Please explain the answer provided below in such a way as to be understood by someone who is really struggling in linear algebra. THANKS!!! Let N be a 2x2 complex matrix such that N2 = 0. Prove that either N = 0 or N is similar over C to Letf: C2 rightarrow C2 be the linear mapping defined by f(x) = Nx. Then f(f(x)) = N(Nx) = 0 for allx. Hence the image off is a subspace of the kernel off. If dim im(f) = 2, then f is bijective and fLambda2 cannot be the zero mapping, contradiction. Hence dim im(f) is 0 or 1. If dim im(f) = 0, then f is the zero mapping and N is the zero matrix. If dim im(f) = 1, then dim ker(f) = 1, and hence im(f)=ker(f). Denote this subspace by V. Then there is a vector y in C2 which is not in V. Then f(y) is in V and is not zero, hence {y, f(y)} is a basis. With respect to this basis, the matrix representation off is [00; 10]. Hence N is similar to the matrix [0 0; 1 0].
Show transcribed image text Please explain the answer provided below in such a way as to be understood by someone who is really struggling in linear algebra. THANKS!!! Let N be a 2x2 complex matrix such that N2 = 0. Prove that either N = 0 or N is similar over C to Letf: C2 rightarrow C2 be the linear mapping defined by f(x) = Nx. Then f(f(x)) = N(Nx) = 0 for allx. Hence the image off is a subspace of the kernel off. If dim im(f) = 2, then f is bijective and fLambda2 cannot be the zero mapping, contradiction. Hence dim im(f) is 0 or 1. If dim im(f) = 0, then f is the zero mapping and N is the zero matrix. If dim im(f) = 1, then dim ker(f) = 1, and hence im(f)=ker(f). Denote this subspace by V. Then there is a vector y in C2 which is not in V. Then f(y) is in V and is not zero, hence {y, f(y)} is a basis. With respect to this basis, the matrix representation off is [00; 10]. Hence N is similar to the matrix [0 0; 1 0].