Let $A,B$  be two $n \times n $ matrices with real entries and $\alpha \in \mathbb{R}, \alpha \neq 0$. It is given that $$A^2-B^2=\alpha(AB-BA)$$   Prove that $(AB-BA)^n=0$.   At first, I tried to prove that $A(AB-BA)=(AB-BA)A$, as it would imply that $(AB-BA)$ is  and the problem would be solved. But this didn't work.  Then, after writing $\det(A-B)(A+B)=\det(A+B)(A-B)$ and using the given relation I obtained $\det (AB-BA)=0$. But I got stuck and I tried to make a substitution: \begin{align*} X=\frac{1}{2}(A+B) \\ Y=\frac{1}{2}(A-B) \end{align*} The given relation then becomes $$(\alpha+1)XY=(\alpha-1)YX \quad (1)$$ so $XY$ and $YX$ commute. Since $(AB-BA)^n=2^n(YX-XY)^n$, we can expand the latter using the binomial theorem. Combined with $(1)$, I obtained $$(AB-BA)^n=\frac{4^n}{(\alpha-1)^n}(XY)^n=\frac{2^n(-2^n)}{(\alpha+1)^n}(YX)^n$$ and I don't see how to finish this.
121588
I have matrices $A,B$ of dimension $n$ with real coefficients which satisfy the following: $A^2-B^2=c(AB-BA)$ where $c$ is a real number. If $c\neq0$ , prove that $(AB-BA)^n = 0$.   So far, I've been able to show that $AB-BA$ is singular. Can someone help?
76818
Let $A$ and $B$ be $n\times n$ matrices. Suppose $A$ is similar to $B$. Prove $\operatorname{trace}(A) = \operatorname{trace}(B)$.  I'm not sure where to go on this. So far I have this:  If $A$ is similar to $B$, then      $B=P^{-1}AP$ and $A=PBP^{-1}$   This implies that:     $\operatorname{trace}(B) = \operatorname{trace}(P^{-1}AP)$ and $\operatorname{trace}(A) = \operatorname{trace}(PBP^{-1})$   Not sure where to go from here
26750
I am trying to show that $[A,B^n] = nB^{n-1}[A,B]$ where A and B are two Hermitian operators that commute with their commutator. However, I am running into a little problem and would like a hint of how to proceed.  If A and B commute then $[A,B] = ABA^{-1}B^{-1} = e$ where e is the identity element of the group.  $$\therefore AB=BA$$  $$n=1; [A,B^1] = (1)B^0[A,B] = e$$ This statement is certainly true. however moving on to $n = 2$ I find...  $$[A,B^2] = AB^2A^{-1}B^{-2} = ABBA^{-1}B^{-1}B^{-1} = BBAA^{-1}B^{-1}B^{-1}$$  Where in the last step I have used the fact that A and B commute to rearange the terms. However, it is plain to see that this last term simply reduces to the identity as well and for the n = 2 case we have:  $$[A,B^2] = e \ne 2B[A,B] = 2Be = 2B$$  Clearly I have assumed something I shouldn't have. The fact that there is a multiplicative factor of n implies I should be adding things, but I thought if I kept it as general as possible, the answer should just fall out naturally. I don't want an answer please, only guidance.
68316
Let $A, B$ matrices $n\times m$ and $m\times n$ respectively so that $AB=I_n$ and $BA=I_m$. Show that $n=m$.
16747
I'm having problems with an exercise from Apostol's Introduction to Analytic Number Theory.     Given $x$ and $y$, let $m=ax+by$, $n=cx+dy$, where $ad-bc= \pm 1$. Prove that $(m,n)=(x,y)$.   I've tried to give a proof, but I suspect it's wrong (or at least not very good). I would be very thankful for any hints/help/advice!  My proof:     We observe that since $ad-bc= \pm 1$, $ad=bc \pm 1$, and $(ad,bc)=1$.   Now, $(a,b) \mid m$ and $(c,d) \mid n$ but      $$(a,b) = ((d,1)a,(c,1)b)=(ad,a,bc,b)=(ad,bc,a,b)=((ad,bc),a,b)=(1,a,b)=1.$$      Similarly, we determine $(c,d)=1$. So, $1=(a,b) \mid m$ and $1=(c,d) &gt; \mid n$. But $(x,y)$ also divide $m$ and $n$. Since $(x,y) \geq (a,b)=(c,d)=1$, this implies that $(x,y)=m,n$. Hence   $(m,n)=((x,y),(x,y))=(x,y)$.
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.
12345
Matrix Question  If $A^2 = I$, $B^2=I$ and $(AB)^2=I$, then $AB = BA$  Basically, got up to $A(BA-AB)B = 0$ (by cancelling and equating terms from $I^2 = I$ and to $A^2B^2 = A^2B^2$ and using distributive laws), but that doesn't work out too well!  Thanks for help in advance!
62412
Let $A$ be a $n \times n$ real matrix such that $$ A + A^t = 2I, $$ where $I$ is the $n \times n $ identity matrix. Prove that $\det(A) \geq 1$.  It is obvious that tr$(A) = n$. Furthermore, we also have that $$A - 2I = -A^t, $$ so we get that $$\det(A-2I) = (-1)^n \cdot \det(A).$$ Now, we let $\lambda_1, \lambda_2, \cdots, \lambda_n \in \mathbb{C}$ be the  eigenvalues of $A$, so we get that $$(\lambda_1 - 2)(\lambda_2 - 2) \cdots (\lambda_n - 2) = (-1)^n\lambda_1\lambda_2 \cdots \lambda_n, $$ but I couldn't derive anything about the product $\lambda_1\lambda_2 \cdots \lambda_n = \det(A)$ (the only known thing is $\lambda_1 + \cdots + \lambda_n = n$). Also, I tried the same approach for $2 \times 2$ and $3 \times 3 $ matrices, but it didn't lead to anything (especially since the eigenvalues can be complex numbers).
112968
How to prove the  determinant of $$A=(a_{ij})_{n\times n}=b_i^2-b_j^2$$ where $b_1,b_2,...,b_n$ are some distinct real numbers , $n\geq 3$  and  $i,j=1,2,...,n$  is zero?  Here,  $A$ is skew symmetric and I know odd order skew symmetric matrix has determinant zero. How about the even case? or Is there any other method to show $\det(A)=0$ ?
101802
Show that if $A,B \in M_{n \times  n}(K)$, where $K=\mathbb{R}, \mathbb{C}$, then the matrices $AB$ and $BA$ have same eigenvalues.   I do that like this:  let $\lambda$ be the eigenvalue of $B$ and $v\neq 0$  $ABv=A\lambda v=\lambda Av=BAv$  the third equation is valid, because $Av$ is the eigenvector of $B$. Am I doing it right?
96205
Let $A, B\in M_n(\mathbb{C} ) $ such that $B^2=B$. Prove that $\operatorname{rank} (AB-BA) \le \operatorname{rank} (AB+BA) $.    The problem above is the one that appeared in the SEEMOUS 2019, which is a contest for university students. In a contest for 11th graders in my country (here we study both linear algebra and real analysis in the 11th and the 12th grade), the same problem appeared, only that $A, B\in M_n(\mathbb{R})$. Now, I have seen solutions to the SEEMOUS problem which use pretty advanced techniques, which are not taught in high school here, but I want to find one that is appropriate for my level. I suppose there should be one since it appeared in that contest for 11th graders, yet I can't find one and there is also no official solution.  To sum up, I am looking for a solution to this which may only use basic rank inequalities, linear independence, diagonalization, block matrices and other things like these.         EDIT: From $B^2=B$ I deduced that $B$ is diagonalizable and its eigenvalues are $0$ or $1$ (I doubt that eigenvalues help here). If $B$ is invertible we have that $B=I_n$ and the conclusion follows. In the other case, I think that we may use that $B$ is diagonalizable, but I don't know if this helps me express $AB$ and $BA$ somehow.
75546
Assume that the matrices $A,\: B\in \mathbb{R}^{n\times n}$ satisfy  $$ A^k=0,\,\, \text{for some $\,k\in \mathbb{Z^+}$}\quad\text{and}\quad AB=BA. $$  Prove that $$\det(A+B)=\det B.$$
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
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:  $$A^2+B^2=AB\implies\\ A(A^2+B^2)=A^2B \implies\\ A^3+AB^2=A^2B \implies \\ A^3+(A^2+B^2)B=A^2B \implies \\  A^3+A^2B+B^3=A^2B \implies \\ A^3+B^3=0 $$
29786
Let ${a} = \begin{pmatrix}x_a\\y_a\\z_a\end{pmatrix}$, ${b} = \begin{pmatrix}x_b\\y_b\\z_b\end{pmatrix}$, and ${c} = \begin{pmatrix}x_c\\y_c\\z_c\end{pmatrix}$. Show that $(x_a,y_a,z_a)$, $(x_b,y_b,z_b)$, and $(x_c,y_c,z_c)$ are collinear if and only if ${a} \times {b} + {b} \times {c} + {c} \times {a} ={0}.$  Hello,  Is there any other way to do this problem without bashing?  I can't seem to find a nice and slick solution to this.  Thanks in advance!
28988
Let $A$ and $B$ be real square matrices of the same size. Is it true that $$\det(A^2+B^2)\geq0\,?$$  If $AB=BA$ then the answer is positive: $$\det(A^2+B^2)=\det(A+iB)\det(A-iB)=\det(A+iB)\overline{\det(A+iB)}\geq0.$$
78367
(a) Let $x$ and $y$ be $n\times 1$ matrices, $n \ge 1$, and let $A=xy^T$. Show that $\det(A)=0$.  (b) Explain why the statment in part (a) is false if $n=1$.