Can the $n-$th power of a non-upper triangular matrix be an upper triangular matrix?
123496
I have a question regarding upper triangular matrix.  I know that if AB is upper triangular then the |AB| equals to the diagnoal multiplication, but It doesn't seem to help me here.  If AB is upper triangular and non-singular then A and B both upper triangular?  Notice that A and B are both square.  I can't choose AB = 0 becuase its signular....
20497
How can I show that if the cube power of a matrix is the null matrix, then the matrix itself is singular?
109439
Let $\mathbf A = (a_{ij})$ be an $n\times n$ real or complex matrix with eigenvalues $\lambda_j$, for $j=1,...,n$. It is known that $\max|\lambda_j|$ is bounded above by the maximum row sum of $A$ (using entrywise absolute values).  Does the following related bound also hold?  Question: Is it true that $$ \sum_{j=1}^n|\lambda_j| \le \sum_{i,j=1}^{n}|a_{ij}|, $$ where the right hand side is summed over all matrix elements$\,$?  For example, this inequality is obvious for triangular matrices as well as positive definite matrices and slightly less obvious for unitary matrices (using the fact that the rows are unit vectors).  It can also be proved in other special cases but is it true in general?  Thanks.
108447
Prove that an upper triangular matrix is invertible if and only if every diagonal entry is non-zero. I have proved that if every diagonal entry is non-zero, then the matrix is invertible by showing we can row reduce the matrix to an identity matrix. But how do I prove the only if part?
71606
Let $A \in {M_n}$ is nonsingular and each eigenvalue of $A$ is either $+1$ or $-1$.Why $A$ is similar to ${A^{ - 1}}$?
72619
Suppose $A$ is invertible $n\times n$ matrix. Show that $A^{-1}$ can be written as a polynomial of degree at most $n-1$, i.e.  $$A^{-1} = c_{n-1}A^{n-1} + \dots + \ c_{1}A + c_{0}I$$  Are there any tips on how may this proof be approached?
2421
As a part of an exercise I have to prove the following:  Let $A$ be an $(n \times m)$ matrix. Let $A^T$ be the transposed matrix of $A$. Then $AA^T$ is an $(n \times n)$ matrix and $A^TA$ is an $(m \times m)$ matrix. $AA^T$ then has a total of $n$ eigenvalues and $A^TA$ has a total of $m$ eigenvalues.  What I need to prove is the following:  $AA^T$ has an eigenvalue $\mu \not = 0$ $\Longleftrightarrow$ $A^TA$ has an eigenvalue $\mu \not = 0$  In other words, they have the same non-zero eigenvalues, and if one has more eigenvalues than the other, then these are all equal to $0$.  How can I prove this?  Thanks and regards.
1817
There is an invertible $n×n$ matrix with each entry $0$ or $1$. Now we negate ($x\to 1-x$) the diagonal of the matrix, and it is still invertible. How many such matrices exist for a given $n$?  Can we derive a general formula? If $n=2$ there are $2$ such matrices, if $n=3$ there are $48$. But I have no idea whether a formula exists for higher $n$.
72172
Let $a$ belong to a group and $|a|=m$. If $n$ is relatively prime to $m$ show that $a$ can be written as the $n^{th}$ power of some element in the group.   We need to show that if $a\in G$ and  $a^m=e\implies \exists \ b\in G $ such that $b^n=a$ i.e.$b^{nm}=e=({b^m})^n$ i.e. $\exists$ an element of order $mn$ in $G$.   How to show this?
16359
I remember reading somewhere about the following properties of non-negative definite matrix. But I don't know how to prove it now.  Let $A$ and $B$ be two non-negative definite matrices. If $A^2\succ B^2$, then it necessarily follows that $A\succ B$, but $A\succ B$ doesn't necessarily leads to $A^2\succ B^2$.  How can you prove it? Thanks!
47255
Consider the matrix $$ M=\left(\begin{matrix} 0&amp;0&amp;0&amp;1\\ 0&amp;0&amp;0&amp;0\\ 0&amp;0&amp;0&amp;0\\ 0&amp;0&amp;0&amp;0\\ \end{matrix}\right). $$ Is there a matrix $P\in{\Bbb C}^{4\times 4}$ such that $P^n=M$ for some $n&gt;1$?    One obvious fact is that if such $P$ exists, then $P$ must be nilpotent. However, I have no idea how to deal with this problem. Furthermore, what if $M$ is an arbitrary nilpotent matrix with index $k$?
68240
Let $A$ be an $m\times n$ matrix. Prove that all eigenvalues of $A^*A$ are non-negative.
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 $m$.
24140
Let $S$ be the set of $3 \times 3$ matrices $\rm A$ with integer entries such that $$\rm AA^{\top} = I_3$$ What is $|S|$ (cardinality of $S$)?   The answer is supposed to be 48. Here is my proof and I wish to know if it is correct.  So, I am going to exploit the fact that the matrix A in a set will be orthognal, so if the matrix is of the form \begin{bmatrix}         a_{11} &amp; a_{12} &amp; a_{13} \\         a_{21} &amp; a_{22} &amp; a_{23} \\         a_{31} &amp; a_{32} &amp; a_{33} \\         \end{bmatrix}  Then each column and row will have exactly one non-zero element which will be +1 or -1. Thus, I have split possibilities for the first column into three cases and counted the possibilities in each case as follows :- $$a_{11} \neq 0$$ or $$ a_{21} \neq 0$$ or $$ a_{31} \neq 0$$  In case 1), we obviously have two possibilities(+1 or -1) so we consider the one where the entry is +1. Now, notice that the moment we choose the next non-zero entry, all the places for non-zero entries will be decided because of the rule 'each column and row will have exactly one non-zero element'. Meaning, if b and c are remaining two non-zero entries, we only have two possibilities left  \begin{bmatrix}         1 &amp; 0 &amp; 0 \\         0 &amp; b &amp; 0 \\         0 &amp; 0 &amp; c \\         \end{bmatrix}  or  \begin{bmatrix}         1 &amp; 0 &amp; 0 \\         0 &amp; 0 &amp; c \\         0 &amp; b &amp; 0 \\         \end{bmatrix}  Using the fact that b and c are simply $$\pm1$$  In each of the above matrices, we get 4 possibilities for each of the matricies. Thus, 8 possibilities in totality. Basically, we are getting 8 possibilities on the assumption that  $$a_{11} = 1$$  Thus, we get 16 possibilities on the case that $$a_{11} \neq 0$$  Following, the second and third cases analogously, we get a total of 16 possibilities in each of them and 48 possibilities in total.    Source :- Tata Institute of Fundamental Research Graduate School Admissions 2016
48795
find a $3 \times 3$ matrix $A$ such that $A \neq  I_3$ and $A^{197}=I_3$.
112568
Let $T:V \to V$ be linear. $V$ is a complex vector space of dimension $k$. Then there exists a basis so that the matrix generated by $T$ under that basis is upper triangular. The proof is by induction on $k$.   But how to generate such a basis. The first step is that because $v$ is a complex vector space, so there exists a non-zero vector $v$ such that $Tv=\lambda v$ for some $\lambda \in \Bbb F$. So, take $e_1=v$ because then $T(e_1)=\lambda e_1$. So, in the first column of the matrix the rows 2 to n will be zero. Now if I extend $e_1$ to any basis of $v$ that basis may not be a basis for which the matrix is upper triangular. Then how to construct it ?
28670
I have problems to show that if $A$ is a $2 \times 2$ matrix and if there exists some positive integer such that $A^{n}=0$ then $A^{2}=0$. I only showed that $A$ is a singular matrix but nothing else. Thanks any help will be appreciated.