How to find the $\textbf{S}$ in $\textbf{SQS}^\text{T}$ in order to apply Sylvester's Law of Inertia?

Question

I'm trying to use Sylvester's Law of Inertia to find the the values of $k$ that make $$(k+4)x^2_1+2kx_1x_2+2x_2^2$$ a semi-definite positive form. Assuming $\textbf{Q}$ is the coefficients matrix of the quadratic form, I don't know any methodical procedure to find the $\textbf{S}$ that diagonalizes $\textbf{Q}$ through $\textbf{SQS}^\text{T}$.

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I've read the whole Wikipedia article, I looked for it in different forums, YouTube and Google, and still haven't got a clue about how to procede, yet I have the feeling it must be quite simple.

If you don't want to give me a full explanation, you can of course link me to a a comprehensive explanation somewhere else. But, really, I searched for hours and nothing seems satisfying. Is it a matter of simple intuition?

Answer 1

Fortunately, you're in $\Bbb R^2$, so it's not hard.

1. Write down the matrix $Q$: $$ Q = \pmatrix{k+4 & k \\ k & 2}. $$
2. Find its eigenvalues. $$ p(x) = \det(Q - xI) = \det \pmatrix{k+4-x & k \\ k & 2-x} = (k+4 - x) (2-x) - k^2. $$

3. Write this out as a quadratic in $x$, and compute the two eigenvalues using the quadratic formula. (You'll have to do this yourself. But I'll do it for $k = 0$, for instance. In that case, $p(x) = (4-x)(2-x) = x^2 - 6x + 8$, so the roots are $$ x_{1,2} = \frac{6 \pm \sqrt{36 - 32}}{2} = \frac{6 \pm 2}{2} = 4, 2 $$

   4. For each eigenvalue, find the corresponding eigenvector, i.e., solve $$ (Q - x_1 I) \pmatrix{u\\v} = \pmatrix{0\\0} $$ to find the eigenvector. Then normalize it (i.e., divide it by its length) and call those vectors $v_{1,2}$

4. Put the two eigenvectors into a matrix as its columns; that matrix is the $S$ that you want. (Or maybe you need to use the rows..do it and try both!)

Answer 2

here is one with very unpleasant eigenvalues. We don't bother with them.

$$ H = \left( \begin{array}{rrrr} 8 & 11 & 4 & 3 \\ 11 & 12 & 4 & 7 \\ 4 & 4 & 7 & 12 \\ 3 & 7 & 12 & 17 \\ \end{array} \right) $$ $$ D_0 = H $$ $$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = Q_j P_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$

$$ H = \left( \begin{array}{rrrr} 8 & 11 & 4 & 3 \\ 11 & 12 & 4 & 7 \\ 4 & 4 & 7 & 12 \\ 3 & 7 & 12 & 17 \\ \end{array} \right) $$

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$$ E_{1} = \left( \begin{array}{rrrr} 1 & - \frac{ 11 }{ 8 } & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrrr} 1 & - \frac{ 11 }{ 8 } & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrrr} 1 & \frac{ 11 }{ 8 } & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrrr} 8 & 0 & 4 & 3 \\ 0 & - \frac{ 25 }{ 8 } & - \frac{ 3 }{ 2 } & \frac{ 23 }{ 8 } \\ 4 & - \frac{ 3 }{ 2 } & 7 & 12 \\ 3 & \frac{ 23 }{ 8 } & 12 & 17 \\ \end{array} \right) $$

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$$ E_{2} = \left( \begin{array}{rrrr} 1 & 0 & - \frac{ 1 }{ 2 } & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{2} = \left( \begin{array}{rrrr} 1 & - \frac{ 11 }{ 8 } & - \frac{ 1 }{ 2 } & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrrr} 1 & \frac{ 11 }{ 8 } & \frac{ 1 }{ 2 } & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrrr} 8 & 0 & 0 & 3 \\ 0 & - \frac{ 25 }{ 8 } & - \frac{ 3 }{ 2 } & \frac{ 23 }{ 8 } \\ 0 & - \frac{ 3 }{ 2 } & 5 & \frac{ 21 }{ 2 } \\ 3 & \frac{ 23 }{ 8 } & \frac{ 21 }{ 2 } & 17 \\ \end{array} \right) $$

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$$ E_{3} = \left( \begin{array}{rrrr} 1 & 0 & 0 & - \frac{ 3 }{ 8 } \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{3} = \left( \begin{array}{rrrr} 1 & - \frac{ 11 }{ 8 } & - \frac{ 1 }{ 2 } & - \frac{ 3 }{ 8 } \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{3} = \left( \begin{array}{rrrr} 1 & \frac{ 11 }{ 8 } & \frac{ 1 }{ 2 } & \frac{ 3 }{ 8 } \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{3} = \left( \begin{array}{rrrr} 8 & 0 & 0 & 0 \\ 0 & - \frac{ 25 }{ 8 } & - \frac{ 3 }{ 2 } & \frac{ 23 }{ 8 } \\ 0 & - \frac{ 3 }{ 2 } & 5 & \frac{ 21 }{ 2 } \\ 0 & \frac{ 23 }{ 8 } & \frac{ 21 }{ 2 } & \frac{ 127 }{ 8 } \\ \end{array} \right) $$

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$$ E_{4} = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & - \frac{ 12 }{ 25 } & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{4} = \left( \begin{array}{rrrr} 1 & - \frac{ 11 }{ 8 } & \frac{ 4 }{ 25 } & - \frac{ 3 }{ 8 } \\ 0 & 1 & - \frac{ 12 }{ 25 } & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{4} = \left( \begin{array}{rrrr} 1 & \frac{ 11 }{ 8 } & \frac{ 1 }{ 2 } & \frac{ 3 }{ 8 } \\ 0 & 1 & \frac{ 12 }{ 25 } & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{4} = \left( \begin{array}{rrrr} 8 & 0 & 0 & 0 \\ 0 & - \frac{ 25 }{ 8 } & 0 & \frac{ 23 }{ 8 } \\ 0 & 0 & \frac{ 143 }{ 25 } & \frac{ 228 }{ 25 } \\ 0 & \frac{ 23 }{ 8 } & \frac{ 228 }{ 25 } & \frac{ 127 }{ 8 } \\ \end{array} \right) $$

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$$ E_{5} = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & \frac{ 23 }{ 25 } \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{5} = \left( \begin{array}{rrrr} 1 & - \frac{ 11 }{ 8 } & \frac{ 4 }{ 25 } & - \frac{ 41 }{ 25 } \\ 0 & 1 & - \frac{ 12 }{ 25 } & \frac{ 23 }{ 25 } \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{5} = \left( \begin{array}{rrrr} 1 & \frac{ 11 }{ 8 } & \frac{ 1 }{ 2 } & \frac{ 3 }{ 8 } \\ 0 & 1 & \frac{ 12 }{ 25 } & - \frac{ 23 }{ 25 } \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{5} = \left( \begin{array}{rrrr} 8 & 0 & 0 & 0 \\ 0 & - \frac{ 25 }{ 8 } & 0 & 0 \\ 0 & 0 & \frac{ 143 }{ 25 } & \frac{ 228 }{ 25 } \\ 0 & 0 & \frac{ 228 }{ 25 } & \frac{ 463 }{ 25 } \\ \end{array} \right) $$

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$$ E_{6} = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & - \frac{ 228 }{ 143 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{6} = \left( \begin{array}{rrrr} 1 & - \frac{ 11 }{ 8 } & \frac{ 4 }{ 25 } & - \frac{ 271 }{ 143 } \\ 0 & 1 & - \frac{ 12 }{ 25 } & \frac{ 241 }{ 143 } \\ 0 & 0 & 1 & - \frac{ 228 }{ 143 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{6} = \left( \begin{array}{rrrr} 1 & \frac{ 11 }{ 8 } & \frac{ 1 }{ 2 } & \frac{ 3 }{ 8 } \\ 0 & 1 & \frac{ 12 }{ 25 } & - \frac{ 23 }{ 25 } \\ 0 & 0 & 1 & \frac{ 228 }{ 143 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{6} = \left( \begin{array}{rrrr} 8 & 0 & 0 & 0 \\ 0 & - \frac{ 25 }{ 8 } & 0 & 0 \\ 0 & 0 & \frac{ 143 }{ 25 } & 0 \\ 0 & 0 & 0 & \frac{ 569 }{ 143 } \\ \end{array} \right) $$

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$$ P^T H P = D $$ $$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ - \frac{ 11 }{ 8 } & 1 & 0 & 0 \\ \frac{ 4 }{ 25 } & - \frac{ 12 }{ 25 } & 1 & 0 \\ - \frac{ 271 }{ 143 } & \frac{ 241 }{ 143 } & - \frac{ 228 }{ 143 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 8 & 11 & 4 & 3 \\ 11 & 12 & 4 & 7 \\ 4 & 4 & 7 & 12 \\ 3 & 7 & 12 & 17 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & - \frac{ 11 }{ 8 } & \frac{ 4 }{ 25 } & - \frac{ 271 }{ 143 } \\ 0 & 1 & - \frac{ 12 }{ 25 } & \frac{ 241 }{ 143 } \\ 0 & 0 & 1 & - \frac{ 228 }{ 143 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 8 & 0 & 0 & 0 \\ 0 & - \frac{ 25 }{ 8 } & 0 & 0 \\ 0 & 0 & \frac{ 143 }{ 25 } & 0 \\ 0 & 0 & 0 & \frac{ 569 }{ 143 } \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ \frac{ 11 }{ 8 } & 1 & 0 & 0 \\ \frac{ 1 }{ 2 } & \frac{ 12 }{ 25 } & 1 & 0 \\ \frac{ 3 }{ 8 } & - \frac{ 23 }{ 25 } & \frac{ 228 }{ 143 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 8 & 0 & 0 & 0 \\ 0 & - \frac{ 25 }{ 8 } & 0 & 0 \\ 0 & 0 & \frac{ 143 }{ 25 } & 0 \\ 0 & 0 & 0 & \frac{ 569 }{ 143 } \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & \frac{ 11 }{ 8 } & \frac{ 1 }{ 2 } & \frac{ 3 }{ 8 } \\ 0 & 1 & \frac{ 12 }{ 25 } & - \frac{ 23 }{ 25 } \\ 0 & 0 & 1 & \frac{ 228 }{ 143 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 8 & 11 & 4 & 3 \\ 11 & 12 & 4 & 7 \\ 4 & 4 & 7 & 12 \\ 3 & 7 & 12 & 17 \\ \end{array} \right) $$