For $n$ by $n$ matrix A, prove if dimension of row space > $n/2$, then $A^2 \ne 0$

Question

I'm trying to prepare for an exam and came across the following question:

Given $n \times n$ matrix A, let U represent row space and W represent column space.

A) Prove: $W \subseteq$ $U^{\perp} $ if and only if $A^2=0$

B) Prove that if $dim(U)>n/2$, then $A^2 \ne 0$.

I was able to do part A pretty easily as follows:

$\Rightarrow$: Let rows of A be {$ \vec{r}_1, \vec{r}_2 ,...,\vec{r}_n $} and columns be {$ \vec{c}_1, \vec{c}_2 ,...,\vec{c}_n $}. Assume $W \subseteq$ $U^{\perp} $. Entry $a_{ij}$ (in row i, column j of matrix $A^2=AA$) equals the dot product of $\vec{r}_i \cdot \vec{c}_j $, and by assumption $\vec{c}_j$ is orthogonal to $ \vec{r}_1, \vec{r}_2 ,...,\vec{r}_n $, so it follows that $a_{ij}=0$ for all i,j. Thus $A^2=0$

$\Leftarrow$: Assume $A^2=0$. We can represent the columns of $A^2$ as

$$ \pmatrix{\vec{r}_1 \\ \vec{r}_2 \\ \vdots \\ \vec{r}_n} \cdot \vec{c}_1 , \pmatrix{\vec{r}_1 \\ \vec{r}_2 \\ \vdots \\ \vec{r}_n} \cdot \vec{c}_2 , ... , \pmatrix{\vec{r}_1 \\ \vec{r}_2 \\ \vdots \\ \vec{r}_n} \cdot \vec{c}_n ,$$ But by assumption, $A^2=0$, meaning all columns of $A^2=0$ and all the above dot products are zero. Hence, $\vec{c}_i$ is orthogonal to all vectors in {$\vec{r}_1, \vec{r}_2 ,...,\vec{r}_n $}, and thus all linear combination of those vectors. Since this applies to all $\vec{c}_i$, then $W \subseteq$ $U^{\perp} $.

As for Part B, I'm having trouble and not sure exactly where to begin. Presumably, I would use the results from part A, meaning that if $A^2 \ne 0$, then $W \not\subseteq$ $U^{\perp} $, that is, there is some vector in column space that is not orthogonal to every vector in row space.

Some properties I'm aware of that may be useful but not sure if or how:

• $dim(U)=dim(W)$
• The orthogonal complement of the row space is null space
• the dimension of the row space and nullspace equals $n$
• dimension of column space of A equals dimension of row space of $A^T$

Anyway, any tips would be greatly appreciated!

Answer 1

Hint: compare the dimensions of $W$ and $U^{\perp}$, given that $\dim(U) >\frac n2$. We can't include a vector space inside one of smaller dimension...

Answer 2

Let us regard a matrix as a linear transformation.

$A^2=0$ means Range of $A$ is contained in kernel of $A$. ($A^2v=0$ means $A(Av)=0$ ie $Av\in \ker A$.) This is not possible as the dimension of range of $A$ is the dimension of column (row) space of $A$, which is given to be $>n/2$, and kernel dimension, by rank-nullity theorem, is $<n/2$.