Positive Definite Proof Method I am interested in an effecient way of showing that the $n \times n$ matrix,
\begin{pmatrix}
2&1&1& & 1\\
1&3&1& \cdots & 1\\
1&1&4& & 1\\
&\vdots& &\ddots& \vdots\\
1&1&1& \cdots & n+1
\end{pmatrix}
is positive definite.  
This is an old qualifier question, so it should not require extensive brute force computation.  
Sub-question: Do row operations affect positive-definiteness?
 A: Observe that
\begin{align}
\begin{pmatrix}
2&1&1& \cdots & 1\\
1&3&1& \cdots & 1\\
1&1&4& \cdots & 1\\
\vdots&\vdots&\vdots &\ddots& \vdots\\
1&1&1& \cdots & n+1
\end{pmatrix}
=
\begin{pmatrix}
1&1&1& \cdots & 1\\
1&1&1& \cdots & 1\\
1&1&1& \cdots & 1\\
\vdots&\vdots&\vdots &\ddots& \vdots\\
1&1&1& \cdots & 1
\end{pmatrix}
+
\begin{pmatrix}
1&&&  & \\
&2&&  & \\
&&3&  & \\
&& &\ddots&\\
&&&  & n
\end{pmatrix}
=A_1+A_2,
\end{align}
where $A_1$ and $A_2$ denote the corresponding decomposed matrices.
Now, given ${\bf x}=(x_1,x_2,\ldots,x_n)^\top\in\mathbb{R}^n$, we have
\begin{align}
{\bf x}^\top A_1{\bf x}&=
\begin{pmatrix}
\displaystyle\sum_{k=1}^nx_k
&\displaystyle\sum_{k=1}^nx_k
&\cdots
&\displaystyle\sum_{k=1}^nx_k
\end{pmatrix}
\begin{pmatrix}
x_1\\x_2\\\vdots\\x_n
\end{pmatrix}
=\left(\sum_{k=1}^nx_k\right)^2\ge 0,\\
{\bf x}^\top A_2{\bf x}&=
\begin{pmatrix}
x_1&2x_2&\cdots&nx_n
\end{pmatrix}
\begin{pmatrix}
x_1\\x_2\\\vdots\\x_n
\end{pmatrix}
=\sum_{k=1}^nk\cdot x_k^2.
\end{align}
If ${\bf x}\neq{\bf 0}$, at least one entry $x_k\ne 0$ for
some $k$, and then the value ${\bf x}^\top A_2{\bf x}$ must be positive. Thus
\begin{align}
{\bf x}^\top\begin{pmatrix}
2&1&1& \cdots & 1\\
1&3&1& \cdots & 1\\
1&1&4& \cdots & 1\\
\vdots&\vdots&\vdots &\ddots& \vdots\\
1&1&1& \cdots & n+1
\end{pmatrix}{\bf x}
&={\bf x}^\top(A_1+A_2){\bf x}\\
&={\bf x}^\top A_1{\bf x}+{\bf x}^\top A_2{\bf x}\\
&\ge{\bf x}^\top A_2{\bf x}\\
&>0,
\end{align}
and we conclude that the matrix must be positive definite.
