The Dimension of a Sum of Subspaces Examples 2

# The Dimension of a Sum of Subspaces Examples 2

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 more example problems regarding this important formula for finite-dimensional vector spaces.

## Example 1

Let $V$ be a finite-dimensional vector space and let $U_1$, $U_2$, …, $U_m$ be finite-dimensional subspaces of $V$. Show that $\mathrm{dim} (U_1 + U_2 + ... + U_m) ≤ \mathrm{dim} (U_1) + \mathrm{dim} (U_2) + ... + \mathrm{dim} (U_m)$.

Consider the sum $U_1 + (U_2 + ... + U_m)$. In applying the formula above we see that:

(2)
\begin{align} \quad \mathrm{dim} (U_1 + [U_2 + ... + U_m]) = \mathrm{dim} (U_1) + \mathrm{dim} (U_2 + ... + U_m) - \mathrm{dim} (U_1 \cap [U_2 + U_3 + ... + U_m]) \end{align}

If we apply the dimension formula to the term $\mathrm{dim} (U_2 + [U_3 + ... + U_m])$ we have that:

(3)
\begin{align} \quad \mathrm{dim} (U_1 + U_2 + ... + U_m) = \mathrm{dim} (U_1) + \mathrm{dim} (U_2) + \mathrm{dim} (U_3 + ... + U_m) - \mathrm{dim} (U_2 \cap [U_3 + ... +U_m]) - \mathrm{dim} (U_1 \cap [U_2 + ... + U_m]) \end{align}

Let $S = \mathrm{dim} (U_1 \cap [U_2 + ... + U_m]) + \mathrm{dim} (U_2 \cap [U_3 + ... + U_m]) + ... + \mathrm{dim} (U_{m-1} \cap [U_m]) > 0$. Inductively we have that:

(4)
\begin{align} \quad \mathrm{dim} (U_1 + U_2 + ... + U_m) = \sum_{i=1}^{m} \mathrm{dim} (U_i) - S \\ \quad \mathrm{dim} (U_1 + U_2 + ... + U_m) ≤ \sum_{i=1}^{m} \mathrm{dim} (U_i) \\ \quad \mathrm{dim} (U_1 + U_2 + ... + U_m) ≤ \mathrm{dim} (U_1) + \mathrm{dim} (U_2) + ... + \mathrm{dim} (U_m) \end{align}

## Example 2

Let $V$ be a finite-dimensional vector space such that $\mathrm{dim} (V) = 5$ and let $U$ be a subspace of $V$ such that $\mathrm{dim} (U) = 2$. If $V = U \oplus W$, then what values can $\mathrm{dim} (W)$ be?

Using the formula from above and we have that:

(5)
\begin{align} \quad \mathrm{dim} (U + W) = \mathrm{dim} (U) + \mathrm{dim} (W) - \mathrm{dim} (U \cap W) \\ \quad 5 = 2 + \mathrm{dim} (W) - \mathrm{dim} (U \cap W) \\ \quad 3 = \mathrm{dim} (W) - \mathrm{dim} (U \cap W) \end{align}

Notice that since $V = U \oplus W$ then $U \cap W = \{ 0 \}$ so $\mathrm{dim} (U \cap W) = 0$, and so the equation above implies that $\mathrm{dim} W = 3$.

## Example 3

Consider the vector space $\mathbb{R}^5$, and let $U = \{ (x_1, x_2, x_3, x_4, x_5) \in \mathbb{R}^5 : x_1 = x_2, x_3 = x_4 = x_5 \}$ and $W = \{ (x_1, x_2, x_3, x_4, x_4) \in \mathbb{R}^5 : x_1 + x_2 = x_3 + x_4 + x_5 \}$ both be subspaces of $\mathbb{R}^5$. Determine $\mathrm{dim} (U \cap W)$.

Rearranging the dimension formula, we have that:

(6)
\begin{align} \quad \mathrm{dim} (U \cap W) = \mathrm{dim} (U) + \mathrm{dim} (W) - \mathrm{dim} (U + W) \end{align}

We can first rewrite the subspace $U$ as:

(7)
\begin{align} \quad U = \{ (x_1, x_1, x_3, x_3, x_3) \in \mathbb{R}^5 : x_1, x_3 \in \mathbb{R} \} \end{align}

Therefore $U = \mathrm{span} ((1, 1, 0, 0, 0), (0, 0, 1, 1, 1))$ and so $\mathrm{dim} (U) = 2$.

We can also rewrite the subspace $W$ as:

(8)
\begin{align} \quad W = \{ (x_1, x_2, x_3, x_4, x_1 + x_2 - x_3 - x_4) \in \mathbb{R}^5 : x_1, x_2, x_3, x_4 \in \mathbb{R} \} \end{align}

Therefore $W = \mathrm{span} (1, 0, 0, 0, 1), (0, 1, 0, 0, 1), (0, 0, 1, 0, -1), (0, 0, 0, 1, -1) \in \mathbb{R}^5 : x_1, x_2, x_3, x_4 \in \mathbb{R} \}$ and so $\mathrm{dim} (W) = 4$.

We also see that $\mathrm{R}^5 = U + W$ so $\mathrm{dim} (U + W) = 5$, and so then:

(9)
\begin{align} \quad \quad \mathrm{dim} (U \cap W) = \mathrm{dim} (U) + \mathrm{dim} (W) - \mathrm{dim} (U + W) \\ \quad \quad \mathrm{dim} (U \cap W) = 2 + 4 - 5 \\ \quad \quad \mathrm{dim} (U \cap W) = 1 \end{align}