4.Prove . If A element Mn(R) is nilpotent, then (In-A) is nonsingular.?

Question

3. Definition. An eigenvalue of a matrix A element of Mn(R) is a scalar c such that given a linear transformation f: XY where X and Y are vector spaces over a field F with X=Y, f(x)=cx, x element of X; or alternatively, Ax=cx, and x is called the eigenvector associated with the eigenvalue c. Prove the ff:

i.c=0 is an eigenvalue of A iff A is singular.
ii.If c is an eigenvalue of A, then c^-1 is an eigenvalue of A^-1. Hence, all eigenvalues of A^-1 are the reciprocals of the eigenvalues of A.
iii.If c is an eigenvalue of A, then c is also an eigenvalue of Atranspose.




9. Given A element of Mn(R)

i.Show that det(A) is the product of all the roots of the characteristics polynomial of A.
ii.Show that tr(A) is the sum of the eigenvalues of A.

Answer 1

4. Let A^k = 0, then (I-A)(I + A + A^2 + ... + A^(k-1))
= I + A - A - A^2 + A^2 - A^3 + ... - A^(k-1) + A^(k-1) - A^k
= I. So I-A is invertible (non-singular).

3. i) =>: Suppose c=0 is an eigenvalue of A and the associated vector is x (x not equal to 0). Suppose also that A is non-singular. Then A^-1 exists. But x = Ix = (A^-1A)x = A^-1(Ax) = A^-1(cx) = A^-1(0) = 0, a contradiction. So A must be singular.
<=: Suppose A is singular. Consider the equation Ax = 0. If we form the augmented matrix [A | 0] and put it into row-echelon form, the bottom row must be all zeros (else A is invertible). Hence we are left with n x-component variables and (n-1) or fewer equations; the system is consistent (since x=0 is a solution) and underdetermined, so there must be infinitely many solutions. So there must be some nonzero x with Ax = 0.

ii) If Ax = cx, then A^-1 Ax = A^-1 cx => x = c. A^-1 x => A^-1 x = (1/c) x.


Have to leave it there now, sorry - enjoy Christmas!