Let $A$ and $B$ be two invertible $n\times n$ real matrices. Assume that $A+B$ is invertible. Show that $A^{-1} + B^{-1}$ is also invertible.
122746
If $A$ and $B$ be two invertible $n \times n$ real matrices and $A + B$ is invertible, how to prove that $A^{-1} + B^{-1}$ is also invertible?
40437
The problem is from  Algebra by Artin, Chapter 1, Miscellaneous Problems, Exercise 8. I have been trying for a long time now to solve it but I am unsuccessful.      Let $A,B$ be $m \times n$ and $n \times m$ matrices. Prove that $I_m - AB$ is invertible if and only if $I_n - BA$ is invertible.    Please provide with only a hint.
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 $B$.
48342
Let $A$ be a squared matrix, and suppose there exists an $n\in \Bbb N$ in a way that $A^n=0$. Show that $I-A$ is invertible and that $(I-A)^{-1}=I+A+\cdots+A^{n-1}$    I don't have a clue where to start from.
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?
16715
Let $A\in M_{n\times n}(\mathbb{R})$ symmetric and $b\in \mathbb{R}^n$.  How I can prove that if $b\perp \ker A$ then there exists $x\in \mathbb{R}^n$ such that $Ax=b$?  Many thanks!
114526
Let $n$ be a positive integer. Show that the matrix  $$\begin{pmatrix} 1 &amp; 1/2 &amp; 1/3 &amp; \cdots &amp; 1/n \\ 1/2 &amp; 1/3 &amp; 1/4 &amp; \cdots &amp; 1/(n+1) \\ \vdots &amp; \vdots &amp;  \vdots &amp; \ddots &amp; \vdots\\ 1/n &amp; 1/(n+1) &amp; 1/(n+2) &amp; \cdots &amp; 1/(2n-1) \end{pmatrix}$$  is invertible and all the entries of its inverse are integers. This is an exercise in Hoffman and Kunze's linear algebra book. Any hints will be appreciated!
26512
Let $A$ be an $n \times n$ matrix. Prove that if A is invertible, then there exists a polynomial $p$, such that $A^{-1}=p(A)$  Thus far:  Let $W$ denote the $k$ dimensional A-cyclic subspace spanned by a vector $v$.  Then, $I_n=\sum_{i=0}^{k} a_{i}A^i$ for some scalar $a_i$.  What I want to do is multiply both sides by $A^{-1}$, so that way it just falls into my hands, but Im not sure that is 'legal'.
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.
11622
Given a ring $R$ (with unity), prove that if $a \in R$ is nilpotent, then $a+1,a-1$ are both invertible.   Suppose $a^n=0$. Then $1=1-a^n=(1-a)(1+a+\ldots+a^{n-1})$, so $1-a$ is invertible.  If $n$ is odd we can use $1+a^n=(1+a)(1-a+\ldots+a^{n-1})$.  But what if $n$ is even?
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.
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!
22526
Let $A=[a_{ij}]_{n×n}$ be the matrix defined by letting $a_{ij}$ be the rational number such that $$a_{ij}=\gcd(i,j ).$$ How prove that $A$ is invertible, and compute $\det(A)$? thanks in advance
97671
Let $A$ be a real square matrix en let $I$ be the identity matrix with the same size. Let $n&gt;0$ be the smallest natural number for which there exist  $a_0,a_1,...a_{n-1} \in {R}$ such that $A^n+a_{n-1}A^{n-1}+....+a_1A+a_0I=O$.  Show that A is invertible if and only if $a_0\neq0 $.  So i need to proof    $A$ invertible $\rightarrow$$a_0 \neq 0$ $a_0 \neq0 \rightarrow$ $A$ invertible    I don't really know where to start with 1, and with 2 I tried to prove it with a contradiction but I don't think it's right either.
68316
Let $A, B$ matrices $n\times m$ and $m\times n$ respectively so that $AB=I_n$ and $BA=I_m$. Show that $n=m$.
1337
I can't find out if the product of two invertible matrix is an invertible matrix or if the sum of two invertible matrix is an invertible matrix.  Can anyone suggest an answer/counterexample?  Thanks a lot in advice
90841
How do you prove the following fact about the transpose of a product of matrices? Also can you give some intuition as to why it is so.  $(AB)^T = B^TA^T$