prove that every nilpotent matrix is singular?

Answer 1

Suppose ∃n such that A^n=0 and A^(-1) exists. Consider (A^(-1))^n * A^n. Since A^(-1)*A=I by definition, this product should equal I. However, A^n=0 and any matrix times 0 is 0. Thus, you have that I = 0, which is absurd. Thus our initial assumption, that A is nilpotent and non-singular, must be false. Q.E.D.

Answer 2

If k is an eigenvalue of A, k^n is an eigenvalue of A^n. If A is nilpotent, then A^n is 0 for some A, so all its eigenvalues are 0. Thus k^n = 0 for some n, but this can only be true if k = 0. So all the eigenvalues of A are 0 and A is most certainly singular (a theorem says that a square matrix A is non-singular iff none of its eigenvalues are 0).

Answer 3

One way to prove that a given matrix is singular is to prove that its determinant is zero.

So let's try this way for the given kind of matrices.

If matrix A is nilpotent, it is such that there is some power of it that is equal to zero, or
A^k =0.

There is a property :
det(A^k) = [det(A)]^k

So apply this to the nilpotence condition A^k =0 to get

det(A^k) = [det(A)]^k = det(0) =0
so [det(A)]^k = 0 or finally
det(A) = 0

Answer 4

A is nilpotent if A^n =0 matrix
=> det (A^n) =0
=> det(A) ^n =0
=> det(A)=0 <=> A is singular .