Find the symmetric matrix $A$ for $$x_1^2+x_3^2+8x_1x_2-6x_1x_3+4x_2x_3, (x_1,x_2,x_3)∈\mathbb{R}^3$$
My work so far
$$A=\begin{bmatrix}1&4&-3\\4&0&2\\-3&2&1\end{bmatrix}$$
$$A-\lambda I=\begin{bmatrix}1-λ&4&-3\\4&0-λ&2\\-3&2&1-λ\end{bmatrix}$$
$$λ^3+2λ^2+28λ-68$$
$$ \lambda_1 \approx -5.442, \qquad \lambda_2 \approx 2.559, \qquad \lambda_3 \approx 4.882$$
Is my process correct so far? Also, if no $x_2^2$ value is provided, would $0$ go in its place?
Best Answer
This is an illustration of Sylvester's law of Inertia. The relationship between my $H$ and $D$ is called "congruence." This says that there are two positive and one negative eigenvalue. Which you knew...
$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - 4 & 1 & 0 \\ - \frac{ 1 }{ 2 } & \frac{ 7 }{ 8 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 4 & - 3 \\ 4 & 0 & 2 \\ - 3 & 2 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 4 & - \frac{ 1 }{ 2 } \\ 0 & 1 & \frac{ 7 }{ 8 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 16 & 0 \\ 0 & 0 & \frac{ 17 }{ 4 } \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 4 & 1 & 0 \\ - 3 & - \frac{ 7 }{ 8 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 16 & 0 \\ 0 & 0 & \frac{ 17 }{ 4 } \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 4 & - 3 \\ 0 & 1 & - \frac{ 7 }{ 8 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 4 & - 3 \\ 4 & 0 & 2 \\ - 3 & 2 & 1 \\ \end{array} \right) $$