Transformation of positive semi-definite matrices Let $a,b,c,d,e$ be positive reals such that the following matrix is positive semi-definite:
$$
\begin{pmatrix} a+4b+6c+4d+e & a+3b+3c+d & a+2b+c \\ 
a+3b+3c+d & a+2b+c & a+b  \\ 
a+2b+c & a+b & a \\ \end{pmatrix}
$$ 
Does it follow that also the following matrix is positive semi-definite?
$$
\begin{pmatrix} e & d & c \\ 
d & c & b  \\ 
c & b & a \\ \end{pmatrix}
$$
[By iterated subtraction of rows and columns, the two matrices have the same determinant; but do these operations preserve also this stronger property?]
 A: Let 
$$
A = \begin{pmatrix} a+4b+6c+4d+e & a+3b+3c+d & a + 2b+c \\a+3b+3c+d & a+2b+c & a+b \\ a+2b+c & a+b & a\end{pmatrix}, \quad
B = \begin{pmatrix} e & d & c \\ d & c & b \\ c & b & a \end{pmatrix}.$$
Then $A=UBU^T$ and $B=VAV^T$, where
$$U = \begin{pmatrix} 1 & 2 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}, \quad 
V = U^{-1} = \begin{pmatrix} 1 & -2 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 1\end{pmatrix} $$
Therefore $A$ and $B$ represent the same quadratic form, but in different bases,  hence $A$ is positive semi-definite if and only if $B$ is positive semi-definite.
Added to answer: $x^TBx = (V^Tx)^TA(V^Tx) \geq 0$, hence $B$ is positive semi-definite.
A: Edit: I make a rectification of an erroneous extension to semi-definite matrices (following remarks of @Paolo Leonetti).
I will consider only the particular case of positive definite matrices according to Sylvester criterion (see https://en.wikipedia.org/wiki/Positive-definite_matrix). 
This criteria is applied by checking the positivity of  determinants of "russian dolls matrices" beginning 


*

*by the North-West corner $M(1,1), M(1:2,1:2)$ and $M(1:3,1:3)=M$, or 

*by the South-East corner $M(3,3), M(2:3,2:3)$ and $M(1:3,1:3)=M$. 
It is this latter criteria that we will consider.
$$A=\begin{pmatrix} a+4b+6c+4d+e & a+3b+3c+d & a+2b+c \\ 
a+3b+3c+d & a+2b+c & a+b  \\ 
a+2b+c & a+b & a \\ \end{pmatrix} \ \ \text{and} \ \ B=\begin{pmatrix} e & d & c \\ 
d & c & b  \\ 
c & b & a \\ \end{pmatrix}$$
Let $M_{23}=M(2:3,2:3)$ (we keep in a matrix $M$ the elements which are in column 2 and 3, i.e., we suppress line 1 and column 1). 
Thus it suffices to check the equivalence:
$$det(A)>0 \ \& \ det(A_{23})> 0 \ \& \ a > 0  \ \ \Longleftrightarrow \ \ det(B)\geq0 \ \& \ det(B_{23})\geq 0 \ \& \ a \geq 0 \ \ \ (1)$$
But $det(A)=det(B)$ (you say it in your text), and one can verify that:
$$det(A_{23})=det(B_{23})=ac-b^2$$
Thus (1) is established.
Remark: coefficients $1,2,1$, then $1,3,3,1$, then $1,4,6,4,1$ that appear in matrix $A$ come from Pascal triangle. This remark could open the way to a new theorem and new proof, generalizable to larger matrices of this form.
