The given expression is:

$x^2 - y^2 + 4z^2 + 4xy + 2yz + 6xz$

We are tasked with proving that this expression is indefinite. To determine if a quadratic form is indefinite, we examine its corresponding matrix and check its signature, which will tell us if the form can take both positive and negative values.

Step 1: Express the quadratic form in matrix notation

The general quadratic form in three variables $x$, $y$, and $z$ can be written as:

$Q(x, y, z) = [x, y, z] \begin{bmatrix} a & b & c \\ b & d & e \\ c & e & f \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}$

We want to identify the coefficients $a$, $b$, $c$, $d$, $e$, and $f$ from the given quadratic form:

$x^2 - y^2 + 4z^2 + 4xy + 2yz + 6xz$

• $x^2$ has a coefficient of $1$, so $a = 1$.
• $y^2$ has a coefficient of $-1$, so $d = -1$.
• $z^2$ has a coefficient of $4$, so $f = 4$.
• $xy$ has a coefficient of $4$, so $b = 2$ (since it appears as $4xy$, the factor is shared by both).
• $yz$ has a coefficient of $2$, so $e = 1$.
• $xz$ has a coefficient of $6$, so $c = 3$.

Thus, the corresponding matrix for the quadratic form is:

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

Step 2: Compute the determinant and minors

We calculate the principal minors to check if the matrix is indefinite.

1. First minor (top left element): $\Delta_1 = 1$

2. Second minor (determinant of top-left $2 \times 2$ submatrix): $\Delta_2 = \det \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} = (1)(-1) - (2)(2) = -1 - 4 = -5$

3. Third minor (determinant of the full matrix): $\Delta_3 = \det \begin{bmatrix} 1 & 2 & 3 \\ 2 & -1 & 1 \\ 3 & 1 & 4 \end{bmatrix}$

Using cofactor expansion along the first row:

$\Delta_3 = 1 \times \det \begin{bmatrix} -1 & 1 \\ 1 & 4 \end{bmatrix} - 2 \times \det \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} + 3 \times \det \begin{bmatrix} 2 & -1 \\ 3 & 1 \end{bmatrix}$

First term: $\det \begin{bmatrix} -1 & 1 \\ 1 & 4 \end{bmatrix} = (-1)(4) - (1)(1) = -4 - 1 = -5$

Second term: $\det \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} = (2)(4) - (1)(3) = 8 - 3 = 5$

Third term: $\det \begin{bmatrix} 2 & -1 \\ 3 & 1 \end{bmatrix} = (2)(1) - (-1)(3) = 2 + 3 = 5$

Now substitute back:

$\Delta_3 = 1 \times (-5) - 2 \times 5 + 3 \times 5 = -5 - 10 + 15 = 0$

Step 3: Conclusion

The matrix has minors of mixed signs, indicating that the quadratic form is indefinite. Specifically, the form can take both positive and negative values for different input vectors, satisfying the condition of indefiniteness.

Would you like further details or have any other questions?

Here are 5 related questions to deepen your understanding:

1. What is the significance of the determinant in a quadratic form?
2. How can we classify a quadratic form as positive definite, negative definite, or indefinite?
3. What role do eigenvalues play in determining the definiteness of a quadratic form?
4. How can we geometrically interpret a quadratic form being indefinite?
5. How do the signs of the principal minors help in classifying the definiteness?

Tip: When working with quadratic forms, always try to express them in matrix form to analyze their properties effectively through determinants and eigenvalues.