I need to prove or give a counterexample:     If matrix $A$ is unitary and $B^2 = A$ then $B$ is also unitary   I think the statement is true since the unitary matrix A can only be Identity matrix I or negative identity matrix $-I$; and $B=A^2$ is an identity matrix which makes sure it is unitary.
123979
If matrix $A$ is unitary and the matrix $B$ satisfies $B^2=A$, is $B$ necessarily unitary?
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.
27490
The below statement is a true/false exercise.  Statement: For all square matrices A and B of the same size, it is true that $(A + B)2 = A^2 + 2AB + B^2$.  My thought process: Since it is not a proof, I figure I can show by example and come to a valid conclusion based on such example.  My work:  Come up with a square matrix A and B let both be a 2 by 2 matrix(rows and cols must be same).  Matrix $A$:  $A = \begin{array}{ccc} 3 &amp; 5 \\ 4 &amp; 6 \\ \end{array} $  Matrix $B$:  $B =  \begin{array}{ccc} 5 &amp; 8 \\ 9 &amp; 4 \\ \end{array} $  $A + B =  \begin{array}{ccc} 8 &amp; 13 \\ 13 &amp; 10 \\ \end{array}$  $(A + B)^2 =  \begin{array}{ccc} 233 &amp; 234 \\ 234 &amp; 264 \\ \end{array}$  $A^2 =  \begin{array}{ccc} 29 &amp; 45 \\ 36 &amp; 56 \\ \end{array}$  $(AB) =  \begin{array}{ccc} 60 &amp; 44 \\ 74 &amp; 56 \\ \end{array}$  $2(AB) =  \begin{array}{ccc} 120 &amp; 88 \\ 234 &amp; 112 \\ \end{array}$  $B^2 =  \begin{array}{ccc} 97 &amp; 72 \\ 81 &amp; 88 \\ \end{array}$  $A^2 + 2AB + B^2 =  \begin{array}{ccc} 246 &amp; 205 \\ 265 &amp; 256 \\ \end{array}$  Based my above work, the answer is false.   Is there another way to approach the problem? It seems like a lot of work needed to be done for a true/false question which raised my suspicion about whether there is a better way to look at the problem.
2575
This is a proof question and I am not sure how to prove it. It is obviously true if you start with $A = 0$ and square it.   I was thinking:  If $ A^2 = 0 $  then $ A A = 0 $  $ A A A^{-1} = 0 A^{-1}$  $I\,A = 0 $   but the zero matrix is not invertible and that it was not among the given conditions.  Where's a good place to start?
49815
If $A$ and $B$ two $n\times n$ matrix such that $AB=BA$, then my guess is $\dim(A\cap B)\geq \dim AB$. But I am unable to prove this. Any help in this direction is highly appreciated.  Thanks.
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 $L_A L_B$
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
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.
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.
113463
Why does $a^{b}·a^{c}=a^{b+c}$ ?  I want proof for it, I asked professor and he replied: "I don't know, it's a property anyway that is true and that's all what you need to  know."
91915
Suppose that $A$ and $B$ are two real square matrices and $A^TA=B^TB$. Can we say that $A=QB$ for some orthogonal matrix $Q$?  If they are vectors we have $\|a\|^2=a^Ta=b^Tb=\|b\|^2$, so intuitively clear, since we just have to rotate. But it is hard to picture the matrix case but I have not been able to show.
77211
If $A, B$ and $A + B$ are all $n × n$ invertible matrices. Prove that $A^{−1} + B^{−1}$ is invertible and the inverse is $A(A + B) ^{−1}B$.  I am afraid I am really stuck on this one, and I haven't really tried much because I don't know what to try.  Thanks for all the help guys, I understand now.
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?
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.
76200
a.) Let A and B be $n$ x $n$ matrices. Prove that the matrix products   $AB$  and $BA$ have the same eigenvalues.      b.) Prove that every eigenvalue of a matrix A is also an eigenvalue of   its transpose $A^T$. Also, prove that if v is an eigenvector of A with   eigenvalue $\lambda$ and w is an eigenvector of $A^T$ with a different   eigenvalue $\mu \ne \lambda$, then v and w are orthogonal vectors with   respect to the dot product.   For a, I know that if their eigenvalues are the same then their eigenvectors must relate too.    For the first part of b, is it similar to proving that the $det(A) =det (A^T)$? And, I do not know how to do the second part.
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.
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.