Transformation of positive semi-definite matrices Let $a,b,c,d,e$ be positive reals such that the following matrix is positive semi-definite:
$$
\begin{pmatrix} a+4b+6c+4d+e &amp; a+3b+3c+d &amp; a+2b+c \\ 
a+3b+3c+d &amp; a+2b+c &amp; a+b  \\ 
a+2b+c &amp; a+b &amp; a \\ \end{pmatrix}
$$ 

Does it follow that also the following matrix is positive semi-definite?
$$
\begin{pmatrix} e &amp; d &amp; c \\ 
d &amp; c &amp; b  \\ 
c &amp; b &amp; a \\ \end{pmatrix}
$$

[By iterated subtraction of rows and columns, the two matrices have the same determinant; but do these operations preserve also this stronger property?]
Let 

$$
A = \begin{pmatrix} a+4b+6c+4d+e &amp; a+3b+3c+d &amp; a + 2b+c \\a+3b+3c+d &amp; a+2b+c &amp; a+b \\ a+2b+c &amp; a+b &amp; a\end{pmatrix}, \quad
B = \begin{pmatrix} e &amp; d &amp; c \\ d &amp; c &amp; b \\ c &amp; b &amp; a \end{pmatrix}.$$

Then $A=UBU^T$ and $B=VAV^T$, where

$$U = \begin{pmatrix} 1 &amp; 2 &amp; 1 \\ 0 &amp; 1 &amp; 1 \\ 0 &amp; 0 &amp; 1 \end{pmatrix}, \quad 
V = U^{-1} = \begin{pmatrix} 1 &amp; -2 &amp; 1 \\ 0 &amp; 1 &amp; -1 \\ 0 &amp; 0 &amp; 1\end{pmatrix} $$

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.
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.

$$A=\begin{pmatrix} a+4b+6c+4d+e &amp; a+3b+3c+d &amp; a+2b+c \\ 
a+3b+3c+d &amp; a+2b+c &amp; a+b  \\ 
a+2b+c &amp; a+b &amp; a \\ \end{pmatrix} \ \ \text{and} \ \ B=\begin{pmatrix} e &amp; d &amp; c \\ 
d &amp; c &amp; b  \\ 
c &amp; b &amp; a \\ \end{pmatrix}$$

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:

$$det(A)&gt;0 \ \&amp; \ det(A_{23})&gt; 0 \ \&amp; \ a &gt; 0  \ \ \Longleftrightarrow \ \ det(B)\geq0 \ \&amp; \ det(B_{23})\geq 0 \ \&amp; \ a \geq 0 \ \ \ (1)$$

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.