I'm having trouble trying to find a $2 \times 2$ matrix $R$ in which $R$ is not equal to $I$, but $R^7=I$.  I thought to start with the matrix $$ \begin{bmatrix} \cos x &amp; -\sin x \\ \sin x &amp; \cos x\end{bmatrix}$$ was a good place to start but what I got when trying a sample run on my calculator was that it comes out to the same matrix as $R$. Any suggestions of how I should approach this? Or should I use a very specific value for $x$? The value I used on my calc was $\pi/3$ just to avoid it being equal to the identity matrix.
120486
The question on the homework wants a $2\times2$ matrix $R$, but $R$ not equal to positive or negative identity, that when you raise to seventh power you get the identity.  $R^7=\operatorname{Id}$.  I thought it would have to be some ones and negative ones,  it no luck, then tried some ones and some $.5$.  Just using an online matrix calculator to do brute force.  We just started talking about eigenvalues with visual example, so assuming wouldn't need that since we have not talked about how to find them yet.  Any hints or ideas to get me back on track?
104885
Let $A = (a_{ij}) \in R^{n\times n}$. Find the determinant if:  $$a_{ij}= |i-j|$$  So we have the symmetric matrix \begin{bmatrix} 0 &amp; 1 &amp; 2 &amp; 3 &amp; 4 &amp; \dots &amp; n-1 \\ 1 &amp; 0 &amp; 1 &amp; 2 &amp; 3 &amp; \dots &amp; n-2 \\ 2 &amp; 1 &amp; 0 &amp; 1 &amp; 2 &amp; \dots &amp; n-3 \\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \ddots &amp; \vdots \\ n-1 &amp; n-2 &amp; n-3 &amp; \dots &amp; \dots &amp; \dots &amp; 0 \end{bmatrix}  But i can't find a way to diagonalize the matrix nor find the determinant by Laplace expansion.  Any ideas ??
117088
I'm tasked with showing that matrix $A$ commutes with every $2\times2$ matrix if and only if $A = \begin{bmatrix}a &amp; 0\\0 &amp; a\end{bmatrix}$ for some a.  I was able to prove in the first direction, assuming that $A = \begin{bmatrix}a &amp; 0\\0 &amp; a\end{bmatrix}$, by just multiplying the matrix by $\begin{bmatrix}a &amp; b\\c &amp; d\end{bmatrix}$ in AB = BA format, and showing how AB does equal BA.  $$ (=&gt;)$$  $$\begin{bmatrix}a &amp; 0\\0 &amp; a\end{bmatrix}\begin{bmatrix}a &amp; b\\c &amp; d\end{bmatrix} = \begin{bmatrix}a &amp; b\\c &amp; d\end{bmatrix}\begin{bmatrix}a &amp; 0\\0 &amp; a\end{bmatrix}$$  $$\begin{bmatrix}a^2 &amp; ab\\ac &amp; ad\end{bmatrix} = \begin{bmatrix}a^2 &amp; ab\\ac &amp; ad\end{bmatrix}$$  But now I have to prove in the other direction, assuming that A already does commute with every $2\times 2$ matrix, and I'm not particularly sure how I can do that. Somehow, I have to show that if A commutes with $\begin{bmatrix}a &amp; b\\c &amp; d\end{bmatrix}$, then A will be $\begin{bmatrix}a &amp; 0\\0 &amp; a\end{bmatrix}$, and I don't know how I can carry that out.   Where can I start? Ideas are appreciated.
4810
The $2 \times 2$ matrix ${A}$ satisfies ${A}^2 - 4 {A} - 7 {I} = {0}$ where ${I}$ is the $2 \times 2$ identity matrix. Prove that ${A}$ is invertible.  I have tried to solve it like a quadratic, but that doesn't work.  Any help is appreciated!
41383
Let$A$ be a $3\times3$ real symmetric matrix such $ A^6=I$ . Then $A^2=I$. How can I prove this statement is true or false? As it is given $A$ is symmetric so $A=A^T$. Also $ A^6=I$. But the main problem is that I can't operate $A^{-1}$ on both sides whether it is invertible or not. Can any help me what should I do?
28755
I was reviewing some matrices and found this interesting  if $r = \begin{pmatrix} 0&amp;1\\ -1&amp;0 \end{pmatrix}$ then $rr=-I$, also $$\exp{(\theta r)} = \cos\theta I + \sin\theta r$$ No wonder, the matrix $R(\theta) = e^{\theta r}$ is the 2d rotation matrix, just like $e^{i\theta}$ rotates a vector in the Argand plane. I have a very cursory knowledge of complex analysis, so I would like to know where I can find the details, i.e what is the unifying theme and in which literature can it be found.
64765
How many solutions $X^{10} - I=0$ has in $M_2(\mathbb{R})$? Where $M_2(\mathbb{R})$ denotes the set of $2 \times 2$ real matrices.  I absolutely have no idea of where I should start from. $I$ and $-I$ solve the equation, but are there any other non-trivial solutions? Are there infinitely many solutions?
12785
Let $A$ belongs to $M_3(\mathbb R)$ which is not a diagonal matrix. Pick out the cases  when $A$ is diagonalizable over $\mathbb R$:  a. when $A^2 = A$; b. when $(A - 3I)^2 = 0$; c. when $A^2 + I = 0$  (a) is certainly true as eigen values are distinct. (c) is certainly not true as eigen values are imaginary not real. but i am not sure about (b). can anybody help? thanks.
90042
I have been thinking about this problem:  Suppose we have all $R$ by $C$ matrices, where the values are integers in $[1,n]$. Two matrices are equivalent under interchanging rows and columns. For example,   1 5     0 0       would be equal to itself and to  0 0 1 5   and would be equal to  0 0  5 1   and would be equal to   5 1 0 0   How many unique such matrices are there?  Any ideas how to go about it? Any help much appreciated!
49764
The Problem Let $A$ be the matrix $\bigl(\begin{smallmatrix}a&amp;b\\c&amp;d\end{smallmatrix} \bigr)$, where no one of $a,b,c,d$ is $0$. Let $B$ be a $2\times 2$ matrix such that $AB+BA=\bigl(\begin{smallmatrix} 0&amp;0\\ 0&amp;0 \end{smallmatrix} \bigr)$. Show that either   $a+d=0$, in which case the general solution for $B$ depends on 2 parameters, or $ad-bc=0$, in which case the general solution for $B$ depends on one parameter.   (this is question 22 of the last matrix exercise of Further Pure Mathematics by Bostock et al.)  Comments Writing $B=\bigl(\begin{smallmatrix} e&amp;f\\ g&amp;h \end{smallmatrix} \bigr)$ and multiplying out I get that    $(a+d)(f+g)+(b+c)(e+h)=0$ $ae+bg+cf+dh=0$   but I am unable to get the required restrictions on $a,b,c,d$. Is there a quick way of doing the problem that doesn't require manual computation? I thought of considering invertible and non-invertible cases but couldn't get anywhere. Help would be much appreciated.
29395
I have a question where I have a matrix $$A=\begin{bmatrix}2&amp;-1\\-1&amp;1\end{bmatrix}$$ and I need to find two examples of a non-zero matrix $X$ such that $AX = XA$. My first matrix is \begin{bmatrix}1&amp;0\\0&amp;1\end{bmatrix} the identity matrix but I cant think of another one. Any ideas?
56483
$A=\begin{pmatrix}0&amp;0&amp;1\\ 0&amp;1&amp;0\\ 1&amp;0&amp;0\end{pmatrix}$  How do I show that there are no solutions for $X^2=A$ in $M_3(R)$?  I tried $X^2=\begin{pmatrix}a_{11}&amp;a_{12}&amp;a_{13}\\ a_{21}&amp;a_{22}&amp;a_{23}\\ a_{31}&amp;a_{32}&amp;a_{33}\end{pmatrix}^2=A$, but I couldn't seem to find anything that shows there are no solutions. How should I approach this?
94030
I f $$A= \begin{pmatrix}1&amp; 0 &amp; 0 \\  1 &amp; 0 &amp; 1\\ 0 &amp; 1 &amp; 0 \end{pmatrix}$$  then $ A^{50} $  is   $$ \begin{pmatrix}1&amp; 0 &amp; 0 \\  50 &amp; 1 &amp; 0\\ 50 &amp; 0 &amp; 1 \end{pmatrix}$$ $$\begin{pmatrix}1&amp; 0 &amp; 0 \\  48 &amp; 1 &amp; 0\\ 48 &amp; 0 &amp; 1 \end{pmatrix}$$ $$\begin{pmatrix}1&amp; 0 &amp; 0 \\  25 &amp; 1 &amp; 0\\ 25 &amp; 0 &amp; 1 \end{pmatrix}$$ $$\begin{pmatrix}1&amp; 0 &amp; 0 \\  24 &amp; 1 &amp; 0\\ 24 &amp; 0 &amp; 1\end{pmatrix}$$   I am stuck on this problem. Can anyone help me please...............
14391
Is the following statement true or false? There exists $A \in M_{3,3}(\mathbb{Z})$ with determinant $2$ such that $$A\begin{pmatrix}2\\1\\4\end{pmatrix} = \begin{pmatrix}4\\-8\\16\end{pmatrix}$$   I first thought of eigenvalues but it doesn't look like the second vector is a multiple of the first. What are the conditions on $A$ so that it's "true"? Or is it always false?
38731
If $A$ and $B$ are two $4\times3$ matrices such that $A\mathbb{v}=B\mathbb{v}$ for all $\mathbb{v}\in\mathbb{R}^3$, then $A=B$?  If it's not true, can you give me an example of it?  Thank you.
4932
Let $m &gt; n$ be positive integers. Show that there do not exist matrices $A ∈ R_{m×n}$ and $B ∈ R_{n×m}$ such that $AB = I_m$, where $I_m$ is the $m × m$ identity matrix  If someone could explain it with an example, it will be appreciable.
48360
So I'm studying linear algebra and one of the self-study exercises has a set of true or false questions. One of the questions is this:     If $A^2 = I$ (Identity Matrix), then $A = \pm I$ ?   I'm pretty sure it is true but the answer says it's false. How can this be false (maybe it's a typography error in the book)?