<s>[INST] 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?]
 [/INST] 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.
</s>