# Is this a subspace? If yes what is its dimension?

• Nov 11th 2011, 02:55 PM
gotmejerry
Is this a subspace? If yes what is its dimension?
Let $r(\le min(m,n))$ be the rank of ${\textbf{A} \in \mathbb{F}^{m\times n}$ matrix, and $L= \{ {\textbf{x} \in \mathbb{F}^n\ |\ \textbf{A} \textbf{x}=0}\}$. Show that $L$ is a subspace of $\mathbb{F}^n$ and decide its dimension.

Thank you!
• Nov 11th 2011, 04:44 PM
Drexel28
Re: Is this a subspace? If yes what is its dimension?
Quote:

Originally Posted by gotmejerry
Let $r(\le min(m,n))$ be the rank of ${\textbf{A} \in \mathbb{F}^{m\times n}$ matrix, and $L= \{ {\textbf{x} \in \mathbb{F}^n\ |\ \textbf{A} \textbf{x}=0}\}$. Show that $L$ is a subspace of $\mathbb{F}^n$ and decide its dimension.

Thank you!

This is just the rank nullity-theorem which states that if $T:V\to W$ (and both $V,W$ are finite dimensional $k$-spaces) then $\dim V=\dim\text{im }T+\dim\ker T$. So, the answer to your question is $\dim L=n-r$. Now, how can we prove this? It all depends upon how much machinery you have. If you're brave of heart (i.e. you have some experience with short exact sequences) you can look here. Otherwise, give us an indication of what machinery you have.
• Nov 11th 2011, 05:29 PM
Deveno
Re: Is this a subspace? If yes what is its dimension?
um, the splitting lemma for vector spaces IS, in some sense, the rank-nullity theorem. which, in turn, IS the first isomorphism theorem for vector spaces.

L is just the null space of A, so dim(L) = nullity(A).

it's clear L = ker(A) (as considered as an element of Hom_F(F^m,F^n)). of course, one can just show L is a subspace directly:

suppose x,y are in L:

then A(x+y) = A(x) + A(y) (matrices are linear)

= 0 + 0 = 0 (these are 0-vectors in F^m), so x+y is in L.

similarly, if a is in F: A(ax) = a(A(x)) = a0 = 0, so ax is in L, whenever x is.

finally, L is always non-empty, since A(0) = 0, so at the very least, the 0-vector of F^n is in L.

by column-reducing the matrix A, we get a matrix AP, where P is an invertible matrix representing the product of elementary column-operation matrices.

if {e1,e2,....,en} is the standard basis for F^n, we have AP(ej) ≠ 0, for j = 1,2,..,r,

and AP(ej) = 0 for j = r+1,...,n. thus the vectors P(ej) for j = r+1,...,n are all in L, and linearly independent by the invertibility of P.

since the first r columns of AP are also linearly independent, NONE of the P(ej) for 1 ≤ j ≤ r, are in L, and the only linear combination of them

which is in L, is the 0-vector.

so {P(e(r+1)),...,P(en)} span L, and are thus a basis for L, which therefore has dimension n - r.
• Nov 11th 2011, 06:44 PM
Drexel28
Re: Is this a subspace? If yes what is its dimension?
Quote:

Originally Posted by Deveno
um, the splitting lemma for vector spaces IS, in some sense, the rank-nullity theorem. which, in turn, IS the first isomorphism theorem for vector spaces.

L is just the null space of A, so dim(L) = nullity(A).

it's clear L = ker(A) (as considered as an element of Hom_F(F^m,F^n)). of course, one can just show L is a subspace directly:

suppose x,y are in L:

then A(x+y) = A(x) + A(y) (matrices are linear)

= 0 + 0 = 0 (these are 0-vectors in F^m), so x+y is in L.

similarly, if a is in F: A(ax) = a(A(x)) = a0 = 0, so ax is in L, whenever x is.

finally, L is always non-empty, since A(0) = 0, so at the very least, the 0-vector of F^n is in L.

by column-reducing the matrix A, we get a matrix AP, where P is an invertible matrix representing the product of elementary column-operation matrices.

if {e1,e2,....,en} is the standard basis for F^n, we have AP(ej) ≠ 0, for j = 1,2,..,r,

and AP(ej) = 0 for j = r+1,...,n. thus the vectors P(ej) for j = r+1,...,n are all in L, and linearly independent by the invertibility of P.

since the first r columns of AP are also linearly independent, NONE of the P(ej) for 1 ≤ j ≤ r, are in L, and the only linear combination of them

which is in L, is the 0-vector.

so {P(e(r+1)),...,P(en)} span L, and are thus a basis for L, which therefore has dimension n - r.

This is the old-fashioned way. I wasn't necessarily stating that the splitting lemma IS the rank-nullity theorem, but as can be seen at the bottom of the page I linked to, it easily implies it.
• Nov 11th 2011, 06:52 PM
Deveno
Re: Is this a subspace? If yes what is its dimension?
yes, it IS the old-fashioned way, and it's a bit ugly, in my opinion. it lacks a certain je ne sais quoi, n'est-ce pas? peut-etre d'elegance.

it has, however, the advantage of being at the level that can be assumed most linear algebra courses will have covered, using basic facts about bases and matrices.

the splitting lemma is certainly more general than the rank-nullity theorem, sort of like prime ideals are more general than prime integers (that's why i said "in a sense" without going into the finer details).
• Nov 12th 2011, 02:01 AM
gotmejerry
Re: Is this a subspace? If yes what is its dimension?
Thank you for the answer. I prefer the most simple, "old-fashioned' ways because I am at the beginning of my algebra course, and we havent yet learnt the rank-nullity theorem neither the kernel of a matrix. So I dont really know what our professor has in mind, how could we solve his homeworks if we havent yet learnt the things we would need.