Negative Definite Quadratic Form Lower Bound For Lambda

Hey guys! Let's dive into a fascinating problem from the realm of algebra – determining the values of λ for which the quadratic form q(x) is negative definite. Buckle up, because we're about to embark on a journey through quadratic forms, matrices, and the magic of linear algebra!

Cracking the Code of Negative Definite Quadratic Forms

So, what exactly are we dealing with? We have this quadratic form:

q(x) = -λ x_1^2 - 6 x_1 x_2 + 6 x_1 x_3 - 4 x_2^2 + 2 x_2 x_3 - 7 x_3^2

Our mission, should we choose to accept it, is to find the values of λ that make this quadratic form negative definite. But what does that even mean? A quadratic form is negative definite if q(x) < 0 for all non-zero vectors x. In simpler terms, no matter what values we plug in for x_1, x_2, and x_3 (as long as they're not all zero), the result of the expression will always be negative.

Now, how do we actually figure out these λ values? This is where the power of matrices comes into play. We can represent our quadratic form using a symmetric matrix. Let's call this matrix A. The elements of A are the coefficients of the quadratic form. Remember, the coefficient of the cross-term x_i x_j is split evenly between the (i, j) and (j, i) entries of the matrix.

So, for our quadratic form, the matrix A looks like this:

| -λ  -3   3 |
| -3  -4   1 |
|  3   1  -7 |

Key Concept: Sylvester's Criterion

This is where Sylvester's Criterion comes to the rescue! This powerful theorem gives us a way to determine if a symmetric matrix (and hence its corresponding quadratic form) is negative definite by examining the signs of its leading principal minors.

But what are leading principal minors? These are the determinants of the submatrices formed by taking the first k rows and k columns of the matrix, where k ranges from 1 to the size of the matrix.

For our 3x3 matrix A, we have three leading principal minors:

1. The determinant of the 1x1 submatrix: det([-λ]) = -λ

2. The determinant of the 2x2 submatrix:

   |-λ  -3|
   |-3  -4|
   

   which is (-λ)(-4) - (-3)(-3) = 4λ - 9

3. The determinant of the entire 3x3 matrix A:

   det(A) = -λ((-4)(-7) - (1)(1)) - (-3)((-3)(-7) - (1)(3)) + 3((-3)(1) - (-4)(3)) = -λ(28 - 1) + 3(21 - 3) + 3(-3 + 12) = -27λ + 54 + 27 = -27λ + 81

Sylvester's Criterion states that a symmetric matrix A is negative definite if and only if the signs of its leading principal minors alternate, starting with a negative sign. In other words:

• The first leading principal minor must be negative.
• The second leading principal minor must be positive.
• The third leading principal minor must be negative.
• And so on...

Let's apply this to our problem!

Putting Sylvester's Criterion to Work

For our matrix A to be negative definite, the following conditions must hold:

1. -λ < 0 (The first leading principal minor is negative)
2. 4λ - 9 > 0 (The second leading principal minor is positive)
3. -27λ + 81 < 0 (The third leading principal minor is negative)

Let's solve these inequalities:

1. -λ < 0 => λ > 0
2. 4λ - 9 > 0 => 4λ > 9 => λ > 9/4
3. -27λ + 81 < 0 => -27λ < -81 => λ > 3

So, we have three conditions: λ > 0, λ > 9/4, and λ > 3. For all three conditions to be satisfied, λ must be greater than the largest of these values, which is 3.

Finding the Infimum

The question asks for the infimum (the greatest lower bound) of the set of λ values. In simpler terms, we want the smallest possible value that λ can be greater than. Since λ > 3, the infimum is 3. Guys, this is where it all comes together! We've successfully navigated through the intricacies of quadratic forms and Sylvester's Criterion to pinpoint the exact lower bound for λ.

Wrapping Up Our Quadratic Form Quest

In conclusion, the quadratic form q(x) is negative definite when λ > 3. The infimum of the set of such λ is 3. We've not only solved the problem but also gained a deeper understanding of negative definite quadratic forms and the power of Sylvester's Criterion. Keep exploring the fascinating world of linear algebra, and who knows what other mathematical treasures you'll unearth!

Let's move on to another interesting problem!

Diving Deeper into the Realm of Quadratic Forms

Alright folks, let's really get into the nitty-gritty of quadratic forms and see how we can leverage them to solve a whole bunch of problems. Remember, a quadratic form is essentially a function that takes a vector as input and spits out a scalar value, and that value is calculated using a combination of squared terms and cross-product terms of the vector's components. We saw this earlier with our q(x) example.

But why are these quadratic forms so important? Well, they pop up in all sorts of places in mathematics, physics, and engineering. Think about things like:

• Optimization problems: Finding the maximum or minimum of a function often involves analyzing its quadratic approximation. Quadratic forms help us understand the shape of the function near a critical point.
• Stability analysis: In systems theory, quadratic forms are used to determine the stability of a system. A negative definite quadratic form can indicate that a system is stable, meaning it will return to its equilibrium state after a disturbance.
• Eigenvalue problems: The eigenvalues of the matrix associated with a quadratic form tell us a lot about the form's behavior. For example, the signs of the eigenvalues determine whether the form is positive definite, negative definite, or indefinite.
• Conic sections: Quadratic forms are intimately connected to conic sections like ellipses, hyperbolas, and parabolas. The equation of a conic section can be expressed as a quadratic form.

So, mastering the art of working with quadratic forms is a valuable skill, guys. Let's delve deeper into some key concepts.

The Matrix Representation – Your Quadratic Form's Secret Identity

We touched on this earlier, but it's worth reiterating: every quadratic form has a secret identity – its matrix representation! This is a symmetric matrix that encodes all the information about the quadratic form's coefficients.

Why is this so useful? Because working with matrices is often much easier than working directly with the quadratic form's expression. We can use powerful tools from linear algebra, like eigenvalues, eigenvectors, and determinants, to analyze the matrix and understand the behavior of the quadratic form.

Remember how we constructed the matrix A from our q(x) earlier? The diagonal elements of A are the coefficients of the squared terms (x_1^2, x_2^2, x_3^2), and the off-diagonal elements are half the coefficients of the cross-terms (x_i x_j). This ensures that A is symmetric (A = A^T).

Definiteness – Decoding the Quadratic Form's Personality

The definiteness of a quadratic form tells us about its