Finding all 2x2 nilpotent matrices

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The discussion centers on finding all symmetric 2x2 nilpotent matrices, specifically those where A^2 equals the zero matrix. The calculations show that for a symmetric matrix of the form [a, b; b, d], the equations derived from matrix multiplication lead to conditions that suggest a^2 equals negative b^2. This implies that non-trivial solutions may not exist, leading to the conclusion that only the trivial solution (the zero matrix) is valid. Ultimately, the consensus is that while non-trivial symmetric nilpotent matrices do not exist, the trivial solution is confirmed.
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Homework Statement



If A2 is a zero matrix, find all symmetric 2x2 nilpotent matrices.

Homework Equations





The Attempt at a Solution



So if A2 is nilpotent, then

[a,b;c,d]*[a,b;c,d] is equal to [0,0;0,0].

Since A is symmetric, b=c. Multiplying the two matrices, I get

[ aa + bb, ab + bd; ba +db, bb + dd] = [0,0;0,0]

each element in the matrix must equal zero, so

aa + bb = 0
ab + bd = 0
ba + bd = 0
bb + dd = 0

with the first equation, a2 must equal negative b2, so there is no solution. Is there no 2x2 symmetric nilpotent matrices, or did I mess up somewhere?
 
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brownman said:

Homework Statement



If A2 is a zero matrix, find all symmetric 2x2 nilpotent matrices.

Homework Equations





The Attempt at a Solution



So if A2 is nilpotent
What they're saying is that A is nilpotent. A2 is the 2 x 2 zero matrix.
brownman said:
, then

[a,b;c,d]*[a,b;c,d] is equal to [0,0;0,0].
You're looking for symmetric 2 x 2 matrices, which means they have to look like this:

$$ \begin{bmatrix} a & b \\ b & c\end{bmatrix}$$
brownman said:
Since A is symmetric, b=c. Multiplying the two matrices, I get

[ aa + bb, ab + bd; ba +db, bb + dd] = [0,0;0,0]

each element in the matrix must equal zero, so

aa + bb = 0
ab + bd = 0
ba + bd = 0
bb + dd = 0

with the first equation, a2 must equal negative b2, so there is no solution.
There is a solution.
brownman said:
Is there no 2x2 symmetric nilpotent matrices, or did I mess up somewhere?
 
Oh yeah there is the trivial solution. Thanks for the help in clearing it up :)
 
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