Let A be a 2x2 symmetrical matrix. Find all posibilities for A such that A^2 = 0?

Any 2 x 2 symmetric matrix A can be written in the form
[a b]
[b c] for some real numbers a,b,c.

Since we want A^2 = 0, this yields
[a b][a b]...[a^2+b^2..b(a+c)]....[0 0]
[b c][b c].=.[b(a+c)...b^2+c^2]=[0 0].

Equating like entries,
a^2 + b^2 = 0
b(a+c) = 0
b^2 + c^2 = 0.

The first and last equations imply that a = b = c = 0 (assuming that a,b,c are real numbers).

Hence, the zero matrix is the only such matrix.
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Note: If the entries are allowed to be complex numbers, then there are more solutions.

Starting with a^2 + b^2 = 0, b(a+c) = 0, and b^2 + c^2 = 0:

The second equation implies that b = 0 or c = -a.
(i) If b = 0, then the first equation yields a = 0. In turn, the third equation yields c = 0. Hence, A is the zero matrix.

(ii) If c = -a, then we have a^2 + b^2 = 0 ==> b = ±ai.

Hence, A is of the form
[a...±ai]
[±ai..-a] for some complex number a.
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I hope this helps!

When they say A^2 = 0, the zero is a 2x2 matrix containing only zeros. So find what vales of a, b and c solve the equation below.

[a b][a b] = [0 0]
[b c][b c] = [0 0]

One obvious solution is a = b = c = 0

You will :)
For example, multiply a matrix
(1 -1; 1 -1) with itself.
A possible way to solve the problem is to write the matrix to be found as
(a b; c d) and solve the system a^2+bc=0, ab+bd=0, ac+cd=0, bc+d^2=0.
If You know properties of the eigenvalues, use the fact that eigenvalues of the matrix should be equal to lambda_1=0 and lambda_2=0, too.
Another hint: the rows must be of form
[a b]; [ka kb] or [ka kb]; [a b] since the determinant is equal to 0.