Usbrth

Using the rules for matrix multiplication, I have found four algebraic equations as "constraints" for getting every element in the resulting matrix to be zero. Assuming that the elements in the resulting matrix are $a, b, c$ and $d$ ($a, b$ are the elements in the first row and $c, d$ are the elements in the second). The equations are therefore;

1. $a^2 + bc = 0$


2. $ab + bd = 0$


3. $ac + cd = 0$


4. $d^2 + bc = 0$

I have found from the four equations that $a=-d$ and that $bc=-a^2$ hence $bc$ must be a negative quantity thus $b$ and $c$ have opposite signs but that does not seem to be enough to guarantee that the resulting matrix is always zero. Now, I do not know how to think about this problem;

How do I find more constraints from these equations?

And how do I know that I have found all possible constraints?

Why is that when I square one equation, some how I get a new constraint?

Shouldn't the four equations be enough to determine the exact conditions for $a, b, c$ and $d$?

With

$A = beginbmatrix a & b \ c & d endbmatrix, tag 1$

we have

$A^2 = beginbmatrix a & b \ c & d endbmatrixbeginbmatrix a & b \ c & d endbmatrix = beginbmatrix a^2 + bc & (a + d)b \ (a +d)c & d^2 + bc endbmatrix = 0, tag 2$

whence

$a^2 + bc = 0, tag 3$

$d^2 + bc = 0, tag 4$

$(a + d)b = 0, tag 5$

$(a + d)c = 0; tag 6$

suppose

$a + d ne 0; tag 7$

then via (5) and (6),

$b = c = 0; tag 8$

then via (3) and (4),

$a^2 = d^2 = 0 Longrightarrow a = d = 0 Longrightarrow a + d = 0 Rightarrow Leftarrow a + d ne 0; tag 9$

this contradiction rules out the case (7), so

$a + d = 0 Longrightarrow d = -a; tag10$

now if

$a = d = 0, tag11$

then (3)-(4) imply, assuming $A ne 0$, exactly one of

$b = 0, ; c ne 0, tag12$

$b ne 0, ; c = 0, tag13$

holds. Thus the solutions in the event that (11) binds are one of the forms

$A = beginbmatrix 0 & b \ 0 & 0 endbmatrix, ; beginbmatrix 0 & 0 \ c & 0 endbmatrix; tag14$

when

$d = -a ne 0, tag15$

we find that (3)-(4) imply

$b, c ne 0 tag16$

and

$c = -dfraca^2b; tag16$

therefore

$A = beginbmatrix a & b \ -dfraca^2b & -a endbmatrix. tag17$

We see that (2) implies $A$ takes one of the three forms (14), (17); the former being comprised of two one-parameter families of matrices, and the latter comprised of one two-parameter family.

There's an easier way to do it. A $2times2$ matrix $A$ such that $A^2=0$ must have the single eigenvalue $0$, so its trace and determinant must be zero, because the characteristic polynomial is $X^2-operatornametr(A)X+det(A)$.

This yields, with your notation, $a+d=0$ and $ad-bc=0$, so $d=-a$ and $a^2+bc=0$.

Note that these conditions are sufficient, because the minimal polynomial is a divisor of the characteristic polynomial, so it is either $X$ (null matrix) or $X^2$ (nonzero nilpotent matrix).

Now you can classify the nonzero nilpotent matrices as those of the form
beginalign
&beginbmatrix 0 & b \ 0 & 0 endbmatrix
&&(bne0) \[6px]
&beginbmatrix 0 & 0 \ c & 0 endbmatrix
&&(cne0) \[6px]
&beginbmatrix a & b \ -a^2/b & -a endbmatrix
&&(ane0,bne0)
endalign

If $b=c=0$ then you get the matrIx of the form $$beginpmatrix a & 0 \ 0 & -a \ endpmatrix.$$

Now suppose that both of them are not simultaneously $0.$ WLOG let us assume that $b neq 0.$ Then the matrix is of the form $$beginpmatrix a & b \ -frac a^2 b & -a \ endpmatrix.$$

So this kind of matrices have at most $2$ degrees of freedom.