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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?]

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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.

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  • $\begingroup$ The result generalizes to higher order matrices of those forms -- the rows of $U$ are the rows of the Pascal triangle. $\endgroup$ – Catalin Zara Mar 30 '16 at 1:07
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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.

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  • $\begingroup$ Where did you take such version of Sylvester criterion about positive semi-definite matrices? I believe too the answer is affirmative, but shouldn't we check all principal minors of $A$ and $B$? .. $\endgroup$ – Paolo Leonetti Mar 28 '16 at 21:12
  • $\begingroup$ No, only the leading principal minors are to be checked, as explicitly said in the article of wikipedia I have cited. Here are their terms "The kth leading principal minor of a matrix M is the determinant of its upper-left k by k sub-matrix. It turns out that a matrix is positive definite if and only if all these determinants are positive". $\endgroup$ – Jean Marie Mar 28 '16 at 21:27
  • $\begingroup$ Indeed, here we are not talking about positive definite matrices. $\endgroup$ – Paolo Leonetti Mar 28 '16 at 21:32
  • $\begingroup$ What is the nature of my misunderstanding ? Do you mean positive semi-definiteness versus $definiteness$ (without the "semi") ? or is it the positivity of coefficients you are meaning ? $\endgroup$ – Jean Marie Mar 28 '16 at 21:41
  • $\begingroup$ A matrix $M$ is positive semi-definite if $x^TMx$ is nonnegative for each vector $x$. On the other hand, positive definiteness require strictly positiveness. The version of Sylvester criterion which you are talking about is about the latter case. Of course, there are variants of such criterion also for PSD matrices, but probably your condition is not sufficient. [I am just aware of a version requiring to check all principal minors, which is not quoted in Wikipedia] $\endgroup$ – Paolo Leonetti Mar 28 '16 at 21:47

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