I know that the sum of the eigenvalues will be the same. But can we say something about them individually?  When A is transpose(B) I'm getting something interesting  A = (1  2  0)     (0 -1  2)  B =  (1  0)      (2 -1)      (0  2)  AB has 7,3 as eigenvalues BA has 3,7,0 as eigenvalues
16874
Let $ A, B $ be two square matrices of order $n$. Do $ AB $ and $ BA $ have same minimal and characteristic polynomials?   I have a proof only if $ A$ or $ B $ is invertible. Is it true for all cases?
85866
Given an $m\times n$-matrix $A$ and an $n\times p$-matrix $B$. Prove that $(AB)^T = B^TA^T$.   Here is my attempt:  Write the matrices $A$ and $B$ as $A = [a_{ij}]$ and $B = [b_{ij}]$, meaning that their $\left(i,j\right)$-th entries are $a_{ij}$ and $b_{ij}$, respectively.  Let $C=AB=[c_{ij}]$, where $c_{ij} = \sum_{k=1}^n a_{ik}b_{kj}$, .  We want $(AB)^T = C^T = [c_{ji}]$. That is the element in position $j,i$ is $\sum_{k=1}^n a_{ik}b_{kj}$. For instance, if $i=2, j=3$, then the element in $2,3$ of $C$ is that sum, but the element in position $3,2$ of the transpose is that sum.  I need to get the same value for the element in position $3,2$ of the right side.  The transpose matrices are $B^T=[b_{ji}], A^T=[a_{ji}]$. They are size $p \times n$ and $n \times m$. That is, they switch rows and columns.  Let $D = B^T A^T = [d_{ji}]$. I write the indices backwards because if I want the element in position $3,2$, that is, $i=2, j=3$ just like on the other side.  So I need the summation for $d_{ji}$. But I get as $d_{ji} = \sum_{k=1}^n b_{jk}a_{ki}$, which does not match.
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.
94529
I have a question on determining eigenvalues for a given matrix A:  $$ A=  \begin{bmatrix}  2 &amp; 1 &amp; 2 \\  0 &amp; 2 &amp; -1 \\  0 &amp; 1 &amp; 0 \\  \end{bmatrix} $$  Here's what I have so far:  \begin{align} &amp; det(A-\lambda I_3) = 0\\ &amp; =det \begin{bmatrix}  2-\lambda &amp; 1 &amp; 2 \\  0 &amp; 2-\lambda &amp; -1 \\  0 &amp; 1 &amp; 0-\lambda \\  \end{bmatrix} = 0 \\ &amp; = (2-\lambda) \begin{vmatrix}  2-\lambda &amp; -1 \\  1 &amp; 0-\lambda \\  \end{vmatrix} \\ &amp; = (2-\lambda)(-2\lambda+\lambda^2+1) \\ &amp; = (2-\lambda)(\lambda-1)^2 = 0\end{align}  So my eigenvalues are $2$ and $1$. I am almost postitive this is correct. Here's my confusion/question. I know that the determinant of an upper triangular matrix is the product of the terms along the diagonal.  Some of my classmates used a different approach where from the beginning they row reduced $A$ into an upper triangular matrix by eliminating the $1$ from the $3$rd row, then subtracted $\lambda I$ and then instead of solving the determinant like I did they just multiplied the diagonal terms together.  The problem I see with this is that you can obtain different matrices depending on how you choose to row reduce, which means you can get different eigenvalues. Is it possible to solve for the eigenvalues by using the approach of making $A$ into an upper triangular matrix first? (So that you can solve the determinant by multiplying the diagonal entries) I hope that makes sense. Thanks!
89778
If $A$ and $B$ are two matrices of the same order $n$, then   $$ \operatorname{rank} A + \operatorname{rank}B \leq \operatorname{rank} AB + n. $$   I don't know how to start proving this inequality. I would be very pleased if someone helps me. Thanks!  Edit I.  Rank of $A$ is the same of the equivalent matrix $A' =\begin{pmatrix}I_r &amp; 0 \\ 0 &amp; 0\end{pmatrix}$. Analogously for $B$, ranks of $A$ and $B$ are $r,s\leq n$. Hence, since $\operatorname{rank}AB = \min\{r,s\}$, then $r+s\leq \min\{r,s\} + n$. (This is not correct since $\operatorname{rank} AB \leq \min\{r,s\}$.  Edit II. A discussion on the rank of a product of $H_f(A)$ and $H_c(B)$ would correct this, but I don't know how to formalize that $\operatorname{rank}H_f(A) +\operatorname{rank}H_c(B) - n \leq \operatorname{rank}[H_f(A)H_c(B)]$.
25668
I can't seem to find the answer to this using Google.   Is the transpose of the inverse of a square matrix the same as the inverse of the transpose of that same matrix?
124156
It is a well known fact that if $A,B\in M_{n\times n}(\mathbb C)$ and $AB=BA$, then $e^Ae^B=e^Be^A.$  The converse does not hold. Horn and Johnson give the following example in their Topics in Matrix Analysis (page 435). Let $$A=\begin{pmatrix}0&amp;0\\0&amp;2\pi i\end{pmatrix},\qquad B=\begin{pmatrix}0&amp;1\\0&amp;2\pi i\end{pmatrix}.$$ Then $$AB=\begin{pmatrix}0&amp;0\\0&amp;-4\pi^2\end{pmatrix}\neq\begin{pmatrix}0&amp;2\pi i\\0&amp;-4\pi^2\end{pmatrix}=BA.$$ We have $$e^A=\sum_{k=0}^{\infty}\frac 1{k!}\begin{pmatrix}0&amp;0\\0&amp;2\pi i\end{pmatrix}^k=\sum_{k=0}^{\infty}\frac 1{k!}\begin{pmatrix}0^k&amp;0\\0&amp;(2\pi i)^k\end{pmatrix}=\begin{pmatrix}e^0&amp;0\\0&amp;e^{2\pi i}\end{pmatrix}=\begin{pmatrix}1&amp;0\\0&amp;1\end{pmatrix}.$$   For $S=\begin{pmatrix}1&amp;-\frac i{2\pi}\\0&amp;1 \end{pmatrix},$ we have $$e^B=e^{SAS^{-1}}=Se^AS^{-1}=S\begin{pmatrix}1&amp;0\\0&amp;1\end{pmatrix}S^{-1}=\begin{pmatrix}1&amp;0\\0&amp;1\end{pmatrix}.$$   Therefore, $A,B$ are such non-commuting matrices that $e^Ae^B=\begin{pmatrix}1&amp;0\\0&amp;1\end{pmatrix}=e^Be^A.$  It is clear that $\pi$ is important in this particular example. In fact, the authors say what follows.      It is known that if all entries of $A,B\in M_n$ are algebraic numbers and $n\geq 2,$ then $e^A\cdot e^B=e^B\cdot e^A$ if and only if $AB=BA.$   No proof is given. How does one go about proving that?
86559
I would be very grateful if someone can help me with the following assignment:  I'm given an $n\times n$ matrix $A$.  If $A \neq I,0$ and $A = A^2$, I need to prove that $\lambda=0$ and $\lambda=1$ are $A$'s eigenvalues and that they are $A$'s only eigenvalues.
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
8100
If I have a $2\times 2$ or $3\times 3$ matrix, how should I go about computing the eigenvalues and eigenvectors of the matrix?  NB: I am making this question to provide a unified answer to questions about eigenvalues of small matrices so that all of the specific examples that come up can be marked as duplicates of this post. See .
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 $B$.
78808
Let matrix $A_{n\times n}$, be such that there exists a matrix $B$ for which $$AB-BA=A$$  Prove or disprove that there exists $m\in \mathbb N^{+}$such  $$A^m=0,$$  I know $$tr(A)=tr(AB)-tr(BA)=0$$ then I can't.Thank you
107009
I am trying to find out whether $A = \begin{bmatrix}     3 &amp; 1 \\     0 &amp; 2 \\ \end{bmatrix}$ and $B = \begin{bmatrix}     4 &amp; 1 \\     -2 &amp; 1 \\ \end{bmatrix}$ are similar. Both matrices have the same trace, same rank, same determinant and the same eigenvalues. So I think they are similar. The eigenvalues are $3, 2$. Now I'm not sure how to find if they are indeed similar.   I am trying to find if there is a basis for $\mathbb{R^2}$ that contains only eigenvectors but when trying to solve $(3 I - A)v = 0$ there is no solution... What should I do from here?
30963
Let $A$ be an $n \times n$ matrix.  $i)$Prove that if the sum of each row of $A$ equals $s$, then $s$ is an eigenvalue of $A$.  $ii)$Prove that if the sum of each column of $A$ equals $s$, then $s$ is an eigenvalue of $A$.  I think that being an eigenvalue of $A$ implies that $sv=Av$ for some vector $v$. Furthermore, I know that $[a_i] = s$ if we let $a_i$ denote the i-th row of $A$. However, I do not seem to be able to find a link between these two facts. Could anyone please help me out?
90042
I have been thinking about this problem:  Suppose we have all $R$ by $C$ matrices, where the values are integers in $[1,n]$. Two matrices are equivalent under interchanging rows and columns. For example,   1 5     0 0       would be equal to itself and to  0 0 1 5   and would be equal to  0 0  5 1   and would be equal to   5 1 0 0   How many unique such matrices are there?  Any ideas how to go about it? Any help much appreciated!
78573
Well considering two $n \times n$ matrices does the following hold true: $$\det(A+B) = \det(A) + \det(B)$$ Can there be said anything about $\det(A+B)$?  If $A/B$ are symmetric (or maybe even of the form $\lambda I$) - can then things be said?
89858
I have $3\times3$ matrices such that S=A+B. I know there is an inequality connecting rank of the matrices A, B and its sum S. Could you write down that here? It will be a great help for me.  Means equation  or inequality connecting rank(S), rank(A) and rank(B).