The Dimension of a Sum of Subspaces Examples 1

The Dimension of a Sum of Subspaces Examples 1

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

(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}

We will now look at some example problems regarding this important formula for finite-dimensional vector spaces.

Example 1

Consider the vector space $\mathbb{R}^7$, and suppose that $U$ and $W$ are subspaces of $\mathbb{R}^7$ such that $\mathrm{dim} (U) = 3$ and $\mathrm{dim} (W) = 4$ and that $U + W = \mathbb{R}^7$. Show that $\mathbb{R}^7 = U \oplus W$.

We are already given that $\mathbb{R}^7 = U + W$, so to show that $\mathbb{R}^7 = U \oplus W$ we only need to prove that $U_1 \cap U_2 = \{ 0 \}$.

Note that $\mathrm{dim} (U + W) = \mathrm{dim} (\mathbb{R}^7) = 7$, $\mathrm{dim} (U) = 3$ and $\mathrm{dim}(W) = 4$. Using the formula above, we see that:

(2)
\begin{align} \quad \mathrm{dim} (U + W) = \mathrm{dim} (U) + \mathrm{dim} (W) - \mathrm{dim} (U \cap W) \\ \quad7 = 3 + 4 - \mathrm{dim} (U \cap W) \\ \quad 0 = \mathrm{dim} (U \cap W) \end{align}

So $\mathrm{dim} (U \cap W) = 0$ which implies that $U \cap W = \{ 0 \}$. Thus $\mathbb{R}^7 = U \oplus W$.

Example 2

Consider the vector $\mathbb{R}^{11}$, and suppose that $U$ and $W$ are subspaces of $\mathbb{R}^{11}$ such that $\mathrm{dim} (U) = 7$ and $\mathrm{dim} (W) = 8$. Show that $U \cap W \neq \{ 0 \}$.

We note that since $U$ and $W$ are both subsets of $\mathbb{R}^{11}$ then the sum $U + W$ is also a subset of $\mathbb{R}^{11}$ and so $\mathrm{dim} (U + W) ≤ 11$. Using the dimension formula from above, we see that:

(3)
\begin{align} \quad \mathrm{dim} (U + W) = \mathrm{dim} (U) + \mathrm{dim} (W) - \mathrm{dim} (U \cap W) \\ \quad \mathrm{dim} (U + W) = 7 + 8 - \mathrm{dim} (U \cap W) \\ \quad 15 - \mathrm{dim} (U \cap W) ≤ 11 \\ \quad - \mathrm{dim} (U \cap W) ≤ -4 \\ \quad \mathrm{dim} (U \cap W) ≥ 4 \end{align}

Since $\mathrm{dim} (U \cap W) ≥ 4$ we see that it is not possible for $\mathrm{dim} (U \cap W) = 0$ so $U \cap W \neq \{ 0 \}$.