Let \begin{equation} A =  \begin{pmatrix} \lambda &amp; 0\\ 0 &amp; \frac{1}{\lambda}\\ \end{pmatrix} \end{equation} be a matrix in $SL_2(\mathbb{C}$). I am trying to prove that $SL_2(\mathbb{C})$ is generated by the matrices \begin{equation} \begin{pmatrix} 1 &amp; z\\ 0 &amp; 1\\ \end{pmatrix}, \begin{pmatrix} 1 &amp; 0\\ z &amp; 1\\ \end{pmatrix} \end{equation} for $z\in\mathbb{C}$. I have already handled all the cases expect for matrices like $A$. How can I write it as a product of matrices of this type? Any hints are appreciated.
120694
The book asks you to prove that $SL_n(\mathbb{R})$ is generated by elementary (row operation) matrices in which one nonzero off-diagonal entry is added to the identity matrix.  For example,   $$ \begin{bmatrix} 1 &amp; a \\ 0 &amp; 1 \end{bmatrix} $$  acts by left multiplication on $2\times2$ matrices by adding $a$ times (row 2) to (row 1).   Considering a simple example:  $$ M= \begin{bmatrix} a &amp; 0  \\  0 &amp; 1/a \end{bmatrix}, a \neq 0$$  you can see that the matrix $M$ does belong to $SL_n(\mathbb{R})$.  However, the elementary matrices composing $M$ are of the below type and not of the first type (nonzero off-diagonal entry).  $$ \begin{bmatrix}  c &amp; 0 \\   0 &amp; 1  \end{bmatrix} $$ i.e. one nonzero diagonal entry added to the identity matrix.  So  $$M =  \begin{bmatrix} a &amp; 0 \\ 0 &amp; 1 \end{bmatrix} \begin{bmatrix} 1 &amp; 0 \\ 0 &amp; 1/a \end{bmatrix} $$  So $M$ clearly isn't generated by the first type.  What is going wrong?
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:  $$A^2+B^2=AB\implies\\ A(A^2+B^2)=A^2B \implies\\ A^3+AB^2=A^2B \implies \\ A^3+(A^2+B^2)B=A^2B \implies \\  A^3+A^2B+B^3=A^2B \implies \\ A^3+B^3=0 $$
79300
The question is motivated from the following problem:     Let $I\neq A\neq -I$, where $I$ is the identity matrix and $A$ is a real $2\times 2$ matrix. If $A=A^{-1}$, then the trace of $A$ is   $$ (A) 2 \quad(B)1 \quad(C)0 \quad (D)-1 \quad (E)-2$$   Since $A=A^{-1}$, $A^2=I$. If the converse of  is true, then $\lambda^2=1$ and thus $\lambda=\pm1$. And then $\rm{trace}(A)=1+(-1)=0$.   Here are my questions:   Is  $C$ the answer to the quoted problem?  Is the converse of Cayley-Hamilton Theorem, i.e.,"for the square real matrix $A$, if $p(A)=0$, then $p(\lambda)$ is the characteristic polynomial of the matrix $A$" true? If it is not, then what's the right method to solve the problem above?
24786
Let $A,B,C$ and $D$ be $n \times n$ matrices such that $AC = CA$. Prove that $$\det \begin{pmatrix} A &amp; B\\ C &amp; D\end{pmatrix} = \det(AD-CB)$$   The solution is to first assume that $A$ is invertible and then consider the product $$\begin{pmatrix} I &amp; O\\ -CA^{-1} &amp; I \end{pmatrix}\begin{pmatrix} A &amp; B\\ C &amp; D \end{pmatrix}=\begin{pmatrix} A &amp; B\\ O &amp; D-CA^{-1}B \end{pmatrix}$$ then it is not hard to prove that the claim is true if $A$ in invertible. Finally, we use the fact that the set $GL_n$ form a dense open subset of $M_n$ to get rid of the invertibility assumption. My question is: how to come up with such a weird matrix $\begin{pmatrix} I &amp; O\\ -CA^{-1} &amp; I \end{pmatrix}$? thank you so much. Is there any other problems that uses the technique of assuming invertibility? (one of which i know is to prove $\det (I+AB) = \det(I+BA)$), thanks in advance
34229
Let $GL_2(R)$ be the general linear group of $2\times2$ matrices and $SL_2(R)$ the special linear group of $2\times2$ matrices. Show that the commutator subgroup of $GL_2(R)$ is $SL_2(R)$.  I can show that the commutator subgroup is contained in $SL_2(R)$ as if $A,B \in GL_2(R)$ then $$\operatorname{det}(ABA^{-1}B^{-1})=\operatorname{det}(A)\operatorname{det}(B)\operatorname{det}(A^{-1})\operatorname{det}(B^{-1})=1.$$   But how can I show the reverse inclusion? That is, that $SL_2(R)$ is contained in the commutator subgroup of $GL_2(R)$.  Any help will be appreciated.
1594
Let $N= GL_n(\mathbb{R})$ and $M=SL_n(\mathbb{R})$. Is $M$ normal subgroup in $N$? Why or why not?  I know how to do this with $GL_2(\mathbb{R})$ and $SL_2(\mathbb{R})$ but with $N= GL_n(\mathbb{R})$ and $M=SL_n(\mathbb{R})$ I don't even know all of their elements so I can't check the left and right cosets of $M$ in $N$.
54760
Let $E \in M_n(C)$ be an $n \times n$ matrix with entries in a(n algebraically closed, characteristic 0) field $C$, with eigenvalues $\lambda_1, \ldots, \lambda_n$. Show that the commutator map $M_n(C) \to M_n(C)$ given by $B \mapsto [E,B] = EB - BE$ has eigenvalues $\lambda_i - \lambda_j$.  This is an ingredient a proof of the fact that any regular singular equation $y' = Ay$ over $C((z))$  is equivalent to $v' = Dz^{-1}v$ with $D$ a constant matrix. See exercise (7) (a) (i) of  by Marius van der Put (page 6 of the PDF).  This is homework, so hints rather than full answers would be appreciated.  Things I have found that may or may not be useful:   $EB - BE = (\lambda_i - \lambda_j)B$ can be written $(E-\lambda_i)B = B(E- \lambda_j)$. $EB$ and $BE$ .
102971
Let $GL(n,\mathbb{R})$ be the space of invertible $n \times n$ matrices over $\mathbb{R}$. I have already proved that $GL(n,\mathbb{R})$ is open in $\mathbb{R}^{n^2}$ and that it is not connected with the relative topology from $\mathbb{R}^{n^2}$. How do I prove that the map $f:GL(n,\mathbb{R})\rightarrow GL(n,\mathbb{R})$ given by $f(x):=x^{-1}$ is continuous?  Any help would be greatly apprecaited.   Additionally how do I prove that $g:GL(n,\mathbb{R})\times GL(n,\mathbb{R}) \rightarrow GL(n,\mathbb{R})$ given by $g(x,y):=xy$ (matrix multiplication) is continuous, where we take the relative topology from $\mathbb{R}^{n^2} \times \mathbb{R}^{n^2} \cong \mathbb{R}^{2n^2}$?
109652
Let $SL_n(\mathbb{R})$ be the set of $n\times n$ matrix with determinant one. Where can I find a proof that $[SL_n(\mathbb{R}),SL_n(\mathbb{R})]=SL_n(\mathbb{R})$? I've seen this result mentioned in several places, but couldn't find a proof.  For reference, the commutator $[SL_n(\mathbb{R}),SL_n(\mathbb{R})]$ is defined as the group generated by the set of all elements of the form $a^{-1}b^{-1}ab$, with $a,b\in SL_n(\mathbb{R})$.
85504
Let $n\in\mathbb{N}$ and $A\in M_n(\mathbb{F})$. Assume that the minimal polynomial of the square matrix $A$ is defined as the minimal polynomial of the linear operator $x\mapsto Ax$ for all $x\in\mathbb{F}^n$, on $\mathbb{F}^n$. Now, assume that the minimal polynomial of $A$ splits over $\mathbb{F}$. Show that $A$ and $A^T$ are similar.  Two matrices $A$ and $B$ similar if there exists an invertible matrix $S$ such that $B=SAS^{-1}$. Finding such an $S$ for $B=A^T$ will solve the problem. Since the minimal polynomial of $A$ splits over $\mathbb{F}$, let us denote it by $$m_A(t) = (t-\lambda_1)^{m_1}...(t-\lambda_k)^{m_k}$$ for some $k\in\mathbb{N}$. We know that $m_A(A)$ corresponds to the zero linear operator, i.e. $m_A(A)x=0$, $\forall x\in \mathbb{F}^n$. I quickly noticed that if matrices $P_1,P_2,...P_q$ commute, then $(P_1P_2...P_k)^T = P_1P_2...P_k$. It may be useful to replace each $P_i$ with $(A-\lambda_i)^{m_i}$? I'm not sure how to proceed from here. I'd appreciate any hints!
86386
Let $A\in GL_2\left(\mathbb{Z}\right)$, the group of invertible matrices with integer coefficients, and denote by $\omega(A)$ the order of $A$.  How we prove that $$\left\{\omega(A);A\in GL_2\left(\mathbb{Z}\right)\right\}=\{1,2,3,4,6,\infty\}.$$
4463
Suppose $C$ is an $n\times n$ matrix over complex numbers, with trace $0$.  Are there always $n\times n$ matrices $A,B$ such that $AB - BA = C$?  (Inspired by a recent question which asked for a trace free proof of non-existence of solutions for $C=I$).
50716
Is it true for complex matrices $X,Y$ that $$ (XY)^*=X^* Y^*? $$ where $^*$ refers to complex conjugation.  How can we prove this if so?  Thanks!  Note: I am referring to complex conjugation, not the hermitan transpose. The answer below refers to hermitan tranpose.
96857
Prove that  $GL(n, \mathbb{C})$ is isomorphic to a subgroup of $GL(2n, \mathbb{R})$.  My Proof: For an $A \in GL(2, \mathbb{C})$, $$ A = \begin{bmatrix} a+bi &amp;c+di \\ e+fi &amp; g+hi \end{bmatrix} =  \begin{bmatrix} a &amp;c \\ e &amp; g \end{bmatrix} + i \begin{bmatrix} b &amp; d \\ f &amp; h \end{bmatrix} .$$ It follows that $ GL(n, \mathbb{C})$ is a subset of $GL(2, \mathbb{R}) \oplus GL(2, \mathbb{R})$. (it is a subset because the condition that $\det A \neq 0$ puts more restrictions on the elements of $A$ than the condition that the two little matrices are invertible.) Thus the result holds for $n=2$.  Can this argument be generalized to any $n$? The hint was to use group actions, of $GL(n, \mathbb{C})$ on $GL(2n, \mathbb{R})$, using the fact that $\mathbb{C}^n \cong \mathbb{R}^{2n}$. I don't understand why a group action would be helpful?
102543
I have a problem with this. Actually, still don't have the right way to start :/  Problem :  Let $A$ and $B$ be $n \times n$ complex matrices such that $AB - B^2A^2 = I_n$. Prove that if $A^3 + B^3 = 0$, then $BA - A^2B^2 = I_n$.  Thanks for any help.
95653
In one of my classes we discussed the ring of 2x2 matrices $M_{2}(\mathbb{Z})$. We said that its group of units was $SL_{2}(\mathbb{Z})$ which means that it is the set of 2x2 with determinant equal to $\pm$1. Why can't we have a 2x2 matrix with entries a,b,c, and d such that $\frac{a}{ad-bc}$,$\frac{-b}{ad-bc}$,$\frac{-c}{ad-bc}$, and $\frac{d}{ad-bc}$ are all integers?  I'm sure its a simple contradiction argument, but I couldn't see it. So if anyone knows a quick elementary argument, it'd be greatly appreciated
5906
There is the standard proof using $$\det(A)=\det( A^{T} ) = \det(-A)=(-1)^n \det(A)$$  I would like a proof that avoids this. Specifically, there is the proof that for $A$ a $\bf{real} $ matrix, the transpose is the same as the adjoint, which gives (using the complex inner product) $\lambda \|x\|^2 =\langle  Ax, x \rangle= \langle  x, -Ax \rangle=-\overline{\lambda } \|x\|^{2}$, so any eigenvalue is purely imaginary. Then we conclude that, since any odd-dimensional real matrix has a real eigenvalue, that eigenvalue must be zero.  This argument doesn't work for a general complex skew-symmetric matrix.  Is there something I'm missing, is there a way to modify this argument to get that zero is an eigenvalue for the complex case?  Also, can somebody please give a geometric reason why odd-dimensional skew-symmetric matrices have zero determinant (equiv., a zero eigenvalue)?  Thanks!