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Let \(A=\left[\begin{array}{ll}a & b \\ b & d\end{array}\right]\) be a symmetric \(2 \times 2\) matrix. Prove that \(A\) is positive definite if and only if \(a>0\) and \(\operatorname{det} A>0 .\left[\text {Hint}: a x^{2}+2 b x y+d y^{2}=\right.\) \(\left.a\left(x+\frac{b}{a} y\right)^{2}+\left(d-\frac{b^{2}}{a}\right) y^{2} \cdot\right]\)

Short Answer

Expert verified
Matrix \( A \) is positive definite if \( a>0 \) and \( ad-b^2 > 0 \).

Step by step solution

01

Understanding the Matrix

Consider the given symmetric matrix \( A = \begin{pmatrix} a & b \ b & d \end{pmatrix} \). For a matrix to be positive definite, all its leading principal minors need to be greater than zero.
02

First Condition for Positive Definiteness

The first condition is that the top-left element \( a \) must be greater than zero. So, we have \( a > 0 \).
03

Determinant Condition for Positive Definiteness

The determinant of matrix \( A \) must also be greater than zero for it to be positive definite. Calculate \( \text{det}(A) = ad - b^2 \). Thus, we require \( ad - b^2 > 0 \).
04

Utilizing the Provided Hint

Use the hint: express \( ax^2 + 2bxy + dy^2 \) as \( a \left( x + \frac{b}{a} y \right)^2 + \left( d - \frac{b^2}{a} \right) y^2 \). This indicates that \( d - \frac{b^2}{a} > 0 \) is another way to express \( ad - b^2 > 0 \).
05

Combining Conditions

For \( A \) to be positive definite, we need both conditions: \( a > 0 \) and \( ad - b^2 > 0 \). These ensure all leading principal minors of \( A \) are greater than zero.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Symmetric Matrix
A symmetric matrix is one that is equal to its transpose. In simple terms, it means that the matrix looks the same whether you read it from the top to bottom or left to right along the main diagonal. For a matrix to be symmetric, the element at row i, column j, must equal the element at row j, column i. This results in a mirroring effect across the diagonal.
For example, in the matrix \( A = \begin{pmatrix} a & b \ b & d \end{pmatrix} \), \( a \), \( b \), and \( d \) are such that swapping rows and columns does not change the matrix. It maintains its structure \( (a, b, b, d) \).
  • Symmetric matrices appear frequently in various areas such as statistics, physics, and machine learning due to their desirable mathematical properties.
  • One property is that symmetric matrices always have real eigenvalues.
  • Also, when symmetric matrices are real, they are diagonalizable by orthogonal matrices, simplifying they can be made diagonal by a rotation transformation.
Determinant Condition
The determinant condition is key to determining the positive definiteness of a matrix. For a \(2 \times 2\) matrix \( A = \begin{pmatrix} a & b \ b & d \end{pmatrix} \), the determinant is given by \( \text{det}(A) = ad - b^2 \). The significance of the determinant lies in its geometric interpretation: it represents the area (or volume in higher dimensions) of the parallelogram (or parallelepiped) formed by its column vectors.
To ensure that our matrix is positive definite, the determinant must be positive. Simply put, \( ad - b^2 > 0 \). If this condition is met, it implies that the transformation associated with this matrix does not collapse the space to a lower dimension (such as a line or a point), ensuring that the volume remains positive.
  • A positive determinant indicates that the transformation preserves or increases orientation, unlike a negative determinant which would imply a reflection.
  • In conjunction with the requirement that \( a > 0 \), it ensures the product of all eigenvalues is positive, a necessity for positive definiteness.
Leading Principal Minors
Leading principal minors are integral in analyzing the definiteness of a matrix. For a matrix to be positive definite, all of its leading principal minors must be positive. In the case of a \(2 \times 2\) symmetric matrix \( A = \begin{pmatrix} a & b \ b & d \end{pmatrix} \), you'll check the principal minors, which for this matrix, are quite simple.
  • The first leading principal minor is just the element \( a \), so \( a > 0 \) needs to be true.
  • The second leading principal minor in a \( 2 \times 2 \) matrix is the determinant itself, \( \text{det}(A) = ad - b^2 \).
Being positive definite implies that both these principal minors are positive. Once \( a > 0 \) and \( ad - b^2 > 0 \) are shown to hold, it guarantees that the matrix has the positive definiteness property, meaning the quadratic form \( ax^2 + 2bxy + dy^2 \) is positive for all non-zero vectors \( (x, y) \).
Consider ensuring that each principal minor follows its condition, especially as the matrix size increases, as having all positive leading principal minors becomes increasingly important in higher dimensionality matrices.

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Most popular questions from this chapter

Diagonalize the quadratic forms by finding an orthogonal matrix \(Q\) such that the change of variable \(\mathbf{x}=\) Qy transforms the given form into one with no cross-product terms. Give Q and the new quadratic form. $$x^{2}+z^{2}-2 x y+2 y z$$

Sometimes the graph of a quadratic equation is a straight line, a pair of straight lines, or a single point. We refer to such a graph as a degenerate conic. It is also possible that the equation is not satisfied for any values of the variables, in which case there is no graph at all and we refer to the conic as an imaginary conic. Identify the conic with the given equation as either degenerate or imaginary and, where possible, sketch the graph. $$x^{2}-y^{2}=0$$

Diagonalize the quadratic forms by finding an orthogonal matrix \(Q\) such that the change of variable \(\mathbf{x}=\) Qy transforms the given form into one with no cross-product terms. Give Q and the new quadratic form. $$2 x y+2 x z+2 y z$$

Find the orthogonal projection of v onto the subspace \(W\) spanned by the vectors \(\mathbf{u}_{\mathbf{r}}\) (You may assume that the vectors \(\mathbf{u}_{i}\) are orthogonal. ) $$\mathbf{v}=\left[\begin{array}{r} 3 \\ 1 \\ -2 \end{array}\right], \mathbf{u}_{1}=\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right], \mathbf{u}_{2}=\left[\begin{array}{r} 1 \\ -1 \\ 0 \end{array}\right]$$

Classify each of the quadratic forms as positive definite, positive semidefinite, negative definite, negative semidefinite, or indefinite. $$x_{1}^{2}+x_{2}^{2}-2 x_{1} x_{2}$$

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