Can you guys think of any way to find a 2x2 matrix that when raised to the third power is the identity? I know there is the obvious answer of the identity itself, but I was specifically wondering about matrices that would take more than one elementary row operation to achieve. Thanks in advance to anyone who can give me some insight!
121997
So it's been a while since I've taken Linear Algebra, but my friend asked me a question, that I couldn't answer.  If a matrix $A$ exists such that $A^3 = I$, does $A$ have to equal the identity matrix $I$?  My first instinct was to say no, but... (edited out my incorrect math)  EDIT: thanks guys for the awesome examples  EDIT2: Followup question: Is there a way to solve for all possibilities of A if given A^3 = I?
34491
I guess that I'm missing here something...  How to prove that, if two matrices are similar, then their minimal polynomials are the same one.   Thanks!
63632
How do you prove the identity: a$\cdot($b$\times$c$) =$ c$\cdot($a$\times$b$) = -$b$\cdot$(a$\times$c) I'm thinking that this is because of row swaps in the determinant of a matrix with a,b,c as its row vectors. But I'm not sure how to write out the proof, and why its related to the determinant of that matrix
91044
Suppose the m * n matrix A =$ \begin{bmatrix}     A_1 \\A_2 \end{bmatrix}$ where A1 is a nonsingular square matrix (n*n) and A2 is an arbitrary matrix. Why is $||A^{+}||_2 \leq ||A_1^{-1}||_2$  ?   ($A^{+}$ denotes the psuedoinverse)
85589
I was thinking about the problem that says:     If $A$ and $B$ are $3\times 3$ real matrices such that $\operatorname{rank}(AB)=1$, then $\operatorname{rank}(BA)$ can not be which of the following?   (a) $0$   (b) $1$   (c) $2$   (d) $3$.   My attempt: I have chosen suitable $3 \times 3 $ matrices for $A$ and $B$ keeping in mind that $\operatorname{rank}(AB)=1$. Say for example if I take $A$ and $B$ to be  $$A = \begin{pmatrix} 1 &amp;2  &amp;0 \\  0 &amp; 0 &amp;0 \\  0 &amp; 0 &amp;0  \end{pmatrix}$$  and  $$B = \begin{pmatrix} -2 &amp;1  &amp;0 \\  1 &amp; 0 &amp;0 \\  0 &amp; 0 &amp;0  \end{pmatrix}$$ respectively, then I see $\operatorname{rank}(AB) = \operatorname{rank}(BA) = 1$. So, option (b) can not be correct. Do I have to keep choosing the matrices and then observe which of the option holds good. Is this kind of approach right to tackle the problem? I am looking for a direct way (e.g. application to some theorem) which can give me the result. I have also noticed that $AB$ and $BA$ are similar matrices as we see that $A^{-1}(AB)A=BA$. Is this observation going to help me in any way? Thanks in advance for your time.
40601
Assuming that given a $2\times2$ matrix $A$ with the property $A^3=0$ show that $A^2=0$.  Okay so this question is messing with me. When it says $A^3=0$ do you think it means the determinant? How would one approach this? Any help would be much appreciated
15141
I need to find $|GL_2(F_p)|$, I am very very uncomfortable in counting, so please help here I proceed,  $\begin{pmatrix}a&amp;b\\c&amp;d\end{pmatrix}$ be such an invertible matrix, so we need and sufficient that $ad\neq bc$, so if i chose all non zero $a,b,c$ arbitrarily and chose $d\neq a^{-1}bc$ and form matrices then the cardinality   will be $(q-1)^3\times (q-2)$, well now if one of $a,b,c,d$ be zero then other three can be anything nonzero so in this case cardinality will be $4(q-1)^3$, so finally if two entries are $0$ then they must be opposite to each other i.e diagonally, but other two must be nonzero, so cardinality in that case $2(q-1)^2$, so total number is just sum of three cases.
71172
Find the determinant of the following matrix: $$A = \begin{bmatrix} 1+x_1^2 &amp;x_1x_2  &amp; ... &amp; x_1x_n \\   x_2x_1&amp;1+x_2^2  &amp;...  &amp; x_2x_n\\   ...&amp; ... &amp; ... &amp;... \\   x_nx_1&amp; x_nx_2  &amp;...  &amp; 1+x_n^2 \end{bmatrix}$$  I computed for the case $n=2$, and $n=3$ and guessed that $\det(A)$ should be $ 1+\sum_{i=1}^n x_i^2 $  but not sure how to proceed for any $n$.
78925
Let $A$ be the matrix described when $A \in M_n(\mathbb{R})$.  Let $a,b \in \mathbb{R}$.  Prove that:   Iv'e tried to factor out $a+b$, but got nowhere.  A hint or a solution would be very appreciated.  Thanks a lot!
3063
I'm reading Linear Algebra by Hoffman, Kunze where the authors explained that a $n\times n$ matrix $A$ being invertible is equivalent to the fact that $A$ is row-equivalent to $n\times n$ matrix $R$ which is an identity matrix.  In the proof of the theorem, they wrote:     $$R= E_k\ldots E_2E_1 A$$ where $E_1,\ldots,E_k$ are elementary matrices. Each $E_j$ is invertible, and so $$A = E_1^{-1}\ldots E_k^{-1}~R\,. $$  ... Since, $R$ is a (square) row-reduced echelon matrix, $R$ is invertible if and only if $R=I\,.$  [...]   I couldn't get the conclusion, since any row of $R$ can't be zero, it has to be identity matrix $I\,.$ Why is it so?  Isn't there any other row-reduced echelon matrix other than the identity matrix having no zero row and invertible? Why is it so?
1337
I can't find out if the product of two invertible matrix is an invertible matrix or if the sum of two invertible matrix is an invertible matrix.  Can anyone suggest an answer/counterexample?  Thanks a lot in advice
79397
Can  of finding the determinant be used for finding the determinant of matrices greater than $3\times3$, as I am unable to find any example of a matrix greater than $3\times3$ whose determinant is found by Sarrus' method?  I have tried many questions and Sarrus' method only works for $2\times2$ and $3\times3$ matrices. I have not read about Sarrus' method in any book but only from the internet, so I'm not sure about it.
76063
Find three real orthogonal matrices of order $3$ having all integer entries.   I have no idea to solve the problem. I don't know how to start. If $A$ be such matrix then $AA^T=A^TA=I_3$. Please help me.
68302
Is there square matrix $A$ of size $3$ with real entries such that  $$ \operatorname{tr}(A)=0\text{ and }A^2+A^T=I. $$  I have proved that there is not with size $2$ using definition of  "trace", but for size $3$ it becomes complicated.  Here is the sketch of the proof for $2$. $$ a_{11}+a_{22}=0,\\ a_{11}^2+a_{12}a_{21}+a_{11}=1,\\ a_{11}a_{12}+a_{12}a_{22}+a_{21}=0,\\ a_{21}a_{11}+a_{21}a_{22}+a_{12}=0,\\ a_{12}a_{21}+a_{22}^2+a_{22}=1. $$ Putting $a_{11}=-a_{22}$ it is easy to see that above inequalities can't be true together.   Thanks!
55713
The question is:     Are there square matrices $A,B$ over $\mathbb{C}$ s.t. $A^2+B^2=AB$ and $BA-AB$ is non-singular?   From $A^2+B^2=AB$ one could obtain $A^3+B^3=0$. Can we get something from this?  Edit:  $$A^2+B^2=AB\implies\\ A(A^2+B^2)=A^2B \implies\\ A^3+AB^2=A^2B \implies \\ A^3+(A^2+B^2)B=A^2B \implies \\  A^3+A^2B+B^3=A^2B \implies \\ A^3+B^3=0 $$
118486
Do anybody know how matrices are built into ZFC theory ? I pretty have no idea how to build them from ZFC axioms. I would like a constructive proof of their existence if possible.  Thanks in advance.
55830
$$A=\begin{pmatrix} 1 &amp; -1 \\ -2 &amp; 1 \end{pmatrix}$$      Find a matrix $B$ such that $B^3$ = A   My attempt:  I found $\lambda_1= 1+{\sqrt 2}$ and  $\lambda_2= 1-{\sqrt 2}$  I also found their corresponding eigenvectors $\vec v_1 =\begin{pmatrix} \frac{-\sqrt 2}{2} \\ 1 \end{pmatrix}$ and $\vec v_2 = \begin{pmatrix} \frac{\sqrt 2}{2} \\ 1 \end{pmatrix}$  I know the Power function of a matrix formula $A=PDP^{-1}$  Because it's the cubed root I'm looking for I don't know how to get the cubed root of the eigenvaules and keep the maths neat. Is there another way to solve this problem or an I going the wrong way about doing it ?