My approach, $A$ AND $B$ are $n \times n$ matrices such that, $BA=I$ Now, We can write, $A = AI$ $A= A (BA)$ $A= (AB)A$ $A(I-AB)=0$ $I-AB =0$ Hence, $AB=I$ I don't feel like what I'm doing is the correct way to go. Please help ! Thanks in advance.
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.
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.
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.
31703
If $A$ and $B$ are $n\times n$ matrices, and $AB$ is invertible then $A$ and $B$ are invertible.    I started out by writing that since $AB$ is invertible, then for the equation $ABx=b$ has a unique solution for any $b$, $x=(AB)^{-1}b$.  Then I don't know how to break up the composition of AB to prove they are each individually similar.
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!
101417
Can anyone help me to understand what is suppose to do in this question?  "A matrix that is similar to the identity matrix" I should say something about this but I do not understand what is meant to do.  I know that two n x n matrices A and B are similar if $B=P^{-1}AP $.   They also have some properties such as:  The same:  Rank; Characteristic equation;  Determinant; Trace; Eigenvalues etc.   However I do not understand the question and consequently I do not know how to start.  Can anyone help me on this?  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 $L_A L_B$
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.
57513
Let $A∈\mathscr{M}_{3×2}(\mathbb{R})$ and $B∈\mathscr{M}_{2\times3}(\mathbb{R})$ be matrices satisfying   $AB =\begin{bmatrix} 8 &amp;2 &amp;−2\\ 2 &amp;5 &amp;4\\ −2 &amp;4&amp; 5 \end{bmatrix}$. Calculate $BA$. (Golan, The Linear Algebra a Beginning Graduate Student Ought to Know, Exercise 426.)   Maybe it can be solved by solving a system of equations, but I think there is a shorter way since this problem was in my exam.  Thanks.
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?
49764
The Problem Let $A$ be the matrix $\bigl(\begin{smallmatrix}a&amp;b\\c&amp;d\end{smallmatrix} \bigr)$, where no one of $a,b,c,d$ is $0$. Let $B$ be a $2\times 2$ matrix such that $AB+BA=\bigl(\begin{smallmatrix} 0&amp;0\\ 0&amp;0 \end{smallmatrix} \bigr)$. Show that either   $a+d=0$, in which case the general solution for $B$ depends on 2 parameters, or $ad-bc=0$, in which case the general solution for $B$ depends on one parameter.   (this is question 22 of the last matrix exercise of Further Pure Mathematics by Bostock et al.)  Comments Writing $B=\bigl(\begin{smallmatrix} e&amp;f\\ g&amp;h \end{smallmatrix} \bigr)$ and multiplying out I get that    $(a+d)(f+g)+(b+c)(e+h)=0$ $ae+bg+cf+dh=0$   but I am unable to get the required restrictions on $a,b,c,d$. Is there a quick way of doing the problem that doesn't require manual computation? I thought of considering invertible and non-invertible cases but couldn't get anywhere. Help would be much appreciated.
39380
$A$ is a $n\times n$ matrix such that $ A^m = 0 $ for some positive integer $m$.   Show that $A^n = 0$.   My attempt: For $n &gt; m$, it's obvious since matrix multiplication is associative.  For $n &lt; m$, $A^n\times A^{m-n} = 0$; not sure what to do next. Also I know that $\det A = 0$.
42734
Let $A$ be an $n \times n$ matrix and let $\Lambda$ be an $n \times n$ diagonal matrix. Is it always the case that $A\Lambda = \Lambda A$? If not, when is it the case that $A \Lambda = \Lambda A$?  If we restrict the diagonal entries of $\Lambda$ to being the equal (i.e. $\Lambda = \text{drag}(a, a, \dots, a)$), then it is clear that $A\Lambda = AaI = aIA = \Lambda A$. However, I can't seem to come up with an argument for the general case.
103841
What is the determinant of an $n \times n$ matrix $B=A+I$, where $I$ is the $n\times n$ identity matrix and $A$ is the $n\times n$ matrix  $$ A=\begin{pmatrix} a_1 &amp; a_2 &amp; \dots &amp; a_n \\ a_1 &amp; a_2 &amp; \dots &amp; a_n \\ \vdots &amp; \vdots &amp; \ddots &amp; \vdots \\ a_1 &amp; a_2 &amp; \dots &amp; a_n \\ \end{pmatrix} $$
76818
Let $A$ and $B$ be $n\times n$ matrices. Suppose $A$ is similar to $B$. Prove $\operatorname{trace}(A) = \operatorname{trace}(B)$.  I'm not sure where to go on this. So far I have this:  If $A$ is similar to $B$, then      $B=P^{-1}AP$ and $A=PBP^{-1}$   This implies that:     $\operatorname{trace}(B) = \operatorname{trace}(P^{-1}AP)$ and $\operatorname{trace}(A) = \operatorname{trace}(PBP^{-1})$   Not sure where to go from here
106041
$$\begin{cases} &amp;x  &amp;- &amp;y &amp;+ &amp;2z &amp;= 4 \\ &amp;3x &amp;- &amp;2y &amp;+ &amp;9z &amp;= 14 \\ &amp;2x &amp;- &amp;4y &amp;+ &amp;az &amp;= b \end{cases} $$ I know that $a$ and $b$ has to either equal to something or not in order to satisfy the $4$ conditions stated above.   my matrices looked like $$         \begin{bmatrix}         1 &amp; -1 &amp; 2 \\         0 &amp; 1 &amp; 3 \\         0 &amp; 0 &amp; a+12 \\         \end{bmatrix} \begin{bmatrix}         x &amp; \\         y &amp;\\         z &amp;  \\         \end{bmatrix}= \begin{bmatrix}         6 &amp;  \\         2 &amp; \\         b+8 &amp; \\         \end{bmatrix} $$