If two matrices $A$ and $B$ commute, then $e^{A+B} = e^A e^B$ by rearrangement of the $A$'s and $B$'s in the sum. But would the converse be true?   So far I've tried to find a counterexample by considering the cases where $A$ and $B$ are 2x2 real upper triangular matrices, but those don't work.   Any hint is appreciated, thanks.
124156
It is a well known fact that if $A,B\in M_{n\times n}(\mathbb C)$ and $AB=BA$, then $e^Ae^B=e^Be^A.$  The converse does not hold. Horn and Johnson give the following example in their Topics in Matrix Analysis (page 435). Let $$A=\begin{pmatrix}0&amp;0\\0&amp;2\pi i\end{pmatrix},\qquad B=\begin{pmatrix}0&amp;1\\0&amp;2\pi i\end{pmatrix}.$$ Then $$AB=\begin{pmatrix}0&amp;0\\0&amp;-4\pi^2\end{pmatrix}\neq\begin{pmatrix}0&amp;2\pi i\\0&amp;-4\pi^2\end{pmatrix}=BA.$$ We have $$e^A=\sum_{k=0}^{\infty}\frac 1{k!}\begin{pmatrix}0&amp;0\\0&amp;2\pi i\end{pmatrix}^k=\sum_{k=0}^{\infty}\frac 1{k!}\begin{pmatrix}0^k&amp;0\\0&amp;(2\pi i)^k\end{pmatrix}=\begin{pmatrix}e^0&amp;0\\0&amp;e^{2\pi i}\end{pmatrix}=\begin{pmatrix}1&amp;0\\0&amp;1\end{pmatrix}.$$   For $S=\begin{pmatrix}1&amp;-\frac i{2\pi}\\0&amp;1 \end{pmatrix},$ we have $$e^B=e^{SAS^{-1}}=Se^AS^{-1}=S\begin{pmatrix}1&amp;0\\0&amp;1\end{pmatrix}S^{-1}=\begin{pmatrix}1&amp;0\\0&amp;1\end{pmatrix}.$$   Therefore, $A,B$ are such non-commuting matrices that $e^Ae^B=\begin{pmatrix}1&amp;0\\0&amp;1\end{pmatrix}=e^Be^A.$  It is clear that $\pi$ is important in this particular example. In fact, the authors say what follows.      It is known that if all entries of $A,B\in M_n$ are algebraic numbers and $n\geq 2,$ then $e^A\cdot e^B=e^B\cdot e^A$ if and only if $AB=BA.$   No proof is given. How does one go about proving that?
88515
Let A be some $2x2$ matrix with real entries. Prove that $A^T$$A$ = $I$ if and only if $A$ is the rotation matrix or the reflection matrix.  My Progress: It can be shown that if $A$ is either the rotation or reflection matrix, then $A^T$$A$ = $I$ holds by matrix multiplication. Where I get suck is showing that if $A$ is a $2x2$ orthogonal matrix, then $A$ must either be the Rotation or Reflection Matrix. I suppose that since $A$ is orthogonal, it is distance preserving - and the only $2x2$ matrices that preserve distance are the Rotation and Reflection Matrices, but this isn't really a proof.  Any advice and help would be much appreciated.
47706
From Dr. Pinter's "A Book of Abstract Algebra":     Given $a$ and $b$ are in $G$ and $ab=ba$, we say that $a$ and $b$ commute.         Prove $a^2$ commutes with $b^2$      I tried:  $$ ab=ba $$ $$ a^{2}b^{2}=b^{2}a^{2} \text{ // to prove that $a^2$ commutes with $b^2$} $$ $$ aabb=bbaa $$  Then, knowing that $$a=b^{-1}ab,$$ I substituted that '$a$' into '$aabb=bbaa$'.  But that simplified to '$aabb=bbaa$', i.e. I got nowhere with it.  How can I prove that $$aabb=bbaa$$ per this problem?
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?
20240
Mariano mentioned somewhere that everyone should prove once in their life that every matrix is conjugate to its transpose.  I spent quite a bit of time on it now, and still could not prove it. At the risk of devaluing myself, might I ask someone else to show me a proof?
69606
$A$ and $B$ are $n\times n$ matrices and $A+B=AB$. I have an interesting proof that this implies $A$ commutes with $B$, but the proof only works when $||B|| \lt 1$. I'm looking for a way to salvage the proof so that it works for all $B$.  $$A=AB-B$$ $$A=(A-I)B$$ by repeatedly substituting $A$ with $(A-I)B$ on the RHS, $$A= (A-I)B^N - \sum_{i=1} ^{N-1} B^i$$  Since the space of $n\times n$ matrices is Banach (hence complete) under the standard operator norm, the limit $\sum_{i=1}^{\infty}B^i$ is well defined given $||B|| \lt 1$. Hence, taking the limit as $N \rightarrow \infty$, $$A =  -\sum_{i=1}^{\infty}B^i$$  Now the LHS commutes with B, hence so does the RHS.    Is there some trick to extend the method to work with any $B$ where $||B|| \geq 1$? I know that the result is true due to another simple algebraic proof, but I'd like to see if this method can be salvaged.
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)
122569
Show that the matrices    $$A=\begin{bmatrix} a &amp; b \\ 0 &amp; c \end{bmatrix} \text{and B = $\begin{bmatrix} d &amp; e \\ 0 &amp; f \end{bmatrix}$}$$    commute if and only if    $$\begin{vmatrix} b &amp; a-c \\ e &amp; d-f\end{vmatrix}=0.$$     My approach:  For $AB=BA$, we must have  $AB=\begin{bmatrix} ad &amp; ae+bf \\\ 0 &amp; cf\end{bmatrix}$ and $BA=\begin{bmatrix} ad &amp; bd+ce \\\ 0 &amp; cf\end{bmatrix}$ where $ae+bf=bd+ce$. We need to show that if $ae+bf=bd+ce$ then $\begin{vmatrix} b &amp; a-c \\ e &amp; d-f\end{vmatrix}=0.$   Since $\begin{vmatrix} b &amp; a-c \\ e &amp; d-f\end{vmatrix}=0$, we see that: $$b(d-f)-e(a-c)=0 \iff bd-bf-ae+ce=0 \iff bd+ce=ae+bf.$$   Thus, if  $AB=BA$ then $\begin{vmatrix} b &amp; a-c \\ e &amp; d-f\end{vmatrix}=0.$   (Showing the converse)  If $\begin{vmatrix} b &amp; a-c \\ e &amp; d-f\end{vmatrix}=0$ then we have $bd+ce=ae+bf$ and $AB=BA$ if and only if $ae+bf=bd+ce$, which we have.    Is this a proper approach am I missing something here?
66382
Prove that, if A and B is similar, then $A^T$ is similar to $B^T$.  Attempt: I try to use the defn of similarity, and then the rule to transpose in different order, but get nowhere. This should be simple...  (Note to self: page 92.)
53901
I saw this tucked away in a  and am asking this question to preserve (and advertise?) it. It's a nice problem!  Problem: Suppose $A$ and $B$ are real $n\times n$ matrices with $A^2+B^2=AB$. If $AB-BA$ is invertible, prove $n$ is a multiple of $3$.
39672
I've found this while googling some properties of  matrices. (Unfortunately, I cannot remember where I've discovered it.) If this is true, it'll greatly save my time in my work. Is it true? How can you prove it?    Let's say I have two real symmetric and positive definite matrices $\mathbf A$ and $\mathbf B$ of the same size. Also, assume that $\mathbf A - \mathbf B$ is positive semidefinite too. Then, $\mathbf B^{-1} - \mathbf A^{-1}$ is also positive semidefinite.
15155
So my friend and I are working on this and here is what we have so far.  We want to show that $\exists \, B$ s.t. $(I+A)B = I$. We considered the fact that $I - A^k = I$ for some positive $k$. Now, if $B = (I-A+A^2-A^3+ \cdots -A^{k-1})$, then $(I+A)B = I-A^k = I$. My question is: in matrix $B$, why is the sign for $A^{k-1}$ negative? Couldn't it be positive, in which case we'd get $(I+A)B = I + A^k$?  Thank you.
3188
I've come across a paper that mentions the fact that matrices commute if and only if they share a common basis of eigenvectors. Where can I find a proof of this statement?
77239
Let $a,b$ be two group elements of finite order that commute.   What can be said about the order of $ab$?   I thought that $|ab| = \text{lcm}(|a|,|b|)$. My proof was that $(ab)^n = a^n b^n =e$ if and only if $a^n = b^n = e$ if and only if $|a|,|b|$ both divide $n$. The smallest $n$ such that both orders divide it is the least common multiple of $|a|$ and $|b|$.  By chance I came across . It has an upvote so it's clearly correct. (?)  But it contradicts what I think: It's clear that   $$ (ab)^{\text{lcm}(|a|,|b|)} = e$$  hence $|ab| \mid \text{lcm}(|a|,|b|)$.  It seems to me that this is saying more than $|ab| \mid |a| |b|$. Is it not?     Now my question is: what is the precisest statement that can be made about $|ab|$?  Is there anything more than $|ab| \mid \text{lcm}(|a|,|b|)$ that can be said about $|ab|$?
46445
I see on Wikipedia that the product of two commuting symmetric positive definite matrices is also positive definite. Does the same result hold for the product of two positive semidefinite matrices?  My proof of the positive definite case falls apart for the semidefinite case because of the possibility of division by zero...
85866
Given an $m\times n$-matrix $A$ and an $n\times p$-matrix $B$. Prove that $(AB)^T = B^TA^T$.   Here is my attempt:  Write the matrices $A$ and $B$ as $A = [a_{ij}]$ and $B = [b_{ij}]$, meaning that their $\left(i,j\right)$-th entries are $a_{ij}$ and $b_{ij}$, respectively.  Let $C=AB=[c_{ij}]$, where $c_{ij} = \sum_{k=1}^n a_{ik}b_{kj}$, .  We want $(AB)^T = C^T = [c_{ji}]$. That is the element in position $j,i$ is $\sum_{k=1}^n a_{ik}b_{kj}$. For instance, if $i=2, j=3$, then the element in $2,3$ of $C$ is that sum, but the element in position $3,2$ of the transpose is that sum.  I need to get the same value for the element in position $3,2$ of the right side.  The transpose matrices are $B^T=[b_{ji}], A^T=[a_{ji}]$. They are size $p \times n$ and $n \times m$. That is, they switch rows and columns.  Let $D = B^T A^T = [d_{ji}]$. I write the indices backwards because if I want the element in position $3,2$, that is, $i=2, j=3$ just like on the other side.  So I need the summation for $d_{ji}$. But I get as $d_{ji} = \sum_{k=1}^n b_{jk}a_{ki}$, which does not match.
1769
I have two square matrices: $A$ and $B$. $A^{-1}$ is known and I want to calculate $(A+B)^{-1}$. Are there theorems that help with calculating the inverse of the sum of matrices? In general case $B^{-1}$ is not known, but if it is necessary then it can be assumed that $B^{-1}$ is also known.