Notes - Linear Algebra I MT22, Dimension Formula
Dimension formula
Can you state the full result of the dimension formula (aka Grassmann’s Identity) for $U, W, V$?
Let $U, W$ be subspaces of a finite dimensional vector space $V$. Then
\[\dim(U+W) + \dim(U \cap W) = \dim(U) + \dim(W)\]The dimension formula states:
Let $U, W$ be subspaces of a finite dimensional vector space $V$. Then $\dim(U+W) + \dim(U \cap W) = \dim(U) + \dim(W)$.
If you let $U \cap W$ have a basis $\{v _ 1, \ldots, v _ m\}$, what bases does the Steinitz exchange lemma guarentee for $U$ and $W$?
Let $U, W$ be subspaces of a finite dimensional vector space $V$. Then $\dim(U+W) + \dim(U \cap W) = \dim(U) + \dim(W)$.
- $U = \langle v _ 1, \ldots, v _ m, u _ 1, \ldots, u _ p\rangle$
- $W = \langle v _ 1, \ldots, v _ m, w _ 1, \ldots, w _ q\rangle$
The dimension formula states:
Let $U, W$ be subspaces of a finite dimensional vector space $V$. Then $\dim(U+W) + \dim(U \cap W) = \dim(U) + \dim(W)$.
Given that
- $U \cap W = \langle v _ 1, \ldots, v _ m \rangle$
- $U = \langle v _ 1, \ldots, v _ m, u _ 1, \ldots, u _ p\rangle$
- $W = \langle v _ 1, \ldots, v _ m, w _ 1, \ldots, w _ q\rangle$
What do you need to show is a valid basis for $U + W$ in order to prove the result?
Let $U, W$ be subspaces of a finite dimensional vector space $V$. Then $\dim(U+W) + \dim(U \cap W) = \dim(U) + \dim(W)$.
What two things do you need to show for a set to be a basis of a vector space?
- Spanning
- Linearly independent
The difficult part of proving
\[\dim(U+W) +\dim(U\cap W) = \dim(U) + \dim(W)\]
comes down to proving
\[\\{v_1, \ldots, v_m,u_1,\ldots,u_p,v_1,\ldots,v_q\\}\]
is linearly independent (and spanning, and hence a basis). What do you have to notice once you’ve rearranged
\[\sum^m_{i=1}\alpha_i v_i + \sum_{i=1}^{p}\beta_i u_i = -\sum^q_{i=1}\gamma_iw_i\]
That this vector must in $U \cap W$.