Originally Posted by **Country Boy** (2) Let A be a non-zero matrix that is not the identity matrix. Then if $\displaystyle A^2= A$, the eigenvalues are 0 and 1.
Let $\displaystyle \lambda$ be an eigenvalue of A with corresponding eigenvector v. Then $\displaystyle A^2v= A(Av)=A(\lambda v)= \lambda Av= \lambda (\lambda v)= \lambda^2 v$. That is, the eigenvalues of $\displaystyle A^2$ are the squares of the eigenvalues of A with the same eigenvectors. This is true for all matrices. But since $\displaystyle A^2= A$, we must have $\displaystyle A^2v= Av$ so that $\displaystyle \lambda^2v= \lambda v$, $\displaystyle \lambda^2v- \lambda v= (\lambda^2- \lambda)v= 0$. For v non-zero, we have $\displaystyle \lambda^2- \lambda= 0$ which has only 0 and 1 as solutions.
(I am surprised that they specifically exempt the 0 and identity matrices since they obviously also satisfy this theorem.)
4, 5, 6, and 8 are strange- they give an answer then list possible answers! Was the correct answer added? (4) asks you to show that the dimension of the subspace of 2 by 2 matrices, $\displaystyle \begin{pmatrix}a & b \\ c & d \end{pmatrix}$ with the conditions that a+ b= c, b+ c= d, and c+ a= d is 2 but then gives a list of possible answers, one of which (c) is 2!
In any case, to prove that, would use the given equations to solve for as many of a, b, c, and d, in terms of the others as possible. For example, from a + b = c and b + c = d, I can subtract the second from the first to get a - c = c - d so that a = 2c - d. Then a + b = 2c- d + b = c so b = d - c. The third equation, c + a = d can be written c+ a = c+ 2c - d = 3c - d = d so that 2d = 3c and d = (3/2)c. From that b = d - c = (3/2)c - c = (1/2)c and a = 2c - d = 2c- (3/2)c = (1/2)c. a, b, and d can all be written in terms of c so the matrix can be written as $\displaystyle \begin{pmatrix}a & b \\ c & d \end{pmatrix}= \begin{pmatrix}(1/2)c & (1/2)c \\ c & (3/2)c\end{pmatrix}= c\begin{pmatrix}\frac{1}{2} & \frac{1}{2} \\ 1 & \frac{3}{2}\end{pmatrix}$
That is, every matrix in this set is a number, c, times that matrix so the subspace has dimension 1, not 2. (More simply, the set of 2 by 2 matrices has dimension 4 and putting 3 **independent** conditions on it reduces the dimension to 4 - 3 = 1. The calculations I did were really to show that the conditions were independent.) |