Consider a sequence of discs $U_1, U_2,\dots,U_n \subset \mathbb R^2$. Let $S_{ij}$ denote the area of the intersection of $U_i$ and $U_j$. Prove that $$\begin{vmatrix}  S_{11} &amp; S_{12} &amp; … &amp; S_{1n} \\  S_{21} &amp; S_{22} &amp; ... &amp; S_{2n} \\  … &amp; … &amp; \ddots &amp; … \\ S_{n1} &amp;  S_{n2} &amp; … &amp; S_{nn}  \end{vmatrix} \ge 0$$   For $n = 2$, we have $$\begin{vmatrix} S_{11} &amp; S_{12} \\ S_{21} &amp; S_{22} \end{vmatrix} = S_{11}S_{22} - S_{12}S_{21} \ge 0$$ since the diagonal elements are the largest. So the claim holds. I was planning to go through induction but it seems like to get a big deal to see how the determinant will change after adding a new disc to the collection. What we also have are the matrix is symmetric so that $S_{ij} = S_{ji}$ and therefore the eigenvalues are real (I don't see any use of this though).
123220
Consider disks $s_1, \cdots, s_n$ in the plane and let $a_{ij}$ be the area of $s_i\cap s_j$. Is it true that for any real numbers $x_1,\cdots, x_n$ we have  $$ \sum_{i,j=1}^n x_ix_j a_{ij} \geq 0$$    Equivalent formulation: one can put $a_{ij}$ into a matrix $A$ and ask whether it is positive semidefinite. For $n=2$ this is true since  $$a_{12}^2\le \min(a_{11},a_{22})^2 \le a_{11}a_{22} $$
95653
In one of my classes we discussed the ring of 2x2 matrices $M_{2}(\mathbb{Z})$. We said that its group of units was $SL_{2}(\mathbb{Z})$ which means that it is the set of 2x2 with determinant equal to $\pm$1. Why can't we have a 2x2 matrix with entries a,b,c, and d such that $\frac{a}{ad-bc}$,$\frac{-b}{ad-bc}$,$\frac{-c}{ad-bc}$, and $\frac{d}{ad-bc}$ are all integers?  I'm sure its a simple contradiction argument, but I couldn't see it. So if anyone knows a quick elementary argument, it'd be greatly appreciated
57037
This problem seems to be easy yet I have no idea to deal with.  If $\mathbb{M} ^n$ is the set of all real square matrices of order n, identified here with $\mathbb{R} ^{n^2}$ equipped with its usual Euclidean norm which induce an usual topology on the space $\mathbb{M}^n$. Show that:   The set of all invertible real matrices of order n is open in $\mathbb{M} ^n$ with respect to the Euclidean norm topology; The set $\mathbb{S} ^n$ which consists of all real symmetric matrices of order n is closed in $\mathbb{M} ^n$ with respect to the Euclidean norm topology.   For question #1, by the hint which is given by @Brent Kerby, that $det:\mathbb{M}^n \rightarrow \mathbb{R}$ is continuous.
76941
Let $n \geq 2$. Let $a_1, a_2, \ldots, a_n$ be $n$ integers such that $\gcd\left(a_1, a_2, \ldots, a_n\right) = 1$. Prove that there exists a matrix in $\operatorname{SL}_n\left(\mathbb{Z}\right)$ whose first row is $\left(a_1, a_2, \ldots, a_n\right)$.   Since the gcd of the integers $a_1,\ldots, a_n$ is $1$, there exists weights $x_i \in \mathbb{Z}$ such that $a_1x_1+\cdots+ a_nx_n=1$. My two ideas are (a) to brute force construct an $n\times n$ matrix with first row $a_1,\ldots ,a_n$ and to construct the remaining rows such that the determinant is $\sum a_ix_i=1$ or (b) to use induction.  (a) (Constructive) This is tedious since once I find a way to construct the remaining $n-1$ rows to ensure that $a_1x_1$ appears in the determinant, I am not sure how to modify these $n-1$ rows to ensure  that only the terms $a_ix_i$ appear in the cofactor expansion. If such a matrix exists, I'd like to see it.  (ii) (Non-constructive) If I proceed by induction then the base case $n=2$ is settled since I can choose the 2nd row to be $-x_2, x_1$ so that the determinant is $a_1x_1-a_2(-x_2)=1$. However, I'm not sure how to use the inductive hypothesis to show that if I can construct such an $n\times n$-matrix then I can construct an $\left(n+1\right) \times \left(n+1\right)$-matrix with the desired property. In particular, if the gcd $(a_1,\ldots ,a_{n+1})$ is $1$, it is not necesarry that the gcd of any $n$ of these terms is $1$, so induction may not even apply here.  How can I construct such a matrix or prove that one exists (without necessarily constructing it)?
52902
If $A \in M_{n}(\Bbb{R})$ is an Upper triangular matrix with diagonal entries $1$ such that  $A \neq I$, then what can we say about the diagonalizability of $A$ ?  I know that if the matrix has distinct eigenvalues or the set of eigenvectors are linearly independent then the matrix is diagonalizable.  And that the eigenvalues of the triangular matrices are given by the diagonal elements like  and , but they work nocely if we had distinct elements on the main diagonal.  But in my case I have same value 1 on the main diagonal, how can I approach about the diagonalizability of the matrix?
114675
Consider the $n\times n$ matrix over the field $\mathbb F_2$ formed by creating the circulant matrix of the vector consisting of $k$ ones followed by $n-k$ zeroes. E.g., for $n=4$ and $k=2$, the resulting matrix is  $$\begin{bmatrix}1 &amp; 1 &amp; 0 &amp; 0\\ 0 &amp; 1 &amp; 1 &amp; 0\\0 &amp; 0 &amp; 1 &amp; 1\\1 &amp; 0&amp;0 &amp;1\end{bmatrix}.$$  What is the rank of this matrix?  I suspect the answer is $n-d+1$, where $d=\gcd(n,k)$. I can prove this is an upper bound on the rank, since the following vectors are linearly independent and lie in the kernel of the matrix: $$\begin{bmatrix}1 &amp; (i\ \textrm{zeroes}) &amp; 1 &amp; (d-2-i\ \textrm{zeroes}) &amp; 1 &amp; (i\ \textrm{zeroes}) &amp; \cdots \end{bmatrix}$$ with $d-1$ different choices for $i$. Is there an easy way to see that these completely describe the kernel?
68302
Is there square matrix $A$ of size $3$ with real entries such that  $$ \operatorname{tr}(A)=0\text{ and }A^2+A^T=I. $$  I have proved that there is not with size $2$ using definition of  "trace", but for size $3$ it becomes complicated.  Here is the sketch of the proof for $2$. $$ a_{11}+a_{22}=0,\\ a_{11}^2+a_{12}a_{21}+a_{11}=1,\\ a_{11}a_{12}+a_{12}a_{22}+a_{21}=0,\\ a_{21}a_{11}+a_{21}a_{22}+a_{12}=0,\\ a_{12}a_{21}+a_{22}^2+a_{22}=1. $$ Putting $a_{11}=-a_{22}$ it is easy to see that above inequalities can't be true together.   Thanks!
83142
Prove that the determinant of an upper triangular matrix is the product of its diagonal entries.  What I have so far:  We will prove this by induction for an $n$ $\times$ $n$ matrix. For the case of a $2 \times 2$ matrix, let A=$ \left( \begin{array}{ccc} a_{11} &amp; a_{12} \\ 0 &amp; a_{22}  \end{array} \right)$. So det($A$)=$a_{11}a_{22}$ and the statement is true for the case of a $2 \times 2$ matrix. Now suppose that this statement is true for an $n$ $\times$ $n$ matrix. We will show that it also is true for an $(n + 1)$ $\times$ $(n + 1)$ matrix. Let A =$ \left( \begin{array}{ccc} a_{11} &amp; a_{12} &amp; \cdots &amp; a_{1(n+1)}\\ 0 &amp; a_{22} &amp; \cdots &amp; a_{2(n+1)} \\ \vdots &amp; \cdots &amp;  &amp; \vdots\\ 0 &amp; 0 &amp; \cdots &amp; a_{(n+1)(n+1)}\end{array} \right)$.  I don't know what to do after this.
107767
Prove that $$\begin{vmatrix} 1 &amp; 1 &amp; \cdots &amp; 1 \\ a_0 &amp; a_1 &amp; \cdots &amp; a_n \\ a_0^2 &amp; a_1^2 &amp; \cdots &amp; a_n^2 \\ \vdots &amp; \vdots &amp; \ddots &amp; \vdots \\ a_0^n &amp; a_1^n &amp; \cdots &amp; a_n^n  \\\end{vmatrix}=\prod_{j&lt;i}(a_i-a_j)$$. I have trouble with this, I thought it'd be doable with Laplace's theory but I fail to understand it.
85227
I have been wrestling with an exercise concerning latin squares from the textbook A First Course in Discrete Mathematics by Ian Anderson. The exercise is formulated thus:     A set $S$ of $mn$ elements is partitioned into $m$ sets of size $n$ in     two different ways: $S = A_1 \cup ... \cup A_m = B_1 \cup ... \cup   B_m$. Show that the sets $B_i$ can be relabelled so that $A_i \cap B_i   \neq \emptyset$ for each $i = 1, ..., m$. [Hint: consider the sets   $S_i = \{ j : A_i \cap B_j \neq \emptyset\}$.]   I am not sure how to prove this. Here is an example I have completed:  Let $S = \{1,...,25\}, m = 5, n = 5$ and let $$(s_{mn}) =  \begin{pmatrix} 1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 \\ 6 &amp; 7 &amp; 8 &amp; 9 &amp; 10 \\ 11 &amp; 12 &amp; 13 &amp; 14 &amp; 15 \\ 16 &amp; 17 &amp; 18 &amp; 19 &amp; 20 \\ 21 &amp; 22 &amp; 23 &amp; 24 &amp; 25 \end{pmatrix}$$.  Furthermore, let $$A_1 = \{1,6,11,16,21\}, A_2 = \{2,7,12,17,22\}, A_3 = \{3,8,13,18,23\}, A_4 = \{4,9,14,19,24\}, A_5 = \{5,10,15,20,25\}. $$  $$B_1 = \{1,2,3,4,5\}, B_2 = \{6,7,8,9,10\}, B_3 = \{11,12,13,14,15\}, B_4 = \{16,17,18,19,20\}, B_5 = \{21,22,23,24,25\}.$$  Furthermore, $S_1 = S_2 = S_3 = S_4 = S_5 = \{1,2,3,4,5\}$.  I can e.g. relabel the $B_i$s so that $B_1$ becomes $B_2$ etc. and $B_5$ becomes $B_1$ and thus get:  $$(s_{mn}) =  \begin{pmatrix} 21 &amp; 22 &amp; 23 &amp; 24 &amp; 25 \\ 1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 \\ 6 &amp; 7 &amp; 8 &amp; 9 &amp; 10 \\ 11 &amp; 12 &amp; 13 &amp; 14 &amp; 15 \\ 16 &amp; 17 &amp; 18 &amp; 19 &amp; 20 \\ \end{pmatrix}$$.  Then the condition $A_i \cap B_i \neq \emptyset$ still holds for each $i = 1,...,m$.  I guess I in some kind of way have to use Hall's theorem:     The sets $A_1,...,A_m$ possess an SDR if and only if they satisfy the Hall condition.
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
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 $d_1^2+\cdots+d_k^2$.
77656
Let $S$ be the subspace , of  $M_{n \times n}(\mathbb R)$ (the vector space of all $n \times n$ real matrices ) , generated by matrices of the form $AB-BA$ , where $A,B \in M_{n \times n}(\mathbb R)$ , then how do we prove that $\dim S=n^2-1$ ? The only thing that I can determine is that the trace of all matrices of $S$ is $0$ . Please help
4624
$V_n(a_1,a_2\dots, a_n)$ is a $n\times n$ Vandermonde matrix =   $$\left|\begin{array}[cccc]  11&amp;a_1&amp;\cdots&amp;a^{n-1}_1\\ 1&amp;a_2&amp;\cdots&amp;a^{n-1}_2\\ 1&amp;a_3&amp;\cdots&amp;a^{n-1}_3\\ \vdots&amp;\vdots&amp;\ddots&amp;\vdots\\ 1&amp;a_n&amp;\cdots&amp;a_n^{n-1} \end{array}\right|$$  Replace $a_1$ by $x$ so that the first row is $1,x, \dots ,x^{n-1}$  $$V_n(x,a_2\dots, a_n) = \left|\begin{array}[cccc]  11&amp;x&amp;\cdots&amp;x^{n-1}\\ 1&amp;a_2&amp;\cdots&amp;a^{n-1}_2\\ 1&amp;a_3&amp;\cdots&amp;a^{n-1}_3\\ \vdots&amp;\vdots&amp;\ddots&amp;\vdots\\ 1&amp;a_n&amp;\cdots&amp;a_n^{n-1} \end{array}\right|$$  Let $P(x) = V_n(x,a_2\dots, a_n)$.  (a) Show that $P(x)$ is a polynomial in $x$ of degree $\leq n-1$.   $z = 1\det() -x\det() + x^2\det() -\cdots+ (-1)^{n-1}x^{n-1} \cdot \det()$, where each $\det()$ stands for the determinant of a smaller matrix after removing the appropriate columns, rows.   Is this right? But isn't $z$ a polynomial of degree $n$?  (b) Show that $P(x)$ has $n-1$ distinct roots $a_2, \dots a_n$ and therefore has factorization $P(x) = k \prod_{i=2}^n(x-a_i)$ where the constant factor $k$ is the coefficient of $x^{n-1}.$  I'll be honest, I have no clue how to do this. Not even sure where to start. Nor do I know where to start on the rest.  (c) Show that $k = (-1)^{n-1}V_{n-1}(a_2,\dots a_n)$.  (d) Use parts (b) and (c) to deduce the recursion formula $$V_n(a_1, \dots a_n) = \left(\prod_{i=2}^n(a_i-a_1)\right)V_{n-1}(a_2,\dots a_n)$$  (e) Use part (d) to deduce that $V_n(a_1,a_2,\dots a_n) = \prod^n_{1\leq i&lt; j\leq n} (a_j - a_i)$
93148
Prove that: $\forall a,b\in\mathbb{C} \exists c,d \in \mathbb{C}: c+d = a \land cd = b$   We just learned about the characteristic/minimal polynomial and diagonalization but I am not sure if it has something to do with this question.  I'd be glad for guidance!
46151
I have to prove that determinant of skew- symmetric matrix of odd order is zero and also that its adjoint doesnt exist. I am sorry if the question is duplicate or already exists.I am not getting any start.I study in Class 11 so please give the proof accordingly. Thanks!
35912
Is it true that if $A$ is an $m \times n$ matrix and $B$ is an $n \times m$ matrix, with $m &gt; n $ then $\det(AB)=0$?