Observe the reel matrix
$$ A=\begin{pmatrix} 2 & -1 & -1 & 0\\ -1 & 3 & -1 & -1\\ -1 & -1 & 3 & -1\\ 0 & -1 & -1 & 2 \end{pmatrix} $$
and let $q: \mathbb{R}^4 → \mathbb{R}$ be the corresponding quadratic form defined by: $q(v)=v^TAv$ for $v\in\mathbb{R}^4$.
Find all the positive and negative index of inertia for $q$.
I had to varify that 0,2, and 4 are the only eigenvalues of A, which I have shown.
When I have to find the positive and negative index of inertia, I know that the positive index of inertia is the number: $$ i_+q = \sum_{\lambda \gt 0} Alg_A(\lambda) $$ and the negative index of inertia is the number: $$ i_-q = \sum_{\lambda \lt 0} Alg_A(\lambda) $$
It seems to be difficult to calculate the Algebraic multiplicities of A, since it is a $4x4$ matrix, but is it possible to use these definitions to calculate the postive and negative index of inertia or do I have to use Sylvesters law of inertia?
I hope someone can give a hint.
Positive semidefinite. This is Sylvester's Law of Inertia Three positive eigenvalues (repeats are permitted) and one zero. The actual eigenvalues are $(4,4,2,0).$
$$ P^T H P = D $$ $$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ \frac{ 1 }{ 2 } & 1 & 0 & 0 \\ \frac{ 4 }{ 5 } & \frac{ 3 }{ 5 } & 1 & 0 \\ 1 & 1 & 1 & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 2 & - 1 & - 1 & 0 \\ - 1 & 3 & - 1 & - 1 \\ - 1 & - 1 & 3 & - 1 \\ 0 & - 1 & - 1 & 2 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & \frac{ 1 }{ 2 } & \frac{ 4 }{ 5 } & 1 \\ 0 & 1 & \frac{ 3 }{ 5 } & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 2 & 0 & 0 & 0 \\ 0 & \frac{ 5 }{ 2 } & 0 & 0 \\ 0 & 0 & \frac{ 8 }{ 5 } & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ - \frac{ 1 }{ 2 } & 1 & 0 & 0 \\ - \frac{ 1 }{ 2 } & - \frac{ 3 }{ 5 } & 1 & 0 \\ 0 & - \frac{ 2 }{ 5 } & - 1 & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 2 & 0 & 0 & 0 \\ 0 & \frac{ 5 }{ 2 } & 0 & 0 \\ 0 & 0 & \frac{ 8 }{ 5 } & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & - \frac{ 1 }{ 2 } & - \frac{ 1 }{ 2 } & 0 \\ 0 & 1 & - \frac{ 3 }{ 5 } & - \frac{ 2 }{ 5 } \\ 0 & 0 & 1 & - 1 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 2 & - 1 & - 1 & 0 \\ - 1 & 3 & - 1 & - 1 \\ - 1 & - 1 & 3 & - 1 \\ 0 & - 1 & - 1 & 2 \\ \end{array} \right) $$