positive definite of a matrix(diagonal decay) Is this matrix positive definite:
$$\begin{bmatrix}a & b &c & d\\b & b & c & d\\c & c & c & d\\d&d&d&d\end{bmatrix}$$
when $a>b>c>d>0$?
I tried several $a,b,c,d$, the result seems to be true, but I cannot prove it. Thanks!
 A: Thanks to @Friedrich Philipp and @Omnomnomnom's comments, I notice huge defects in my original proof. Instead, I would like to present an another
proof followed by @Omnomnomnom's method.
Denote the given matrix by $A$. Observe that
\begin{align}
A=\begin{bmatrix}a & b &c & d\\b & b & c & d\\c & c & c & d\\d&d&d&d\end{bmatrix}
&=
\begin{bmatrix}a-b & 0 &0 &0 \\0 &  0& 0 &0 \\ 0& 0 & 0 & 0\\0& 0& 0&0\end{bmatrix}
+
\begin{bmatrix}b-c & b-c &0 &0 \\b-c &  b-c& 0 &0 \\ 0& 0 & 0 & 0\\0& 0& 0&0\end{bmatrix}\\
&+
\begin{bmatrix}c-d & c-d &c-d &0 \\c-d &  c-d& c-d &0 \\ c-d& c-d & c-d & 0\\0& 0& 0&0\end{bmatrix}
+\begin{bmatrix}d & d &d &d\\d&  d& d &d \\ d& d & d & d\\d& d& d&d\end{bmatrix}\\
&=A_1+A_2+A_3+A_4,
\end{align}
where $A_i$ denotes the above corresponding decomposed matrix. Then given
$x=(x_1,x_2,x_3,x_4)\in{\sf R}^4$, we see that
\begin{align}
x^tAx&=x^tA_1x+x^tA_2x+x^tA_3x+x^tA_4x\\
     &=(a-b)x_1^2+(b-c)(x_1+x_2)^2+(c-d)(x_1+x_2+x_3)^2+d(x_1+x_2+x_3+x_4)^2.
\end{align}
That is, $x^tAx$ is the sum of the four non-zero terms. Now, if $x\neq{\it 0}$, we discuss the following cases.


*

*$x_1\neq 0$: It is clear that
$x^tAx\geq(a-b)x_1^2>0$.

*$x_1=0$ and $x_2\neq 0$: It follows that $x^tAx\geq(b-c)x_2^2>0$.

*$x_1=x_2=0$ and $x_3\neq 0$: It follows that $x^tAx\geq(c-d)x_3^2>0$.

*$x_1=x_2=x_3=0$ and $x_4\neq 0$: It follows that $x^tAx= dx_4^2>0$.


Hence we conclude that $x^tAx>0$ for all $x\neq{\it 0}$, that is, $A$ is positive definite.
A: Note that this matrix is definitely positive semidefinite, since it is the sum of positive semidefinite matrices.  In particular, we can write
$$
A = (a-b)A_1 + (b-c)A_2 + (c-d)A_3 + dA_4
$$
Where the $A_i$ are rank 1 matrices that I claim are "obviously positive semidefinite".  To show that it's invertible, row-reduce.
A: By Sylvester's criterium $A$ is positive definite if all main determinants are positive. They are $D_1=a>0$, $D_2=b(a-b)>0$, $D_3=c(a-b)(b-c)>0$ and $D_4=d(a - b)(b - c)(c - d)>0$. Hence the matrix is positive definite.
