I need to find a non-null matrix A in which A² = 0. It must be an order 3 matrix.
What I tried to do was split it in a table:
[A, B, C] * [A, B, C]
[D, E, F] * [D, E, F]
[G, H, I] * [G, H, I]
But didn't get very far, I got way too many equations to equal to 0, and got completely lost.
I must find a non-trivial answer, how can I do it?
On
Matrix multiplication codes what are called linear transformations. Square of a matrix will correspond to iterating the transformation once more.
Can you think of some operation that produces zero when repeated? Of course this is vague. One should also specify the domain.
The operation of differentiation makes a degree one polynomial into constant, and continued once more makes the constant into zero. So the differentiation operation on polynomials of degree 1 or less fits the bill. Note that derivative of $f(x)+g(x)$ is the sum of derivatives individually.
In this specific case this will lead to a $2\times2$ matrix. To get examples in bigger order we can try the following modification: Think of calculating second derivative as the main operation: i.e Consider $T(x^n)= n(n-1)x^{n-2}$. Now $T\circ T$ will produce zero on all polynomials of degree 3 or less. This would lead to a $4\times4$ matrix whose square would be zero.
Sometimes it is important to try and simplify your question, using your intuition and what you already know.
What does your intuition tell you? "This matrix could have a lot of zeros" comes to mind. Try the first matrix that comes to mind: One non-zero entry.
Now, think about this matrix: $$ \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\0 & 0 & 0 \\\end{pmatrix} $$
What happens when you multiply this matrix with itself? $$ \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\0 & 0 & 0 \\\end{pmatrix} \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\0 & 0 & 0 \\\end{pmatrix} $$
Every non-zero entry in this matrix, gets multiplied by a zero entry, so you see that: $$ \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\0 & 0 & 0 \\\end{pmatrix}\begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\0 & 0 & 0 \\\end{pmatrix}=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\0 & 0 & 0 \\\end{pmatrix} $$
Unfortunately, you will find problems when you try to add another non-zero entry (try it!), but at least you have produced an example of a (nilpotent) matrix.
Something more non-trivial than this example doesn't immediately come to mind, but it will involve lots of non-zero entries and cancellations, that is likely to happen.
If $A$ is a $3 \times 3$ matrix then $A$ is the matrix (in any basis), of a linear transformation from $\mathbb R^3$ to $\mathbb R^3$.
Therefore, $A$ has a kernel and a range. If $A^2 = 0$, then note that $\ker A$ contains the range of $A$, since $x =Ay \implies Ax = A^2 y = 0$.
Therefore, to construct any $A$ such that $A^2 = 0$, all we need to do is the following :
decide the kernel of $A$, $K$. This will be a subspace of $\mathbb R^3$. If we want $A$ to be non-trivial, we desire $K \neq \mathbb R^3$.
Find a basis for $K$, and extend this basis to the whole space. Call this basis $B$.
Define $A$ , a linear transformation, as follows : it will have kernel exactly $K$ : so send every element of $B$ belonging to $K$, to zero. Next, ensure that every other basis element goes to some non-zero element of the kernel. Then $K$ is exactly the kernel of $A$, and $A^2 = 0$.
For example, take the standard basis $e_1,e_2,e_3$, and let us say we want the kernel equal to the span of $e_1,e_2$. Define the linear tranformation $A$, which takes $e_1,e_2$ to zero and $e_3 \to e_1$. Then, $A$ is non-zero, but $A^2 = 0$. In standard basis, the matrix of $A$ is exactly the first matrix in this answer!