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0
votes
Cofactor an geometrical mean in $\mathit{SPD}_3$: a Gårding-like inequality
Because of the lack of answers, I continued my investigations, and eventually got it !
Notation : because some expressions are too long for the command widehat, I'll sometimes denote ${\rm Cof}A$ for …
0
votes
Cofactor matrices and positive semi-definiteness
After further thinking, I believe that the answer is positive for every $n\ge2$. Here is a tantative proof.
Given an arbitrary vector $v$, we want to show that $q(v):=v^TP_{n-1}(A_1,,\ldots,A_{n-1})v …
5
votes
On primitive type matrix ranks
Denote this smallest $k$ by $k(A)$. At least, one has (obvious)
$$({\rm rk}(A)=1)\Longrightarrow(k(A)=1).$$
On the other hand, Wielandt's Theorem says that for every primitive matrix $A\ge0_n$, one ha …
7
votes
Signed factors of harmonic polynomials
S. Kharlamov pointed out to me that it is a consequence of the diagonalization of the Laplacian $\Delta_S$ over the unit sphere. Its eigenvalues are the integers $\lambda_d=d(d+n-2)$, and the correspo …