# $3 \times 3$ matrix such that $A^2=0$

I'm asked to to find a $3\times3$ matrix, in which no entry is $0$ but $A^2=0$.

The problem is if I I brute force it, I am left with a system of 6 equations (Not all of which are linear...) and 6 unknowns. Whilst I could in theory solve that, is there more intuitive way of solving this problem or am I going to have to brute force the solution?

Any suggestions would be greatly appreciated.

• Such a matrix must have rank $1$. Take a simple example and rotate the co-ordinate axes. Commented Feb 12, 2018 at 8:49
• I have to use solely the properties of inverses, rank and inverses. I'm not supposed to use eigenvectectors, rotations or diagonolizaton. Commented Feb 12, 2018 at 9:02
• Whatever you do amounts to choosing a basis for a vector space. Here the challenge is to find a basis in which the matrix is deliberately not in its simplest form. My method suggests reverse engineering what you would do to simplify. If you are using inverses, what do you want to take the inverse of? - not of $A$, which is singular. It is about making intelligent choices. Commented Feb 12, 2018 at 9:39

There's an idea. We can try to put $A = \alpha \beta'$, where $\alpha,\beta$ are $3 \times 1$ matrix.

Then, $A^2 = \alpha \beta' \alpha \beta'$, so if $\alpha,\beta \ne 0$(all the components are not $0$),$\beta' \alpha = 0$ things will be solved.

So,we could let

$\alpha = \begin{pmatrix} 1 \\ 1\\ 1 \end{pmatrix}, \beta = \begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix}$

and $A = \begin{pmatrix} 1 & 1 & -2 \\ 1 & 1 & -2 \\ 1 & 1 & -2 \end{pmatrix}$

• This is not a $3 \times 3$ matrix though. Commented Feb 12, 2018 at 9:05
• Why? $\alpha 3 \times 1,\beta' 1 \times 3$,so $\alpha \beta'$ IS $3 \times 3$ Commented Feb 12, 2018 at 9:09
• Ohh your edit helped. I was confused for a while. Let me try to reason it out now... Commented Feb 12, 2018 at 9:35
• I'm glad to help Commented Feb 12, 2018 at 9:39
• Note that with $\alpha^T=(1, d/a, e/a); \beta^T=(a, b, c)$ this gives Henry's solution. Commented Feb 12, 2018 at 9:50

Force the first element of $A^2$ to be $0$, for example by finding $b,c,d,e$ with $bd+ce \lt 0$ and let $a=\pm{\sqrt{-(bd+ce)}}$

Then consider

\begin{bmatrix}a&b&c\\d&db/a&dc/a\\e&eb/a&ec/a\end{bmatrix}

• How did you reach this conclusion? Just intuition? Commented Feb 12, 2018 at 8:52
• @FutureMathperson - I forced one cell of $A^2$ to be $0$ and then found an easy way to make the others a multiple of it Commented Feb 12, 2018 at 8:54
• Form @MarkBennet 's comment, this looks to me as if it may be the general solution Commented Feb 12, 2018 at 9:20
• Note that this is equivalent to the solution by Focus, given the comment I made there. Commented Feb 12, 2018 at 9:52
• @FutureMathperson. Note that once $a,b,c,d,e$ are fixed the other entries are determined by the rank $1$ condition and the $A^2=0$ condition implies that $a^2+bd+ce=0$. Commented Feb 12, 2018 at 9:56

Here is a geometric approach. Think about a matrix $P$ which orthogonally projects all of $\Bbb R^3$ onto a one-dimensional subspace spanned by a vector $n$:

$$P_{ij}=n_i n_j$$

After projection, rotate your space by $90^\circ$ around some axes $v$ orthogonal to $n$ by using a rotation matrix $R$. Your desired matrix then can be

$$M:=RP.$$

If you choose appropriate $n$ and $v$, your matrix will not have zero-entries.

Example

Choose $n=(1,2,3)$ and $v=(-2, 1, 0)$. We have $n\cdot v=0$. The resulting matrix is

$$M=\frac1{\sqrt{5}} \begin{pmatrix} 3& 6& 9&\\ 6& 12& 18\\ -5& -10& -15 \end{pmatrix}.$$

You can also drop the $1/\sqrt 5$ in front of the matrix. This will not change the result.

Why does it work

If you project your space onto a line $\ell$ and then rotate this line by $90^\circ$ degree, all vectors are now orthogonal to the line $\ell$. So by applying the matrix again, the complete rotated line will be projected into the zero-vector. Because all vectors were previously projected into the rotated line, everything gets mapped into zero on the second run.

• And we don't need the scalar factor, i.e., we can take $A=\sqrt{5}M$, then we have even integer coefficients. Commented Feb 12, 2018 at 9:19

To do this, first consider a trivial matrix with lots of zeros that does satisfy this condition. One easy one is the matrix with a single $1$ on the diagonal above the principal. That is: $$\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0& 0 \\ 0 & 0 & 0 \end{pmatrix}$$

Now, given any matrix $A$, consider for any invertible matrix $B$, the matrix $B^{-1}AB = C$. If $A^2 =0$, then $C^2 = B^{-1}ABB^{-1}AB = B^{-1}A^2B =0$. So we can modify the matrix above, which we'll call $D$, to get a matrix with no zeros.

Actually, I urge you to take the formula which you have for the inverse of a matrix, and compute $B^{-1}DB$. You will get that each entry of the matrix $B^{-1}DB$ actually looks like some entry(in fact, an entry of the first column) of $B^{-1}$ times some entry of $B$.(PLEASE NOTE : I expect this to be a simple task for you. If not doable, please tell me)

But entries of $B^{-1}$ are just the cofactors of $B$, with some sign, divided by the determinant. That is, all we need to do is to ensure that the determinant, all cofactors and all entries of $B$ are non-zero. This I expect to be a simple task by experiment : start with some fixed first row of $B$, say $1,2,3$ or something, then find non-zero entries of the second rows so that all cofactors are non-zero, and then do this for the third rows. This can be done by simple trial and error, and will give you a matrix $B$ such that $B^{-1} DB$ has non-zero entries.

This is the simplest approach if you are not keen on rotation matrices or anything of that ilk, mentioned in some other answers.

There are standard matrices which satisfy the condition required for $B$ above, like Hankel, Vandermonde and Toeplitz matrices, which in fact satisfy strict positivity of all minors. But even avoiding these, coming up with one such matrix should be the challenge if you are at a level where you cannot use rotation matrices etc.

• To be extremely fair, I am still looking for a simpler way than trial and error to get a matrix of the kind I mentioned. I think I will find it, and if I do I will post it. Commented Feb 12, 2018 at 9:40