m} \mathrm{dim} (U_i)$ elements, and so:
(3)
\begin{align} \quad \mathrm{dim} (V) = \mathrm{dim} (U_1) + \mathrm{dim} (U_2) + ... + \mathrm{dim} (U_m) \end{align}