Lecture 12 - Dimension

Review from last time our proof for Lecture 11 - Starting Infinite Dimensional Spaces!#^54e316. We proved that for any subspace $U$ of $V$ , you can find subspace $W$ such that:

U \oplus W = V

We continue on defining dimension of vector spaces.

Dimension

We know that $R^{n}$ is $n$ -dimensional, but for other spaces it's harder to determine. The process is:

Find a basis for $V$
Determine the number of vectors in the basis. That's the dimension.

As an example, consider $F^{n, n}$ is the vector space of $n \times n$ matrices with entries from $F$ .
Further, let $U^{n}$ and $L^{n}$ denote the subspaces of upper and lower triangular matrices respectively, so these are subspaces. What is the dimension of our vector space? It is:

dim (F^{n, n}) = n \cdot n = n^{2}

This is because you can make each basis vector be some matrix with a 1 in the $i, j$ position, and 0's everywhere else. This would construct a basis with $n^{2}$ vectors:

\underset{n \times n vectors}{\underset{⏟}{{[\begin{matrix} 1 & 0 & \dots & 0 \\ 0 & 0 & \dots & 0 \\ ⋮ \\ 0 & 0 & \dots & 0 \end{matrix}], [\begin{matrix} 0 & 1 & \dots & 0 \\ 0 & 0 & \dots & 0 \\ ⋮ \\ 0 & 0 & \dots & 0 \end{matrix}], \dots, [\begin{matrix} 0 & 0 & \dots & 0 \\ 0 & 0 & \dots & 0 \\ ⋮ \\ 0 & 0 & \dots & 1 \end{matrix}],}}}

Recall that upper triangular allows numbers on the diagonal. Thus, we get:

dim (U^{n}) = \frac{n (n + 1)}{2}

comes from the triangular numbers. It's essentially the top row of $n$ entries plus the next entry of $n - 1$ entries, and so on until 1.

Consequently, it's the same for the lower triangular subspace.

dim (L^{n}) = \frac{n (n + 1)}{2}

Now a good question is what is $U^{n} + L^{n}$ . It's just the entire space $F^{n, n}$ . But how could I show this? Clearly the LHS is a $\subseteq$ to the RHS (it's trivial), as these both are already subspaces. Now we need to show that any element from the RHS belongs in the LHS. Let $x \in F^{n, n}$ is arbitrary:

x = [\begin{matrix} a_{11} & a_{12} & \dots & a_{1 n} \\ a_{21} & a_{22} & \dots & a_{2 n} \\ ⋮ & ⋱ \\ a_{n 1} & a_{n 2} & \dots & a_{n n} \end{matrix}] = [\begin{matrix} a_{11} & a_{12} & \dots & a_{1 n} \\ 0 & a_{22} & \dots & a_{2 n} \\ ⋮ & ⋱ \\ 0 & 0 & \dots & a_{n n} \end{matrix}] + [\begin{matrix} 0 & 0 & \dots & 0 \\ a_{21} & 0 & \dots & 0 \\ ⋮ & ⋱ \\ a_{n 1} & a_{n 2} & \dots & 0 \end{matrix}]

Thus $x$ is the sum of two vectors from $U^{n}$ and $L^{n}$ , so then $F^{n, n} = U^{n} + L^{n}$ .

Notice that since the diagonals are included for each subspace, then this choice of vectors wasn't unique:

x = [\begin{matrix} a_{11} & a_{12} & \dots & a_{1 n} \\ a_{21} & a_{22} & \dots & a_{2 n} \\ ⋮ & ⋱ \\ a_{n 1} & a_{n 2} & \dots & a_{n n} \end{matrix}] = [\begin{matrix} 0 & a_{12} & \dots & a_{1 n} \\ 0 & 0 & \dots & a_{2 n} \\ ⋮ & ⋱ \\ 0 & 0 & \dots & 0 \end{matrix}] + [\begin{matrix} a_{11} & 0 & \dots & 0 \\ a_{21} & a_{22} & \dots & 0 \\ ⋮ & ⋱ \\ a_{n 1} & a_{n 2} & \dots & a_{n n} \end{matrix}]

which shows that $F^{n, n} \neq U^{n} \oplus L^{n}$ .

But this brings up an interesting point. What's the dimension of the sum of both spaces? We already know that:

dim (F^{n, n}) = n^{2} = dim (U^{n} + L^{n})

But notice that this operation itself isn't linear:

dim (U^{n}) + dim (L^{n}) = n^{2} + n \neq dim (F^{n, n})

so it appears that that the dimension of a vector space is the sum of the dimensions of the subspaces minus their dimension of the intersection. We'll prove this:

Dimensions of subspaces

If $U, W$ are subspaces of $V$ , then:

dim (U + W) = dim (U) + dim (W) - dim (U \cap W)

Proof
We won't really do the entire proof, but the main gist is as follows.

Obviously we require that $U, W$ are finite dimensional. Note that we don't require that $V$ be finite dimensional, as here $dim (V)$ isn't needed anywhere.

Let $v_{1}, . . ., v_{k}$ be a basis from $U \cap W$ . Let's call it $B$ , for basis of $U \cap W$ with $k$ vectors.

If you consider just $U$ , notice that $U \cap W$ is a subspace of $U$ , so then we can extend the vectors $v_{1}, . . ., v_{k}$ to make it a basis for $U$ . Let's say we add $j$ vectors, giving a new extended basis $B^{'}$ as $v_{1}, . . ., v_{k}, u_{1}, . . ., u_{j}$ .

In the exact same way, we do the same for $W$ , giving a basis $B_{2}^{'}$ as $v_{1}, . . ., v_{k}, w_{1}, . . ., w_{i}$ with $i$ more vectors.

We claim (and gloss over) that if we concatenate all vectors found above in $B^{'}$ and $B_{2}^{'}$ and you obtain a basis for $U + W$ . Thus, we get that:

dim (U + W) = k + j + i

and:

dim (U) = k + j

and:

dim (W) = k + i

and:

dim (U \cap W) = k

and thus if you plug into the theorem above, you get the expected result.
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As a corollary, if we have instead a direct sum $U \oplus W$ then we get just the standard sum of the dimension:

Corollary

$dim (U \oplus W) = dim (U) + dim (W)$

This is since $dim ({\vec{0}}) = 0$ .

Now, suppose that $U_{1}, U_{2}$ are subspaces of $P_{4} (R)$ , where $dim (U_{1}) = dim (U_{2}) = 3$ . The question is, is $U_{1} + U_{2}$ a direct sum? We know $dim (P_{4} (R)) = 5$ but adding the subspace dimensions makes 6, so then we can't have a direct sum:

dim (U_{1} + U_{2}) = 6 - dim (U_{1} \cap U_{2})

we also have:

dim (U_{1} + U_{2}) \leq 5

since we compare to the dimension of the $V$ , so then:

dim (U_{1} \cap U_{2}) \geq 1

so then $U_{1} \cap U_{2} \neq {\vec{0}}$ , so it can't be a direct sum.