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April 26th, 2017, 04:07 AM   #11
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 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.)
Sorry to confuse you! I have provided the answers at the end. I just asked to confirm me if they were right or not Analysis helps me too! May 18th, 2017, 12:11 AM   #12
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 Originally Posted by Joppy And typed, if it helps (let me know if any typos). 2015 Paper 2) $\text{Let A be an$n \times n$nonzero matrix where A is not an identity matrix. If$A^2 = A$, then the eigenvalues of A are given by 0 and 1.}$ 3) $\text{Let A be a$7 \times 5$matrix over$\mathbb{R}$having at least 5 linearly independent rows: Then the dimension of the null space of A is at least 2.}$ 4) $\text{The dimension of the vector subspace$w$of$M_2(c)$given by }$ $w = \left\lbrace \begin{pmatrix} a & b \\ c & d \end{pmatrix} : a, b, c, d \in c, \ a + b = c, \ b + c = d, \ c + a = d \right\rbrace$ $\text{is equal to 2}$. $\mathbf{a}: 4$ $\mathbf{b}: 3$ $\mathbf{c}: 2$ $\mathbf{d}: 1$ 5) $\text{If$|a - b| = |c - d|$then$a = b + c - d$.}$ $\mathbf{a}: \ a = b + c - d$ $\mathbf{b}: \ a = b - c + d$ $\mathbf{c}: \text{$a = b + c - d$and,$a = b - c + d$}$ $\mathbf{d}: \text{$a = b + c - d$or,$a = b - c + d$}$ 6) $\text{The set of all real numbers$x$for which there is some positive real number$y$such that$x < y$is equal to$\lbrace 0 \rbrace$}$ $\mathbf{a}: \mathbb{R}$ $\mathbf{b}: \text{The set of all negative real numbers}$ $\mathbf{c}: \lbrace 0 \rbrace$ $\mathbf{d}: \text{The empty set}$ 8.) $\text{If$\bar{\gamma} = x \bar{i} + y \bar{j} + z \bar{k}$,$\gamma = \sqrt{x^2 + y^2 + z^2}$and$n \in \mathbb{N}$then$\nabla \gamma^n$is equal to$n\gamma^{(n - 2)} \bar{\gamma}$}$ $\mathbf{a}: n\gamma^{(n - 1)} \bar{\gamma}$ $\mathbf{b}: (n - 1)\gamma^{(n - 2)} \bar{\gamma}$ $\mathbf{c}: n\gamma^{n - 2} \bar{\gamma}$ $\mathbf{d}: (n - 1)\gamma^{n} \bar{\gamma}$ P.S., Anyone know how to circle letters without the tikz package?
Can anyone help me with the other questions ?
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