Span and Matrices

Question

I am currently working on two algebra questions that ask me to answer the following questions on matrix $A$, a $m \times n$ matrix given the following conditions:

a) $n > m$

b) $m > n$

1) There always exists an $x$ not equal to zero such that $Ax = 0$

2) It is possible for the columns of $A$ to span $\mathbb{R}^m$

For question (1) my plan was to provide an example proving the statement or a reason why it is not true, but I am really struggling to find any combinations of $A$ and $x$ that equal zero given the two conditions.

And for the second question I think I am completely missing the point of this question, because I don't see why it would not be possible for the columns of $A$ to span $\mathbb{R}^m$.

Answer 1

The columns of an $m\times n$ matrix are elements of the $m$-dimensional space $\mathbb{R}^m$ (or over any other field), and they are $n$ in number, so if $n>m$ it´s possible, that they span $\mathbb{R}^m$ and there is a nontrivial linear combination of the vector $0$, since a maximal linearly independant set has $m$ elements, which is essentially the same as the existence of a vector $x$ such that $Ax=0$.If $m>n$ they might be linarly independant, so there does not always exist such a vector, and they can´t span the whole space , but just a subspace of dimension $d\leq n$.

Answer 2

For the first question if n>m then for Ax=0 to have a solution we are going to need some free variables. Remember that these can take any real value, and as such Ax=0 can have a non-trivial solution.

For m>n we are not going to have free variables unless the systems has infinite solutions. Therefore the Ax=0 will only occur if all values are equal to zero.

The second part I am not so sure about myself, so you are best to look at the above question.

p.s. I take it you go to Adelaide Uni ;)

Answer 3

The keys are the concept of rank and nullity of a matrix.

The rank of an $m\times n$ matrix $A$ is the dimension of the column space $C(A)$ of $A$, that is, the subspace of $\mathbb{R}^m$ spanned by the columns of $A$.

The nullity of $A$ is the dimension of the subspace $N(A)$ of $\mathbb{R}^n$ defined by $$ N(A)=\{x\in\mathbb{R}^n:Ax=0\} $$

Note: $\mathbb{R}^k$, in these contexts, is considered as consisting of column vectors with $k$ rows.

The link to these concepts is the rank-nullity theorem:

for every $m\times n$ matrix $A$, we have $$\dim C(A)+\dim N(A)=n$$

Moreover, the very definition of rank tells us that $$ \dim C(A)\le m,\qquad \dim C(A)\le n $$ The first inequality is because $C(A)$ is a subspace of $\mathbb{R}^m$, that has dimension $m$; the second one is because a subspace spanned by $n$ vectors can have at most dimension $n$.

Thus the column space of $A$ can be $\mathbb{R}^n$ only if $m\le n$: indeed, if $n<m$, the fact that $\dim C(A)\le n$ implies that $C(A)$ is a proper subspace of $\mathbb{R}^m$.

The existence of $x\ne0$ such that $Ax=0$ entails $\dim N(A)>0$ because in this case the linearly independent set $\{x\}$ can be extended to a basis of $N(A)$. Conversely, $\dim N(A)>0$ implies the existence of at least $\dim N(A)$ non zero vectors $x$ such that $Ax=0$, the elements of a basis. (Actually infinitely many of them.)

Since $\dim N(A)=n-\dim C(A)$ (by the rank-nullity theorem) and $n-\dim C(A)\ge n-m$ (because $\dim C(A)\le m$), if $n>m$ we surely have $\dim N(A)>0$.