Question

Check sings definiteness
Y=f(X1, X2) =2X12- 4X1X2+3X22

Answer 1

Answer on Question #56927 – Math – Linear Algebra

Question

Check signs definiteness

$$y = f(x_1, x_2) = 2x_1^2 - 4x_1x_2 + 3x_2^2$$

Solution

First method

The matrix of quadratic form $y = f(x_1, x_2) = 2x_1^2 - 4x_1x_2 + 3x_2^2$ is

$$M = \begin{pmatrix} 2 & -2 \\ -2 & 3 \end{pmatrix}$$

In mathematics, Sylvester’s criterion is a necessary and sufficient criterion to determine whether a Hermitian matrix is positive definite.

Sylvester criterion states that a Hermitian matrix $M$ is positive definite if and only if all the following matrices have a positive determinant:

- the upper left 1-b-1 corner of $M$

- the upper left 2-by-2 corner of $M$

...

$M$ itself.

In other words, all of the leading principal minors must be positive.

Calculate $\Delta_1 = 2 > 0, \quad \Delta_2 = \begin{vmatrix} 2 & -2 \\ -2 & 3 \end{vmatrix} = 2 \cdot 3 - (-2) \cdot (-2) = 6 - 4 = 2 > 0.$

Because all of the leading principal minors are positive, a matrix and the quadratic form are positive definite according to Sylvester’s criterion.

Second method

If $M$ is the symmetric matrix that defines the quadratic form, and $S$ is any invertible matrix such that $D = SMS^T$ is diagonal, then the number of negative elements in the diagonal of $D$ is always the same, for all such $S$, and the same goes for the number of positive elements according to Sylvester’s law of inertia.

It holds true that the symmetric matrix $M$ is positive definite if and only if all its eigenvalues are strictly positive.

To transform the given quadratic form into a diagonal form, find eigenvalues of the matrix

$$M = \begin{pmatrix} 2 & -2 \\ -2 & 3 \end{pmatrix}$$

by solving the equation

$$\left| \begin{array}{cc} 2 - \lambda & -2 \\ -2 & 3 - \lambda \end{array} \right| = 0$$

We have $(2 - \lambda)(3 - \lambda) - 4 = \lambda^2 - 5\lambda + 2 = 0$

This quadratic equation has two roots

$$\lambda_1 = \frac{5 - \sqrt{17}}{2} \quad \text{and} \quad \lambda_2 = \frac{5 + \sqrt{17}}{2}.$$

Since $\sqrt{17}$ is greater than 4 and less than 5, both roots are positive. So coefficients in diagonal form are strictly positive. Hence the given quadratic form is also positive definite.

**Answer:** the form $f(x_{1},x_{2}) = 2x_{1}^{2} - 4x_{1}x_{2} + 3x_{2}^{2}$ is positive definite.

**Third method**

$y = f(x_{1},x_{2}) = 2x_{1}^{2} - 4x_{1}x_{2} + 3x_{2}^{2} = 2(x_{1}^{2} - 2x_{1}x_{2} + x_{2}^{2}) + x_{2}^{2} = 2(x_{1} - x_{1})^{2} + x_{2}^{2} > 0$ for $(x_{1};x_{2}) \neq (0;0)$, hence the form is positive definite by definition.

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