Let $A,B$ matrix on $\mathbb{R}$ size $nxn$. How can I prove that $det(xI - AB) = det(xI - BA)$ if $A$ and $B$ are singular matrix
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?
62412
Let $A$ be a $n \times n$ real matrix such that $$ A + A^t = 2I, $$ where $I$ is the $n \times n $ identity matrix. Prove that $\det(A) \geq 1$.  It is obvious that tr$(A) = n$. Furthermore, we also have that $$A - 2I = -A^t, $$ so we get that $$\det(A-2I) = (-1)^n \cdot \det(A).$$ Now, we let $\lambda_1, \lambda_2, \cdots, \lambda_n \in \mathbb{C}$ be the  eigenvalues of $A$, so we get that $$(\lambda_1 - 2)(\lambda_2 - 2) \cdots (\lambda_n - 2) = (-1)^n\lambda_1\lambda_2 \cdots \lambda_n, $$ but I couldn't derive anything about the product $\lambda_1\lambda_2 \cdots \lambda_n = \det(A)$ (the only known thing is $\lambda_1 + \cdots + \lambda_n = n$). Also, I tried the same approach for $2 \times 2$ and $3 \times 3 $ matrices, but it didn't lead to anything (especially since the eigenvalues can be complex numbers).
108128
For any matrix $A_{n\times n}$ with $\mathrm{tr}(A)=0$ show that there exist two matrices $B$ and $C$ such that $$A=BC-CB.$$   I know to prove this: if $A=BC-CB$, then we have $\mathrm{tr}(A)=0$ because $$\mathrm{tr}(BC)=\mathrm{tr}(CB)$$ so $$\mathrm{tr}(A)=\mathrm{tr}(BC-CB)=\mathrm{tr}(BC)-\mathrm{tr}(CB)=0.$$ But my problem is that I can't prove it.
21358
I have a qualifying exam coming up in a couple days and I am just trying to understand some pathological examples I have in my notes. I will list a similar problem which I know the solution to and then the question.  True or False:   let $A,B \in \mathbb{Q}^{n \times n }$  Suppose $xI-A$ and $xI-B$ are equivalent then $\det(xI-A) = \det(xI-B)$.     I wrote in my notes this is true because any two matrices are equivalent in $\mathbb{Q[x]}^{n \times n}$ if and only if they have the same invariant factors.  Since the characteristic polynomial is a product of the invariant factors it follows that $\det(xI-A) = \det(xI-B)$.        True or False:   let $A,B \in \mathbb{Q}^{n \times n }$. If $\det(xI-A) = \det(xI-B)$ (in $\mathbb{Q}[x]$) then $xI-A$ and $xI-B$ are equivalent  in $\mathbb{Q}[x]^{n \times n}$   I think this is false because of the fact that equality of determinants is not strong enough to guarantee all invariant factors are equal but I do not have an example.
75546
Assume that the matrices $A,\: B\in \mathbb{R}^{n\times n}$ satisfy  $$ A^k=0,\,\, \text{for some $\,k\in \mathbb{Z^+}$}\quad\text{and}\quad AB=BA. $$  Prove that $$\det(A+B)=\det B.$$
25677
I have to prove/disprove this:     If $\det(A+X) = \det(B + X)~ \forall X \in M_{n \times n} (\mathbb F) \rightarrow A = B$   I believe it is true but I can not think of a direct way to prove it. Any help is 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?
117739
Let $A$ be an $n\times n$ diagonal matrix with characteristic polynomial $(x-c_1)^{d_1}\cdots(x-c_k)^{d_k}$, where $c_1,\ldots,c_k$ are distinct. Let $V$ be the space of $n\times n$ matrices $B$ such that $AB=BA$. Prove that the dimension of $V$ is $d_1^2+\cdots+d_k^2$.
73332
Let $A\in M_{n\times n}(\Bbb R)$ so that $I\notin span(A,A^2,...,A^n)$. Prove that $\det(A)=0$.   I was thinking of showing $A$ is not invertible, meaning it has an eigenvalue of $0$. Since no matter the power you give to $A$, it is still not the identity, you can deduce that at least one eigenvalue is indeed $0$. However this doesn't work if you have different eigenvalues in the Matrix, so I got stuck.  Help will be most appreciated.
94698
I'd like a hint rather than a full solution.  The problem I am considering is the following:     $X$ is an $n\times m$ matrix $Y$ is $m\times n$      Show that      $(I - XY)^{-1}\cdot X = X\cdot(I - YX)^{-1}$   The first $I$ is of dimension $n$ and the second $I$ is of dimension $m$.
89348
Let $A,B \in M_2(\mathbb{R})$ be two matrices such that $(A-B)^2=O_2$.      Prove $\det(A^2 - B^2)=(\det(A) - \det(B))^2$.   OBS. $O_2$ is the zero matrix    Let $D=A-B$ then $D^2=O_2$ therefore, using Cayley Hamilton theorem, we get $tr(D)=\det(D)=0$. It follows $tr(A)=tr(B)=t$ and, from here, $A^2 -B^2=t(A-B) + (\det(A) - \det(B))I_2$  This is all I could get.  UPDATE  The matrices are from $M_2(\mathbb{R})$. Sorry for misleading you.
45003
$\newcommand{\adj}{\operatorname{adj}}$Let $A\in \mathbb{M}_n$ ($n \geq\ 2$) be a regular matrix and $\adj(A)$ its adjoint. Prove that if A is regular then $\adj(\adj(A)) = (\det A)^{n-2} A$ (where $adj(adj(A))$ is the adjoint of $\adj(A)$).  For now, I know, or think I do, that $A^{-1} = \adj(A)/\det(A)$  from where we can see that $\adj(A) = A^{-1} \det(A)$  And that's all I know
71805
I need to find the prove for any rectangular matrix $A$ and any non-singular matrices $P$, $Q$ of appropriate sizes,  rank($PAQ$)= rank($A$)  I know that when singular matrix multiply by non-singular the result will be singular matrix. However, It is not relevant to the rank of the matrix. If A is singular with rank 2, why rank($PAQ$)= 2 not any numbers.
94994
Let $M$ be a (real) $n \times n$ matrix. For $1 \leq i, j \leq n$ we denote by $M_{ij}$ the $(n-1) \times (n-1)$ matrix that we get when the $i$th row and $j$th column of $M$ are removed. Now, consider fixed $i$ and $j$ with $i\neq j$. Let $N$ be the $(n-2) \times (n-2)$ matrix that we get when removing both the $i$th and $j$th row and the $i$th and $j$th column from $M$. Then the following identity holds: $$ \det M \det N = \det M_{ii}\det M_{jj} - \det M_{ij} \det M_{ji}. $$  We were able to prove this looking at all different terms that can occur on both sides when evaluating the determinant as a sum of $n!$ terms (for instance, terms containing $a_{ji}a_{ij}$ are counted both $1$ time in the LHS and $1$ time in the RHS).  We are looking for a slick proof that does not involve writing out the determinant.  Any suggestions or approaches to this problem are appreciated!
119041
Suppose that $A$ is a general $n \times n$ matrix and $B$ is obtained by interchanging the first two rows of $A$. Prove that $\det(B) = -\det(A)$.  By general $n \times n$ matrix, I mean $$\left[\begin{matrix} a_{11} &amp; a_{12} &amp; \cdots &amp; a_{1n} \\ a_{21} &amp; a_{22} &amp; \cdots &amp; a_{2n} \\ \vdots &amp; \vdots &amp; \ddots &amp; \vdots \\ a_{n1} &amp; a_{n2} &amp; \cdots &amp; a_{nn} \end{matrix} \right]$$ I'm thinking of using the cofactor expansion of the 2nd row of the first matrix, yielding $$ \det(A) = c_{21} a_{21} + c_{22} a_{22} + \cdots+ c_{2n} a_{2n}  $$ which will be positive or negative depending on $n$ and the values of $a_{21}, \ldots,a_{2n}$, and using the same on the first row in matrix $B$ to get $c_{11} a_{21}$ etc. and then use the fact that the power of the $c$ subscripts will alter $\det(B)$ to be the opposite sign, but I'm not sure how to express all of this mathematically in a proof directly. Help please?
39803
This is the fourth part of a four-part problem in Charles W. Curtis's book entitled Linear Algebra, An Introductory Approach (p. 216).  I've succeeded in proving the first three parts, but the most interesting part of the problem eludes me.  Part (a) requires the reader to prove that $\operatorname{Tr}{(AB)} = \operatorname{Tr}{(BA)}$, which I was able to show by writing out each side of the equation using sigma notation.  Part (b) asks the reader to use part (a) to show that similar matrices have the same trace.  If $A$ and $B$ are similar, then  $\operatorname{Tr}{(A)} = \operatorname{Tr}{(S^{-1}BS)}$  $= \operatorname{Tr}(BSS^{-1})$  $= \operatorname{Tr}(B)$,  which completes part (b).  Part (c) asks the reader to show that the vector subspace of matrices with trace equal to zero have dimension $n^2 - 1$.  Curtis provides the hint that the map from $M_n(F)$ to $F$ is a linear transformation.  From this, I used the theorem that $\dim T(V) + \dim n(T) = \dim V$ to obtain the dimension of the null space.  Part (d), however, I'm stuck on.  It asks the reader to show that subspace described in part (c) is generated by matrices of the form $AB - BA$, where $A$ and $B$ are arbitrary $n \times n$ matrices.  I tried to form a basis for the subspace, but wasn't really sure what it would look like since an $n \times n$ matrix has $n^2$ entries in it, but the basis would need $n^2 - 1$ matrixes.  I also tried to think of a linear transformation whose image would have the form of $AB - BA$, but this also didn't help me.  I'm kind of stuck...  Many thanks in advance!
25941
Let $A,B\in M_{n}(\mathbb{R})$ be symmetric, with $A&gt;0$ and $B&gt;0$.  I need to prove that $\det (A+B)&gt;\max (\det(A), \det(B))$.  I want to use Sylvester theorem of having a matrix $D$ so that $D=\operatorname{diag}(1,1,\ldots, 1,-1,-1, \ldots-1,0,0,\ldots,0)$.  Do I need to use it? How do I use it here?  Thank you.