06 - Dimension Formula

Dimension Formula

Let $U,W$ be subspaces of a finite-dimensional vector space $V$ over $\mathbf{F}$ then

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

Proof

Take a basis $v_1,v_2,\cdots ,v_m$ of $U\cap W$
As $U\cap W\le U$ and $U\cap W\le W$ then by Spanning Sets containing a Basis
We can separately extend this set to a basis

$$v_1,\cdots ,v_m,u_1,\cdots ,u_p\text{ of }U$$ $$v_1,\cdots ,v_m,w_1,\cdots ,w_q\text{ of }W$$

Hence we can see that

$$\dim (U\cap W)=m,\dim U=m+p,\dim W=m+q$$

(Now we need to show that $S=v_1,\cdots ,v_m,u_1,\cdots ,u_p,w_1,\cdots ,w_q$ is a basis of $U+W$)
Looking at the number of elements in $S$ we know that

$$\begin{array}{rl}m+p+q& =(m+p)+(m+q)-m\\ & \\ & =\dim U+\dim W-\dim (U\cap W)\end{array}$$

$S$ is spanning:
For $x\in U+W$ with $x=u+w$ for some $u\in U,w\in W$ then

$$\begin{array}{rl}u& ={\alpha }_1{v}_1+\cdots +{\alpha }_m{v}_m+{\alpha }_1^{\prime }{u}_1+\cdots +{\alpha }_p^{\prime }{u}_p\\ w& ={\beta }_1{v}_1+\cdots +{\beta }_m{v}_m+{\beta }_1^{\prime }{w}_1+\cdots +{\beta }_q^{\prime }{w}_q\end{array}$$

for some scalars ${\alpha }_i,{\alpha }_i^{\prime },{\beta }_i,{\beta }_i^{\prime }\in \mathbf{F}$ then

$$x=u+w=({\alpha }_1+{\beta }_1)v_1+\cdots +({\alpha }_m+{\beta }_m)v_m+{\alpha }_1^{\prime }{u}_1+\cdots +{\alpha }_p^{\prime }{u}_p+{\beta }_1^{\prime }{w}_1+\cdots +{\beta }_q^{\prime }{w}_q\in ⟨S⟩$$

hence showing that $⟨S⟩=U+W$

$S$ is linearly independent
Take ${\alpha }_1,\cdots ,{\alpha }_m,{\beta }_1,\cdots ,{\beta }_p,{\gamma }_1,\cdots ,{\gamma }_q\in \mathbf{F}$ such that

$${\alpha }_1{v}_1+\cdots +{\alpha }_m{v}_m+{\beta }_1{u}_1+\cdots +{\beta }_p{u}_p+{\gamma }_1{w}_1+\cdots +{\gamma }_q{w}_q=0$$

Then

$${\alpha }_1{v}_1+\cdots +{\alpha }_m{v}_m+{\beta }_1{u}_1+\cdots +{\beta }_p{u}_p=-({\gamma }_1{w}_1+\cdots +{\gamma }_q{w}_q)$$

The LHS vector is in $U$ , and RHS vector is in $W$, hence they are both in $U\cap W$

As $v_1,\cdots ,v_m$ form a basis of $U\cap W$, there are ${\lambda }_1,\cdots ,{\lambda }_m\in \mathbf{F}$ such that

$$-({\gamma }_1{w}_1+\cdots +{\gamma }_q{w}_q)={\lambda }_1{v}_1+\cdots +{\lambda }_m{v}_m$$

which rearranges to

$${\gamma }_1{w}_1+\cdots +{\gamma }_q{w}_q+{\lambda }_1{v}_1+\cdots +{\lambda }_m{v}_m=0$$

But $\{v_1,\cdots ,v_m,w_1,\cdots ,w_q\}$ is linearly independent (basis for $W$) so each ${\gamma }_i=0$
Hence

$${\alpha }_1{v}_1+\cdots +{\alpha }_m{v}_m+{\beta }_1{u}_1+\cdots +{\beta }_p{u}_p=0$$

But $\{v_1,\cdots ,v_m,u_1,\cdots ,u_p\}$ is linearly independent (basis for $U$) so that ${\alpha }_i=0,{\beta }_i=0$

Hence $S$ is linearly independent and so $S$ is a basis for $U+W$