Quadratic form problem 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? 
 A: 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.
A: 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 $(v_i)_i=1$.
A: $q(\boldsymbol{x})=\boldsymbol{x}^{t}A\boldsymbol{x}$, 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.
