I don't recall this well from linear algebra so I just need some quick confirmation. If $A$ is an $n \times n$ matrix and B is also $n \times n$ and $AB=I_n$ does it follow that also $BA=I_n$? I think no but then what do we call inverse matrix: the left inverse, or the right one, or only when the left and right are the same we talk about inverse matrix of $A$? Also, do we define inverse matrices of non-square matrices?
4880
If $A$ and $B$ are square matrices such that $AB = I$, where $I$ is the identity matrix, show that $BA = I$.    I do not understand anything more than the following.   Elementary row operations. Linear dependence. Row reduced forms and their relations with the original matrix.   If the entries of the matrix are not from a mathematical structure which supports commutativity, what can we say about this problem?  P.S.: Please avoid using the transpose and/or inverse of a matrix.
99172
The simple proof goes:  Let B be the left inverse of A, C the right inverse.  C = (BA)C = B(AC) = B  This proof relies on associativity yet does not give any insight as to why this surprising fact about matrices is true.  AC means a bunch of linear combinations of of columns of A. CA means a bunch of linear combinations of rows of A. Completely different numbers get multiplied in each case.  The proof above is just a bunch of syntactical steps not having much to do with matrices directly, I cannot see how CA=AC=I.  Can anyone shed some light on this?
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 $3$.
108668
Is it possible for a matrix $\mathbf{A}\in\mathbb{R}^{3\times3}$,  $$\mathbf{A}^2=-\mathbf{I}$$  I know that It is possible for $2\times2$ matrix, but is it possible for $3\times3$ matrix ?
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!
113708
It is well known that for invertible matrices $A,B$ of the same size we have $$(AB)^{-1}=B^{-1}A^{-1} $$ and a nice way for me to remember this is the following sentence:     The opposite of putting on socks and shoes is taking the shoes off, followed by taking the socks off.   Now, a similar law holds for the transpose, namely:  $$(AB)^T=B^TA^T $$  for matrices $A,B$ such that the product $AB$ is defined. My question is: is there any intuitive reason as to why the order of the factors is reversed in this case?   [Note that I'm aware of several proofs of this equality, and a proof is not what I'm after]  Thank you!
16539
I have two $2\times 2$ matrices, $A$ and $B$, with the same determinant. I want to know if they are similar or not.  I solved this by using a matrix called $S$: $$\left(\begin{array}{cc} a&amp; b\\ c&amp; d \end{array}\right)$$ and its inverse in terms of $a$, $b$, $c$, and $d$, then showing that there was no solution to $A = SBS^{-1}$. That worked fine, but what will I do if I have $3\times 3$ or $9\times 9$ matrices? I can't possibly make system that complex and solve it.  How can I know if any two matrices represent the "same" linear transformation with different bases?  That is, how can I find $S$ that change of basis matrix?  I tried making $A$ and $B$ into linear transformations... but without the bases for the linear transformations I had no way of comparing them.  (I have read that similar matrices will have the same eigenvalues... and the same "trace" --but my class has not studied these yet. Also, it may be the case that some matrices with the same trace and eigenvalues are not similar so this will not solve my problem.)  I have one idea. Maybe if I look at the reduced col. and row echelon forms that will tell me something about the basis for the linear transformation? I'm not really certain how this would work though? Please help.
114634
How can I prove that $\operatorname{adj}(AB) = \operatorname{adj} B \operatorname{adj} A$, if $A$ and $B$ are any two $n\times n$-matrices. Here, $\operatorname{adj} A$ means the  of the matrix $A$.   I know how to prove it for non singular matrices, but I have no idea what to do in this case.
87488
We know that in matrices AB=BA.Why in this Matrices $A=\begin{bmatrix} -1 &amp; 3\\  2 &amp; 0\end{bmatrix}$, $B=\begin{bmatrix} 1 &amp; 2\\ -3 &amp; -5\end{bmatrix}$ are not equal to $AB=BA$. WHY? This is matrix  of order $2\times 2$ for both $A$ and $B$.
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?
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.
25061
I have this problem :  $A,B \in M_n(\Bbb R)$ matrices while $n$ odd number.  Proof if $AB+BA=0$ then atleast one of $A,B$ is singular.  I assume that $A,B$ invertible.  Its clear that $\det(AB) \neq 0$ and  $\det(BA) \neq 0$.  If I could show that, $\det(AB) \neq -\det(BA)$ then I can conclude that $\det(AB)+\det(BA) \neq 0$.  But I don't seem to find a way to show that, I guess it has something to do with $n$ odd number.  Any ideas? Thanks!
108649
Let $A$ and $B$ be $m \times n$ and $n \times m$ real matrices.  I proved that if $\lambda$ is a nonzero eigenvalue of the $m \times m$ matrix $AB$ then it is also an eigenvalue of the $n \times n$ matrix $BA$. But I am having trouble finding an example showing that this need not be true if $\lambda=0$  Thank you in advance
72446
If $A$ and $B$ are $n\times n$ matrices such that $AB = BA$ (that is, $A$ and $B$ commute), show that      $$ e^{A+B}=e^A e^B$$    Note that $A$ and $B$ do NOT have to be diagonalizable.
89030
Let $A,B$ be $m\times m$ matrices such that $AB$ is invertible. Show $A,B$ invertible.   Here is what I have attempted:  We have $AB$ invertible so there exists an $m\times m$ matrix $C$ such that $C(AB)=(AB)C=I_{m}$. Since matrix multiplication is associative we immediately have a right and left inverse respectively: $(CA)B=A(BC)=I_{m}$. I know the fact that I have square matrices needs to play a roll and gaurentees me that the product will be, in general, defined. I have tried to multiply $CA$ and $BC$ on other sides but that didn't get me anywhere.   I've asked myself what would happen if they weren't square and its obvious, by definition, that they would not be invertible but I am not seeing a way to use the fact that they are to gaurentee ourselves a general inverse.
35912
Is it true that if $A$ is an $m \times n$ matrix and $B$ is an $n \times m$ matrix, with $m &gt; n $ then $\det(AB)=0$?
103141
Find all $2\times 2$ matrices that commute with  $$\left( \begin{array}{cc} 2 &amp; 3 \\ 1 &amp; 4 \end{array} \right)$$  My progress:  I know that a square matrix commutes with itself, the identity matrix of that order, the null matrix of that order and any scalar matrix of that order.  The answer has been given as:   $$\left( \begin{array}{cc} m &amp; 3n \\ n &amp; m+2n \end{array} \right)$$  I don't understand how they're getting that form. Can someone please explain?