This is a study question. Let $q(x)=\sum_{i&lt;j}(x_i-x_j)^2$ be a quadratic form over $R^n$. Find the inertia of $q(x)$. I have tried to expand the sum and represent $q(x)$ as a matrix, however I am left then with trying to calculate the determinant of a matrix of size $n\times n$. What is the correct way to begin solving this problem? A hint would be great!
121482
I'm reading Further Linear Algebra by by T. S. Blyth, E. F. Robertson.  I'm somewhat stuck on Problem 9.14.  In particular, they want us to establish that the quadratic form $$ Q(x_1,\ldots,x_n) = \sum_{r &lt; s} (x_r - x_s)^2 $$ has both rank and signature equal to $n-1$.  The hint they give us is to show that the matrix for $Q$ with respect to the standard basis has $n-1$ on the diagonal, and $-1$ elsewhere.  Then they ask us to show that this matrix is congruent to the diagonal matrix with entries $n-1, n-2, \ldots, 2, 1, 0$. I'm assuming there is some clever basis transformation that gives us $P^t A P = \text{diag}(n-1, n-2, \ldots, 2, 1, 0)$, but I can't figure it out.  Any help?
87041
How do we prove that $$ \sum_{i=1}^{n}(x_i-\bar{x})^2\lt\sum_{i=1}^{n}(x_i-a)^2 $$  where $a$ is any value other than $\bar{x}$, the arithmetic mean.  My Attempt: $$ \sum_{i=1}^{n}(x_i^2-2x_i\bar{x}+\bar{x}^2)\lt\sum_{i=1}^{n}(x_i^2-2x_ia+a^2)\\\sum_{i=1}^{n}x_i^2-2\bar{x}\sum_{i=1}^{n}x_i+\sum_{i=1}^{n}\bar{x}^2\lt\sum_{i=1}^{n}x_i^2-2a\sum_{i=1}^{n}x_i+\sum_{i=1}^{n}a^2\\-2\bar{x}.n\bar{x}+n.\bar{x}^2\lt-2a.n\bar{x}+n.a^2\\-2\bar{x}^2+\bar{x}^2\lt-2a\bar{x}+a^2\\-\bar{x}^2\lt-2a\bar{x}+a^2 $$ But how do I proceed further or is there any better way ?
103341
How   $\frac{1}{n}\sum_{i=1}^n X_i^2 - \bar X^2 = \frac{\sum_{i=1}^n (X_i - \bar X)^2}{n}$   i have tried to do that by the following procedure:  $\frac{1}{n}\sum_{i=1}^n X_i^2 - \bar X^2$  =$\frac{1}{n}(\sum_{i=1}^n X_i^2 - n\bar X^2)$  =$\frac{1}{n}(\sum_{i=1}^n X_i^2 - \sum_{i=1}^n\bar X^2)$  =$\frac{1}{n} \sum_{i=1}^n (X_i^2 - \bar X^2)$  Then i have stumbled.
122123
Let $F$ be a subfield of the complex numbers. We define $n$ linear functionals on $F^n$ ($n \geq 2$) by $f_k(x_1, \dotsc, x_n) = \sum_{j=1}^n (k-j) x_j$, $1 \leq k \leq n$. What is the dimension of the subspace annihilated by $f_1, \dotsc, f_n$?   Approaches I've tried so far:   Construct a matrix $A$ whose $k$th row's entries are the coefficients of $f_k$, i.e., $A_{ij} = i - j$, and compute the rank of the matrix. Empirically, the resulting matrix has rank 2 for $n = 2$ to $n = 6$, but I don't see a convenient pattern to follow for a row reduction type proof for the general case. Observe that $f_k$, $k \geq 2$ annihilates $x = (x_1,\dotsc, x_n)$ iff $$\sum_{j=1}^{k-1} (k-j)x_j + \sum_{j = k+1}^n (j-k)x_j = 0$$ iff $$k\left(\sum_{j=1}^{k-1} x_j - \sum_{j=k+1}^n x_j\right) = \sum_{j=1}^{k-1} jx_j - \sum_{j=k+1}^n jx_j,$$ and go from there, but I do not see how to proceed.     Note: This is Exercise 10 in Section 3.5 ("Linear Functionals") in Linear Algebra by Hoffman and Kunze; eigenvalues and determinants have not yet been introduced and so I am looking for direction towards an elementary proof.
88691
True\False?  For every matrix $A\in M_{2}( \mathbb{C}) $ there's  $X\in M_{2}( \mathbb{C})$ such that $X^2=A$.  I know that every complexed matrix has a Jordan form matrix $J$ such that $P^{-1}CP=J$, But it's not diagonalizable for sure.  Thanks
60077
For $n$ variables $x_1,\ldots,x_n$, the determinant $$  \det\left((x_i^{j-1})_{i,j=1}^n\right) =   \left|\begin{matrix}      1&amp;x_1&amp;x_1^2&amp;\cdots &amp; x_1^{n-1}\\      1&amp;x_2&amp;x_2^2&amp;\cdots &amp; x_2^{n-1} \\      \vdots&amp;\vdots&amp;\vdots&amp;\ddots&amp;\vdots\\      1&amp;x_{n-1}&amp;x_{n-1}^2&amp;\cdots&amp;x_{n-1}^{n-1}\\      1 &amp;x_n&amp;x_n^2&amp;\cdots&amp;x_n^{n-1}   \end{matrix}\right| $$ can be computed by induction; the image below says it shows that. I have done this before, if I submit this will I get marks?   MORE IMPORTANTLY how do I do it by induction? The "hint" is to get the first row to $(1,0,0,...,0)$.  I think there are the grounds of induction in there, but I'm not sure how (I'm not very confident with induction when the structure isn't shown for $n=k$, assume for $n=r$, show if $n=r$ then $n=r+1$ is true.  By the way the question is to show the determinant at the start equals the product $$\prod_{1\leq i&lt;j\leq n}(x_j-x_i)$$ (but again, explicitly by induction)
47529
I'm trying to find a closed form solution of the following quadratic form for $x$.  $x^{T}Dx = c$  where $c$ is just a constant placeholder for some terms on the other side.  I know that, because $D$ is positive semi-definite, this equation describes a level curve of an ellipsoid in $x$, which gives me hope that there exists a closed form solution of the form $x = \frac{a\pm  \sqrt{z}}{b}$, for some $a, z, b$ (which may be matrices or vectors).   Do you know if this solution exists and if so how it is determined?
36846
Can someone help me out on how to find the the matrix representation and show proof that it is a linear transformation?    It is the rodrigues roation formula  and the matrix representation should look like    where c and s are cos and sin
46102
Assume M is a $2 \times 2$ real matrix such that $MX = XM$ for all real $2\times2$   matrices $X$. Show that $M$ must be some real multiple $q$ of $I$.   I can see that this is logical and have tried a few examples where I have multiplied $qI$ by some random $2\times2$ matrix from both sides and get the same matrix in both cases. How would I go about showing this though? Surely a few examples aren't actually showing anything for the general case, right?  Also, does the property shown in the question hold for all square matrices, or just $2\times2$ ones?
5469
Question states: "Let $D_n$ denote the determinant of the $n$-by-$n$ matrix whose $(i,j)$th element (the element in the $i$th row and $j$th column) is the absolute value of the difference between $i$ and $j$. Show that $D_n = (-1)^{n-1}(n-1)2^{n-2}$."  It seems like that I can get a recurrence relation (especially the question was from a book chapter dealing with recursion). In trying to get a recurrence relation, here is what I have done so far:  (Let $R_k$ denote the $k$th row and let $(i,j)$ denote the element/spot of $i$th row and $j$th spot)  1) replace $R_1$ with $R_1-(R_2+R_{n-1})$ to get $R_1^*$  2) replace $R_1^*$ with $R_1^*+R_{n-2}$  3) If I am not mistaken, I should have $-2$ on $(1,1)$, $2$ on both $(1,n-1)$ and $(1,n)$, and all other element in the first row as $0$.  4) take out $2$ from the first row, and expand the determinant about the first row  5) I should have $D_n=2(-D_{n-1}+(-1)^n B_{n-1})$, where $B_{n-1}$ is the determinant of a not-easily-reducible mayrix whose last column consists of $n-1$ ones followed by one $-1$.  How should I go about from now on? Once I get a recurrence relation, it would be pretty easy to get the explicit formula for $D_n$ using induction.  P.S. This question apparently is a question taken from 1969 Putnam Competition
13301
Let $ V $ be a vector space of finite dimension on $\mathbb{C} $ and $ \dim(V) \gt 1 $. Show that for every quadratic form $ q : V \to \mathbb{C} $ there exists $ 0 \neq v \in V $ such that $ q(v) = 0 $.  It certainly has something to do with the fact that we are in $\mathbb{C} $ and not $ \mathbb{R} $, but I'm since $ q $ can be represented by any non-singular complex matrix, I don't know how this is even possible.  (edit: V is of finite dimension)
83978
If $A^2 = I$ where $A$ is an $n\times n$ matrix, then why is $\operatorname{rank} (A+I) + \operatorname{rank} (A-I) =n$.   I am not sure why this seems to be true.
79378
Let $A=(a_{ij})$ be a $n \times n$ matrix with $a_{ij}=\gcd(i,j) , \forall i,j=1,2, \cdots, n$ , then how do we prove $\det A=\prod_{i=1}^n \phi(i)$ ? , where $\phi$ is the Euler's phi function
55372
Given a matrix $A=(a_{ij})_{n\times n}$, let $C_{i,j}$ be the cofactor in position $(i,j)$. By the determinant formula, we have  $$\det A=\sum_{i=1}^n a_{i,1}C_{i,1}.$$  What about if we take a different column for the cofactors, that is  $$\sum_{i=1}^n a_{i,1}C_{i,2}$$  Must this evaluate to zero?
28123
Let $A,B\in M_{n\times n}(\mathbb{R})$, prove that $$(\frac{d}{dt}\det(A+tB))_{t=0} = Tr(cof(A)^{T}\cdot B), $$ where $cof(A)$ is the cofactor matrix of $A$. I tried expanding it using row expansion, but it gets messed up. Any help?
78239
I was wondering. If I have a bilinear symmetric form, it is easy to find its matrix. But, when I have a quadratic form, which is the procedure to do that?  I heard that one possibility is: If $q$ is my quadratic form, then $$f(x,y) = \frac{1}{4}q(x+y) - \frac{1}{4}q(x-y)$$ is the bilinear symmetric form associated, so the method reduces to find the matrix of $f(x,y).$   The point is that seems a little noising... How to find, in practice, the matrix of quadratic form?  Thanks in advance.
117353
Prove that $\det(kA) = k^n \det(A)$ for and ($n \times n$) matrix.   I have tried looking at this a couple of ways but can't figure out where to start. It's confusing to me since the equation for a determinant is such a weird summation.