To find an orthonormal basis that diagonalizes the quadratic form $q(x) = 6x_1^2 + 3x_2^2 + 3x_3^2 - 4x_1x_2 + 4x_1x_3 - 2x_2x_3$ on $\mathbb{R}^3$, we can proceed with the following steps:

Step 1: Express $q(x)$ in Matrix Form

The quadratic form can be represented as:

$q(x) = x^T A x$

where $A$ is the symmetric matrix associated with the quadratic form.

Given: $q(x) = 6x_1^2 + 3x_2^2 + 3x_3^2 - 4x_1x_2 + 4x_1x_3 - 2x_2x_3$

The corresponding matrix $A$ is:

$A = \begin{pmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{pmatrix}$

Step 2: Find Eigenvalues and Eigenvectors of $A$

To diagonalize $A$, we find its eigenvalues and eigenvectors. The eigenvalues will form the diagonal entries in the diagonal matrix, and the eigenvectors will form the columns of the orthogonal matrix that diagonalizes $A$.

Eigenvalue Calculation

Let's calculate the eigenvalues by solving the characteristic polynomial $\det(A - \lambda I) = 0$.

After calculation, we find the eigenvalues are:

1. $\lambda_1 = 1$
2. $\lambda_2 = 3$
3. $\lambda_3 = 8$

Eigenvectors

For each eigenvalue, solve $(A - \lambda I)v = 0$ to find the corresponding eigenvector.

• For $\lambda_1 = 1$: Eigenvector $v_1 = \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$
• For $\lambda_2 = 3$: Eigenvector $v_2 = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}$
• For $\lambda_3 = 8$: Eigenvector $v_3 = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}$

Step 3: Orthonormalize the Eigenvectors

Normalize the eigenvectors to make them unit vectors. The orthonormal basis vectors are:

$u_1 = \frac{1}{\sqrt{3}} \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}, \quad u_2 = \frac{1}{\sqrt{2}} \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}, \quad u_3 = \frac{1}{\sqrt{6}} \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}$

Step 4: Diagonal Representation

The diagonal matrix $D$ that represents $q(x)$ in terms of this orthonormal basis is:

$D = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 8 \end{pmatrix}$

Summary

The orthonormal basis that diagonalizes $q$ is: $\left\{ u_1 = \frac{1}{\sqrt{3}} \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}, u_2 = \frac{1}{\sqrt{2}} \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}, u_3 = \frac{1}{\sqrt{6}} \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix} \right\}$

And the diagonal representation of $q(x)$ is: $q(x) = x_1^2 + 3x_2^2 + 8x_3^2$

Would you like further details on any of these steps or an example of applying this basis to a vector?

Related Questions

1. How are the eigenvalues of the matrix related to the shape of the quadratic form?
2. Can any quadratic form in $\mathbb{R}^n$ be diagonalized using eigenvalues and eigenvectors?
3. What is the significance of a positive, negative, or zero eigenvalue in a quadratic form?
4. How does the Gram-Schmidt process relate to finding an orthonormal basis for $\mathbb{R}^3$?
5. How does one interpret the diagonalized form in the context of geometric transformations?

Tip

In quadratic forms, diagonalizing can simplify understanding the behavior of the form by highlighting key directional components (eigenvalues) that determine the "stretching" effects along different axes.