True of False? : If $A$ is a square matrix and $A^2$ has a column of zeros, then A must have a column of zeros.
This was in one of my exercise questions and I'm wondering how to prove it. If it's right, how do you prove it's right? Or if it's wrong, can you only look for counterexamples for T/F statements like this? It's not really easy to come up with counterexamples even if this is wrong, and I'm wondering if there is a better way to solve this rather than resorting to that.
On
Consider (in $\mathbb C$): $$A:=\begin{pmatrix} 1 & i \\ i & -1 \end{pmatrix}$$
and schow that $A^2=0.$
On
Here is a simple somewhat systematic way to search for a counterexample.
You obviously (why) won't find as counterexample any square matrix of size${}\leq1$, so let us look for one of size$~2$. One of the two standard basis vectors should map to zero when multiplying by $A$ twice in succession, but neither of them should when applying $A$ just once. Let us try to make the first standard basis vector $\binom10$ map to zero when multiplying by $A$ twice. After a first multiplication it should not map to zero already, nor should it map to a scalar multiple of either of the standard basis vectors (since these won't be going to zero on the next application). So let $\binom ab$ with both $a$ and $b$ nonzero be the image of $\binom10$ after one application, and the first column of $A$. We want $A\binom ab=\binom00$, and since $\binom10$ and $\binom ab$ form a basis of $\Bbb R^2$ (since $b\neq0$), this will determine the matrix $A$. It remains to express $\binom01$ as linear combination of $\binom10$ and $\binom ab$, and from that its image under multiplication by $A$, w hich gives the second column of $A$. Simple equation solving gives $\binom01=\frac{-a}b\binom10+\frac1b\binom ab$ as linear combination, and therefore $A\binom10=\frac{-a}b\binom ab+\frac1b\binom00=\binom{-a^2/b}{-a}$. So any matrix of the form $$ A=\pmatrix{a&-a^2/b\\b&-a} \quad\hbox{with $a,b\neq0$, for instance $a=b=1$:}\quad A=\pmatrix{1&-1\\1&-1} $$ gives a counterexample. Note that $A^2=0$ in all cases, which is no surprise since the requirements imply that $A$ must have a generalised eigenspace for $\lambda=0$ of dimension (at least) $2$, which in dimension$~2$ forces $A^2=0$.
On
A column of $A^2$ being zero means that the corresponding column of $A$ is in the kernel (nullspace) of $A$. Using this, it is easy to construct a counterexample: Consider the vector $\left[\begin{smallmatrix} 1\\3\\7\end{smallmatrix}\right]$ for example as a candidate of a column that is in the kernel of its own matrix. It is orthogonal to $\left[\begin{smallmatrix} 4\\1\\-1\end{smallmatrix}\right]$ for example and hence in the kernel of the $1\times 3$ matrix $[4\ \ 1\ {-1}]$. Now we can add multiples of this row to obtain the desired column without changing the kernel: $$ A = \begin{bmatrix} 4 & 1 & -1 \\ 12 & 3 & -3 \\ 28 & 7 & -7 \end{bmatrix}. $$ Since this made all columns of $A$ multiples of $\left[\begin{smallmatrix} 1\\3\\7\end{smallmatrix}\right]$, which is in the kernel of $A$, we actually made up a matrix with $A^2=0$.
Instead of adding multiples of the same row, we could have added multiples of any other vector orthogonal to $\left[\begin{smallmatrix} 1\\3\\7\end{smallmatrix}\right]$ of course. For example $\left[\begin{smallmatrix} 3\\-1\\0\end{smallmatrix}\right]$ and $\left[\begin{smallmatrix} 0\\7\\-3\end{smallmatrix}\right]$ are also orthogonal to $\left[\begin{smallmatrix} 1\\3\\7\end{smallmatrix}\right]$ and using scaled versions of them as rows to obtain the desired column we can come up with $$ A' = \begin{bmatrix} 4 & 1 & -1 \\ -9 & 3 & 0 \\ 0 & 7 & -3 \end{bmatrix}. $$ In this case $$ A'^2 = \begin{bmatrix} 7 & 0 & -1 \\ -63 & 0 & 9 \\ -63 & 0 & 9 \end{bmatrix}. $$
It is false. If, for instance,$$A=\begin{bmatrix}1&-1\\1&-1\end{bmatrix},$$then $A^2$ is the null matrix.
A more systematic approach is this: you can easily check that, if $M=\left[\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right]$, then $M^2=\left[\begin{smallmatrix}0&0\\0&0\end{smallmatrix}\right]$. Of course, $M$ is not a counterexample, since it has a column of zeros. However, since $M^2=\left[\begin{smallmatrix}0&0\\0&0\end{smallmatrix}\right]$, the square of any matrix $A$ similar to $M$ will be the null matrix too. But, given any four numbers $a$, $b$, $c$, and $d$ such that $ad-bc\ne0$, you have$$\begin{bmatrix}a&b\\c&d\end{bmatrix}.M.\begin{bmatrix}a&b\\c&d\end{bmatrix}^{-1}=\frac1{ad-bc}\begin{bmatrix}-ac&a^2\\-c^2&ac\end{bmatrix}.\tag1$$So, pick any four numbers $a$, $b$, $c$, and $d$ such that $ad-bc\ne0$ and that both $a$ and $c$ are not $0$, and the RHS of $(1)$ will be a counterexample. For instance, if $a=c=d=1$ and $b=2$, then you will simply get $A=\left[\begin{smallmatrix}1&-1\\1&-1\end{smallmatrix}\right]$.