Positivity of a symmetric $7\times 7$-matrix

Question

Is there any simple method to prove that the following symmetric $7\times 7$-matrix $$M:=\begin{pmatrix} 1 & 0 & 0 & 0 & a_{2} & a_{3} & a_{4}\\ 0 & 1 & 0 & 0 & -a_{1} & -a_{4} & a_{3}\\ 0 & 0 & 1 & 0 & a_{4} & -a_{1} & -a_{2}\\ 0 & 0 & 0 & 1 & -a_{3} & a_{2} & -a_{1}\\ a_{2} & -a_{1} & a_{4} & -a_{3} & a_0 & 0 & 0\\ a_{3} & -a_{4} & -a_{1} & a_{2} & 0 & a_0 & 0\\ a_{4} & a_{3} & -a_{2} & -a_{1} & 0 & 0 & a_0\\ \end{pmatrix}$$ with $\det(M)=(a_0 - a_1^2 - a_2^2 - a_3^2 - a_4^2)^3 >0$, is positive definite, where $a_0>0$ and $a_i$, $i=1,..,4$ are real numbers.

Answer 1

Let us do an $LU$ factorization of the matrix. (We will use Sylvester's theorem eventually.) I will only show the bottom right $3\times 3$ submatrix as none of the rest of the elements change. For time and space purposes, it makes no sense to include the rest of the matrix.

Eliminating down the first column, the bottom right submatrix becomes

\begin{pmatrix} a_0 - a_2^2 & - a_2 a_3 & - a_2 a_4 \\ - a_2 a_3 & a_0 - a_3^2 & -a_3 a_4 \\ -a_2 a_4 & -a_3 a_4 & a_0^2 - a_4^2 \end{pmatrix}

Eliminating down the second column, this becomes

\begin{pmatrix} a_0 - a_1^2 - a_2^2 & -a_2 a_3 - a_1 a_4 & -a_2a_4 + a_1 a_3 \\ -a_2 a_3 - a_1 a_4 & a_0 - a_3^2 - a_4^2 & 0 \\ -a_2 a_4 + a_1 a_3 & 0 & a_0 - a_3^2 - a_4^2 \end{pmatrix}

Eliminating down the third column, this becomes

\begin{pmatrix} a_0 - a_1^2 - a_2^2 - a_4^2 & -a_2 a_3 & a_1 a_3 \\ -a_2 a_3 & a_0 - a_1^2 -a_3^2-a_4^2 & -a_1 a_2 \\ a_1 a_3 & -a_1 a_2 & a_0 - a_2^2 - a_3^2 - a_4^2 \end{pmatrix}

Eliminating down the fourth column, this becomes

\begin{pmatrix} a_0 - a_1^2 - a_2^2 - a_3^2 - a_4^2 & 0 & 0 \\ 0 & a_0 - a_1^2 - a_2^2 - a_3^2 & 0 \\ 0 & 0 & a_0 - a_1^2 - a_2^2 - a_3^2 \end{pmatrix}

So then we can factor $M$ as

$$\begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ a_2 & -a_1 & a_4 & -a_3 & 1 & 0 & 0 \\ a_3 & -a_4 & -a_1 & a_2 & 0 & 1 & 0 \\ a_4 & a_3 & -a_2 & -a_1 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0 & a_2 & a_3 & a_4 \\ 0 & 1 & 0 & 0 & -a_1 & -a_4 & a_3 \\ 0 & 0 & 1 & 0 & a_4 & -a_1 & -a_2 \\ 0 & 0 & 0 & 1 & -a_3 & a_2 & -a_1 \\ 0 & 0 & 0 & 0 & \det M & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & \det M & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & \det M \end{pmatrix}$$

You can check that this works quite easily. Sylvester's theorem says that if $M = LU$, then the $k$th leading principal submatrix has determinant $U_{11}\cdots U_{kk}$. Since all of the diagonal elements of $U$ are positive by assumption (be they $1$ or $\det M$), we see that all of $M$'s leading principal submatrices are positive and thus $M$ is positive.