# Symmetric matrix from the quadratic form

Background

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

$$A=\begin{bmatrix}1&4&-3\\4&0&2\\-3&2&1\end{bmatrix}$$

$$A-\lambda I=\begin{bmatrix}1-λ&4&-3\\4&0-λ&2\\-3&2&1-λ\end{bmatrix}$$

$$λ^3+2λ^2+28λ-68$$

$$λ_1≈-5.442, λ_2≈2.559, λ_3≈4.882$$

Is my process correct so far? Also, if no $$x_2^2$$ value is provided, would 0 go in its place?

• The matrix $A$ is already symmetric. Do you need to diagonalize it? – A.Γ. Aug 2 at 15:49
• @A.Γ. Thanks for the confirmation. Would the 0 be fine in $a_{22}$, if $x_2^2$ is missing from the quadratic form? I also worked out the diagonalising process to see if my process if correct so far, to understand the process better. – Laufen Aug 2 at 15:55
• You can check by yourself if $A$ is correct: compute $\mathbf{x}_T A \mathbf{x}$ and discover if it coincides with the quadratic form you have started with – FormulaWriter Aug 2 at 15:57
• $x_2^2$ is "missing" means $...+0\cdot x_2^2+...$, i.e. $a_{22}=0$. – A.Γ. Aug 2 at 16:01
• @FormulaWriter Thanks for that. Exactly what I was looking for. – Laufen Aug 2 at 16:02

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...
$$P^T H P = D$$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - 4 & 1 & 0 \\ - \frac{ 1 }{ 2 } & \frac{ 7 }{ 8 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 4 & - 3 \\ 4 & 0 & 2 \\ - 3 & 2 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 4 & - \frac{ 1 }{ 2 } \\ 0 & 1 & \frac{ 7 }{ 8 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 16 & 0 \\ 0 & 0 & \frac{ 17 }{ 4 } \\ \end{array} \right)$$ $$Q^T D Q = H$$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 4 & 1 & 0 \\ - 3 & - \frac{ 7 }{ 8 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 16 & 0 \\ 0 & 0 & \frac{ 17 }{ 4 } \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 4 & - 3 \\ 0 & 1 & - \frac{ 7 }{ 8 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 4 & - 3 \\ 4 & 0 & 2 \\ - 3 & 2 & 1 \\ \end{array} \right)$$