Quadratic form problem

Question

let $A=(a_{ij})_{1\leq i,j\leq n}\in M^{\Bbb{R}}_{n\times n}$ and $a_{ij}=\begin{cases} 2t & \text{ if } i=j \\ t & \text{ if } i\neq j \end{cases} $, also in this conditions: $n\geq 2 $ , $t>0$. let the define the function $q: \Bbb{R}^{n}\rightarrow \Bbb{R}$ so: $$q(\mathbf{x})=q(x_{1},...,x_{n})=\sum ^{n}_{i,j=1}a_{ij}x_{i}x_{j}$$ does $q(\mathbf{x})>0$ for all $\mathbf{x}\neq \mathbf{0}$? What I have done so far is that: $$A=\begin{bmatrix} 2t & t& t& ...& t\\ t& 2t& t& ...& t\\ t& t& 2t& & \vdots\\ \vdots & \vdots& & \ddots &t \\ t& t& ...& t& 2t \end{bmatrix}\Rightarrow A-tI=\begin{bmatrix} t & t& t& ...& t\\ t& t& t& ...& t\\ t& t& t& & \vdots\\ \vdots & \vdots& & \ddots &t \\ t& t& ...& t& t \end{bmatrix}\Rightarrow \rho (A-tI)=1$$ and now I know that the dimension for the vector in which is solution for $(A-tI)x=0$ have the dimension of $(n-1)$. if so, $\lambda _{1}=t$ and $\dim V_{\lambda _{1}}=n-1$. How do I proceed?

Answer 1

Your question is equivalent to asking whether the eigenvalues of the $n\times n$ matrix \begin{bmatrix} 2 & 1& 1& ...& 1\\ 1& 2& 1& ...& 1\\ 1& 1& 2& & \vdots\\ \vdots & \vdots& & \ddots &1 \\ 1& 1& ...& 1& 2 \end{bmatrix} are all strictly positive. And yes, they are all strictly positive. Specifically, they are:

$\lambda_1=n+1$
$\lambda_2=\ldots= \lambda_n=1$

A possible set of corresponding eigenvectors is:

$v_1=(1,1,\ldots,1)$

$v_2=(-1,1,0,0,\ldots,0,0)$
$v_3=(-1,0,1,0,\ldots,0,0)$
$v_4=(-1,0,0,1,\ldots,0,0)$
$...$
$v_{n-1}=(-1,0,0,0,\ldots,1,0)$
$v_n=(-1,0,0,0,\ldots,0,1)$

where for $i\ge 2$, all elements $(v_i)_j$ of $v_i$ are zero except for $(v_i)_1=-1$ and $(v_i)_i=1$.

Answer 2

Write down $det(A-\lambda I_n)$ and change the last row by the sum of all rows. That makes the last row to be always the same number $(n+1)t-\lambda$.

From that, you have that $t(n+1)$ is an eigenvalue of your matrix (a positive one).

From your calcuations, you have that $t$ is an eigenvalue of mulyiplicity at least $n-1$.

So the only possibility is that all the eigenvalues are $\lambda=t$ (multipicity $n-1$) and $\lambda=(n+1)t$ (with multiplicity 1).

Hence all eigenvalues are strictly positive and youtr quadratic form is positive.

Answer 3

$q(\boldsymbol{x})=\boldsymbol{x}^{t}A\boldsymbol{x}$, there for I will find the characteristic polynomial of A: $$A=\begin{bmatrix} 2t &t &... & ... &t \\ t& 2t&t & & \\ \vdots& \ddots& \ddots& \ddots& \\ \vdots& & \ddots& \ddots&t \\ t& & & t&2t \end{bmatrix}\Rightarrow P(A-\lambda I)=\begin{vmatrix} 2t-\lambda &t &... & ... &t \\ t& 2t-\lambda&t & & \\ \vdots& \ddots& \ddots& \ddots& \\ \vdots& & \ddots& \ddots&t \\ t& & & t&2t-\lambda \end{vmatrix} \xrightarrow[R_{1}:R_{1}+R_{i}]{\forall i \geqslant 2}\begin{vmatrix} (n+1)t-\lambda & (n+1)t-\lambda& ...& &(n+1)t-\lambda \\ t& 2t-\lambda&t & &t \\ \vdots& \ddots& \ddots& \ddots&\vdots \\ \vdots& & \ddots& \ddots&t \\ t& & & t&2t-\lambda \end{vmatrix} \xrightarrow[C_{j}:C_{j}-C_{j}]{\forall j \geqslant 2}\begin{vmatrix} (n+1)t-\lambda& 0& ...& &0 \\ t& t-\lambda&\ddots & &\vdots \\ \vdots& 0& & & \\ & \vdots& & \ddots& 0\\ t&0 & ...& 0&t-\lambda \end{vmatrix}=((n+1)t-\lambda)(t-\lambda)^{n-1}$$ the eigenvalues are $ \lambda_{1}=t,\; \lambda_{2}=(n+1)t$, and because of $t>0$ than both of the eigenvalues are positive which mean that A is positve.