The complete definition of inverse of a matrix is- if $AB=BA=I$, then $B$ is inverse of $A$.  But l want to  know that from $AB=I$, can we comment $B$ is inverse of $A$? Is it possible that $AB=I$ but $BA$ is not equal to $I$?   [ I personally think it is impossible since the inverse of a matrix is unique. Please clarify if I am wrong.]
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.
108442
Let A be an invertible $n \times n$ matrix whose inverse is itself. Prove that $\det(A)$ is either $1$ or $-1$.    I'm really lost in class. I don't even know where to start. Please help.
107241
Let $A$ and $B$ are $n\times n$ matrices and $x \in C^{n}$.  If $Ax = Bx$ for all $x$ then $A = B$. To prove this I have selected  $x$ from Euclidean basis B = {$e_{1},e_{2},...,e_{n}$}. Then $Ae_{i} = Be_{i}$ implies $i^{th}$ column of A = $i^{th}$ column of B for all $1 \leq i \leq n$. Hence $A = B.$    Is this proof complete?
15525
Let $a,b,c,x$ be elements of a unital non-commutative ring. Assume $c$ is an inverse of $1-ab$: $$ c(1-ab) = 1$$  How can I find an inverse for $1-ba$? What I tried:   Denote the unknown by $x$. Then $x (1-ba) = 1 = x - xba$. I tried to replace $1$ with $c(1-ab)$ but it didn't help because I cannot solve for $x$. I also tried to subtract $x (1-ba) $ from $ c(1-ab)$ but couldn't solve for $x$ either. Any suggestions?
78948
I have shown that if two $n\times n$ matrices $A$ and $B$ are invertible, then $AB$ is also invertible. I am having problems though with showing that $(AB)^{-1}=B^{-1} A^{-1}$. I thought of using their left-multiplication transformations but it seems that this is only associative and not commutative so I can reverse $L_A L_B$
18487
Let  $A$ and $B$  be matrices over $\mathbb C$. Then,          $AB$  and $BA$   always have the same set of eigenvalues.      If  $AB$  and  $BA$  have the same set of eigenvalues then $AB=BA$.           If  $A^{-1}$   exists then  $AB$  and $BA$   are similar.     The rank of  $AB$  is always the same as the  rank of   $BA$ .       Suppose $AB=BA$ Let $x$ be the eigen vector of $A$ corresponding to the eigenvalue $a$. $$ABx=BAx=aBx \implies Bx$$ is the eigen vector of $A$. If the eigen space corresponding to the eigen values of $A$ is one. Then, $Bx=\lambda x \implies x$ is the eigen vector of $B$. So $AB$ and $BA$ have same set of eigen values. statement is false. Am I correct? I don't know, How to judge the statement. I don't know, How to judge the statement. Statement is false, I could obtain the counter examples.   Please check my answers. Please help me.
16289
Show that if $A$, $B$, and $A + B$ are invertible matrices with the same size, then   $A(A^{-1}+B^{-1})B(A+B)^{-1}=I$      What does the result in the first part tell you about the matrix $(A^{-1}+B^{-1})$?   I get the first part. Help me with the second part.
24346
If matrix $A$ is an $n\times n$ matrix such that $A^2 -A -2I=0$.  How can I show that $A$ is invertible and that $A^{-1} = \frac12(A-I)$? Also, how do I show that one of the eigenvalues of $A$ is $2$ or $ -1$? I appreciate any help on this question.
63150
$A$ is defined as an $m\times m$ matrix which is not invertible. How can i show that there is an $m\times m$ matrix $B$ where $AB = 0$ but $B$ is not equal to $0$?   For the solution of this question I think giving an example is not enough because it is too easy to solve this by giving an example, so how can I show that $B$ is not the $0$ matrix?
21751
If $A$ and $B$ are $3\times 3$ matrices and $A$ is invertible, then can we say that there exists an integer $n$ such that $A+nB$ invertible?  I was trying to show this by choosing $n$ such that eignevalues of $A+nB$ are non-zero. In the case where $B = I$ we can find the eigenvalues of $A+nB$ that would be $\lambda + nB$ (though I am not certain about its proof). This choosing of $n$ such that $\lambda$ is not equal to $-n$ times an eigenvalue of $B$ will serve the purpose. But I am not sure about general $B$. What if I take arbitrary matrices $A$ and $B$.
6236
Let A = $\begin{bmatrix}1 &amp; 1 &amp; 4\\0 &amp; 3 &amp; -4\\0&amp;0&amp;-1\end{bmatrix}$. Is the matrix A diagonalizable?  If so find a matrix P that diagonalizes A. Can you write A as a linear combination of rank 1 matrices formed from its eigenvectors? Determine the eigendecomposition $A = PΛP^{-1}$. A = $\begin{bmatrix}3 &amp; 1 &amp; 0\\1 &amp; 2 &amp;1\\0&amp;1&amp;3\end{bmatrix}$. Is the matrix A diagonalizable?  If so find a matrix P that orthogonally diagonalizes A. Can you write A as a linear combination of rank 1 matrices formed from its eigenvectors? (Note that A is real symmetric so that you do not have to compute the inverse of P).   My solution attempts:   det(λI - A) = $\begin{vmatrix}λ - 1 &amp; -1 &amp; -4\\0 &amp; λ - 3 &amp;4\\0&amp;0&amp;λ + 1\end{vmatrix}$ = (λ -1) (λ -3) (λ + 1) = 0   $λ_{1}$ = 1, $λ_{2}$ = 3, $λ_{3}$ = -1  ($λ_{1}I - A)x_{1}$ = 0  $\begin{bmatrix}0 &amp; -1 &amp; -4\\0 &amp; -2 &amp;4\\0&amp;0&amp;2\end{bmatrix}$ $\begin{bmatrix}x_{11}\\x_{12}\\x_{13}\end{bmatrix}$ = $\begin{bmatrix}0\\0 \\0\end{bmatrix}$   $\begin{bmatrix}x_{11}\\x_{12}\\x_{13}\end{bmatrix}$ = $x_{11} \begin{bmatrix}1\\0 \\0\end{bmatrix}$  It follows that $\begin{bmatrix}1\\0 \\0\end{bmatrix}$ is a basis for the eigenspace of A corresponding to $λ_{1}$ = 1.  In a similar way, I found other two basis vectors as $\begin{bmatrix}1/2\\1 \\0\end{bmatrix}$ and $\begin{bmatrix}-2/5\\1 \\1\end{bmatrix}$, and I got $P = \begin{bmatrix}1 &amp; 1/2 &amp; -2/5\\0 &amp; 1 &amp; 1\\0&amp;0&amp;1\end{bmatrix}$. Then my $P^{-1} = \begin{bmatrix}1 &amp; -1/2 &amp; 9/10\\0 &amp; 1 &amp; -1\\0&amp;0&amp;1\end{bmatrix}$. And I said A's left-multiplication by $P^{-1}$ and right-multiplication by $P (P^{-1}AP)$ gives the diagonal matrix Λ that is similar to A. I found Λ as $\begin{bmatrix}1 &amp; 0 &amp; 0\\0 &amp; 3 &amp; 0\\0&amp;0&amp;21/5\end{bmatrix}$. Then I said that if we multiply both sides of the equation $P^{-1}AP = Λ$ by $P$ from the left and by $P^{-1}$ from the right we get $PP^{-1}APP^{-1} = PΛP^{-1} = A$ and that this is the eigendecomposition of A. Here is my first question: Do you think I understood and did correctly what the asker wanted me to do by saying "Determine the eigendecomposition $A = PΛP^{-1}$."?   I started in the same manner and I got three eigenvectors as  $\begin{bmatrix}x_{11}\\x_{12}\\x_{13}\end{bmatrix}$ = $x_{12} \begin{bmatrix}0\\1 \\0\end{bmatrix}$,  $\begin{bmatrix}x_{21}\\x_{22}\\x_{23}\end{bmatrix}$ = $x_{21} \begin{bmatrix}1\\1 \\1\end{bmatrix}$ and  $\begin{bmatrix}x_{31}\\x_{32}\\x_{33}\end{bmatrix}$ = $x_{33} \begin{bmatrix}1\\-2 \\1\end{bmatrix}$.   I said A is not diagonalizable since the eigenvectors are not linearly independent.  Here is my second question: What the hell does the sentence "Can you write A as a linear combination of rank 1 matrices formed from its eigenvectors?" mean?
83554
I'm really struggling with "simple" proofs, and I could only come up with this. I googled for the solution and know how to do it now, but I first want to know if there's anything wrong with my solution? $$ (AB)^{-1}=B^{-1}*A^{-1}. $$ For this equation to hold, $B$ must be invertible and so must $A$. //
61648
Given matrices $A$ and $B$, where that $AB = A + B$, prove $AB = BA$. I keep coming up with AB = AB. It seems like basic algebra, but for the life of me, I'm getting nowhere :/. Someone help please?
104732
This is a problem I've been stuck on for a while. I know that this statement is false if it had $A$ instead of $A^2$, for example, the matrix $A = \begin{bmatrix} 0&amp;1\\0&amp;0 \end{bmatrix}$ can't be written as $X^3$, but since $A^2$ equals $ \begin{bmatrix} 0&amp;0\\0&amp;0 \end{bmatrix}$, it can be written as $X^3$ with $X$ being$ \begin{bmatrix} 0&amp;0\\0&amp;0 \end{bmatrix}$ for example. Other than this, I have no insight into why this statement could be true: nothing can be assumed about diagonalizability, invertibility or anything else. I might be overlooking something simple. Thanks in advance.
99050
Let $A$ and $B$ be real $n\times n$. Is it true that if $ABA=A$ is true $B=BAB$ ?  My reasoning: $$ABA=(ABA)BA=A  \; \Leftrightarrow A(B-BAB)A=0.$$  But as it follows from the latter, that $B-BAB=0$, if in the general case the matrix $A$ is not invertible?
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.
68313
Given a matrix $A$ over a field $F$, does there always exist a matrix $B$ such that $AB = BA$? (except the trivial case and the polynomial ring?)