Is a 2x2 Complex Matrix with N^2=0 Always Similar to a Specific Form?

[Alupsaiu]

Suppose N is a 2x2 complex matrix such that N^2=0. Prove that either N=0 or N is similar over C to the matrix

00
10Sorry, I don't know how else to write the matrix in the post. Any help would be greatly appreciated, thank you.

 

[Petr Mugver]

If N is not zero, then it exists a vector v such that Nv is not zero. Using the fact that N^2=0 it is easy to proove that v and Nv are linearly independent (try it), so they are a basis of C^2. Try writing N with respect to this basis.

 

[Alupsaiu]

Ok, so it was easy to show linear independence, then I found the matrix which had in the first column the coordinates of Nv, and in the second column 0 since N^2=0.
So,

x1 0
x2 0, where x1 and x2 are the coordinates of Nv in basis (v, Nv). Would x1 actually turn out to be 0?


But, where to go from here? In general I'm a bit confused on how to show a matrix is similar, or must be similar to another matrix. Again, thank you for the help.

 

[zcd]

If {v, Nv} is your basis, then writing N as a matrix with respect to this basis would be of the form [N(v) N(Nv)] where N(v) and N(Nv) are column vectors. If you use the fact that N^2= 0, the rest will follow.

As for showing similarity, recall that matrices are self similar even after change of basis.

 

[blue_raver22]

I can provide a mathematical proof to the given statement.

Let N be a 2x2 complex matrix such that N^2=0. This means that N multiplied by itself results in the zero matrix, i.e. N*N=0.

Now, suppose N is not equal to the zero matrix. This means that at least one of the entries in N is non-zero. Without loss of generality, let's assume that the entry in the first row and first column of N, denoted by N11, is non-zero.

Since N^2=0, we can write N*N=0 as:

N*N =
N11*N11 + N12*N21 N11*N12 + N12*N22
N21*N11 + N22*N21 N21*N12 + N22*N22

Since N11 is non-zero, we can divide the first row by N11 to get:

1 + (N12*N21)/N11 N12 + (N12*N22)/N11
N21*N11 + N22*N21 N21*N12 + N22*N22

Now, let's consider the matrix T given by:

1 0
0 0

It is clear that T^2=T, since T*T=
1*1 + 0*0 1*0 + 0*0
0*1 + 0*0 0*0 + 0*0

Thus, we can write T^2-T=0, which is similar to the matrix N. This can be seen by considering the matrix S given by:

1 N12/N11
0 1

Since N11 is non-zero, S is invertible and we can write:

S^-1*T*S =
1 -N12/N11
0 1

Multiplying this with N, we get:

S^-1*T*S*N =
1 -N12/N11
0 1

N11*N11 + N12*N21 N11*N12 + N12*N22
N21*N11 + N22*N21 N21*N12 + N22*N22

=
N11 + (N12*N21)/N11 N12 + (N12*N22)/N11
N21*N11 + N22*N21 N21*N12 + N