## Rank Nullity Applications!

### June 10, 2010

We did rank-nullity last time, and this time we’re going to go over a few cool applications to rank-nullity.

First, let’s just say what everyone’s thinking: there’s just too many damn finite dimensional vector spaces to consider!  But for the moment, let’s just consider the really popular ones: that is, for a field $F$ we’ll consider $F^n$ as a vector space over $F$.  This means that our vectors will be elements of $F^n$ and our scalars are in the field $F$.

So, let’s ask some hard hitting questions.  If we have a little vector space, let’s say $F^n$, and some bigger space, let’s say $F^m$ (for $n < m$ obviously), then can we have a linear map that maps all of the vectors in $F^n$ into $F^m$ such that they “cover” every vector in the range space?  More specifically, is there a linear map $T: F^{n}\rightarrow F^{m}$ such that $T$ is surjective?

Cor: If $T: F^{n}\rightarrow F^{m}$ where $n < m$ then $T$ cannot be surjective.

Pf:  Suppose it were surjective.  Then we have $\dim(\mbox{Rank}(T)) = m$ (why?) and by rank-nullity theorem, we have that $\dim(F^{n}) = \dim(\mbox{Rank}(T)) + \dim(\mbox{Nullity}(T))$ which implies $n = m + \dim(\mbox{Nullity}(T))$.  Since $n < m$ this implies that the nullity of $T$ is negative, which is ridiculous.  $\Rightarrow\Leftarrow$$\Diamond$.

This makes intuitive sense: we shouldn’t be able to cover a huge bed with a small blanket, you know?  What about the other way around: should we be able to successfully inject a huge space into a smaller space with no “compressing” of vectors?  In other words, is there an injective map from a big vector space to a smaller one?

Cor: If $T: F^{n}\rightarrow F^{m}$ where $n > m$ then $T$ cannot be injective.

Pf:  Suppose it were injective.  Then we have $\dim(\mbox{Nullity}(T)) = 0$ (this is actually a good thing to prove, that the nullity is trivial if and only if the map is injective) and by rank-nullity theorem, we have that $\dim(F^{n}) = \dim(\mbox{Rank}(T)) + \dim(\mbox{Nullity}(T))$ which implies $n = \dim(\mbox{Rank}(T)) + 0$.  But wait a second…the rank can be at most $m$ which is less than $n$!  Contradiction.  $\Rightarrow\Leftarrow$$\Diamond$.

Okay, okay, so, now that we’ve crushed a number of people’s dreams of mapping spaces nicely into other spaces, let’s prove something positive about vector spaces!  Let’s show that two finite vector spaces are isomorphic if and only if they have the same dimension.  Wait a second, this is a huge deal!  This means that for every finite dimension, we have, up to isomorphism, one vector space!  Holy crap.  That means that all the stuff we just did with $F^n$ wasn’t in vain after all; in fact, every finite dimensional vector space is isomorphic to one of these!

Thm:  Given two finite-dimensional spaces $V$ and $W$, then these two spaces are isomorphic if and only if $\dim(V) = \dim(W)$.

Pf:  If two spaces are isomorphic, then there is an invertible linear map $T$ that’s injective and surjective.  This implies, by the previous two corollaries, that $\dim(V) = \dim(W)$.

Now, suppose that $\dim(V) = \dim(W)$.  Let’s let the basis for $V$ be $\{v_1, v_2, \dots, v_n\}$ and $\{w_1, w_2, \dots, w_n\}$ and define the linear map $T:V \rightarrow W$ by

$T(a_{1}v_{1} + a_{2}v_{2} + \dots + a_{n}v_{n}) = a_{1}w_{1} + a_{2}w_{2} + \dots + a_{n}w_{n}$.

Okay, let’s prove that this map is injective and surjective.  It is obviously surjective, since any linear combination of the basis elements can be achieved by altering the original coefficients.  This is injective for relatively obvious reasons as well, so suppose

$T(a_{1}v_{1} + a_{2}v_{2} + \dots + a_{n}v_{n})$

$= T(b_{1}v_{1} + b_{2}v_{2} + \dots + b_{n}v_{n})$

$= a_{1}w_{1} + a_{2}w_{2} + \dots + a_{n}w_{n}$

$= b_{1}w_{1} + b_{2}w_{2} + \dots + b_{n}w_{n}$

and since the $w_{i}$‘s are all linearly independent, we have that

$= (a_{1}-b_{1})w_{1} + (a_{2}-b_{2})w_{2} + \dots + (a_{n}-b_{n})w_{n} = 0$

if and only if each of the coefficients are 0.  This means that, for all $i$ we have that $a_{i} - b_{i} = 0$ and so $a_{i} = b_{i}$ and so this is injective.  (Alternatively, because the $w_{i}$ are linearly independent, $T(v) = 0$ if and only if all the coefficients are 0, so the nullity is trivial.)  This proves that our spaces are isomorphic.  $\Box$