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Prove that if \(A=L L^{T}\) with \(L\) real and nonsingular, then \(A\) is symmetric and positive definite.

Short Answer

Expert verified
Matrix A is symmetric because A = A^T, and it is positive definite because x^T A x > 0 for any non-zero x.

Step by step solution

01

- Understand the Given Information

Given the matrix equation \(A = LL^T\) where \(L\) is a real and nonsingular matrix. Need to prove that \(A\) is symmetric and positive definite.
02

- Prove that A is Symmetric

To prove that \(A\) is symmetric, we must show that \(A^T = A\). Start by transposing both sides of the given matrix equation: \(A^T = (LL^T)^T\). Using the property of transposes, \((AB)^T = B^T A^T\), we get:\(A^T = (L^T)^T L^T\). Since \((L^T)^T = L\), it follows that \(A^T = L L^T = A\). Therefore, \(A\) is symmetric.
03

- Prove that A is Positive Definite

To show \(A\) is positive definite, we need to demonstrate that for any non-zero vector \(x\), \(x^T A x > 0\). Substitute \(A = LL^T\):\(x^T A x = x^T (LL^T) x\). Let \(y = L^T x\). Since \(L\) is nonsingular, \(L^T\) is also nonsingular, meaning \(y eq 0\) when \(x eq 0\). Thus, we have:\(x^T A x = x^T L (L^T x) = (L^T x)^T (L^T x) = y^T y\). The expression \(y^T y\) is the dot product of \(y\) with itself, equivalent to \(\text{||y||}^2\), which is always greater than 0 for a non-zero \(y\). Thus, \(x^T A x > 0\), proving that \(A\) is positive definite.

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Key Concepts

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

symmetric matrices
Symmetric matrices play a crucial role in linear algebra, particularly in the study of quadratic forms and eigenvalues. A matrix is symmetric if it is equal to its transpose. In mathematical terms, a matrix \(A\) is symmetric if \(A = A^T\). This implies that across the main diagonal, the entries are mirrored. Symmetric matrices are important because they often appear in physical applications where they represent systems with balanced properties.
For example, in the given problem where \(A = LL^T\), proving that \(A\) is symmetric is straightforward using the matrix transposition property. We do this by showing that \(A^T = (LL^T)^T = L^T L = A\), confirming that \(A\) is indeed symmetric.
matrix transposition property
The transposition property of matrices is essential for understanding symmetric matrices. The transpose of a matrix \(A\), denoted \(A^T\), is obtained by swapping rows with columns. If \(A\) is an \(m \times n\) matrix, \(A^T\) will be an \(n \times m\) matrix. Transposition has several important properties:
  • \((A^T)^T = A\)
  • \((A + B)^T = A^T + B^T\)
  • \((AB)^T = B^T A^T\)
In the context of the given problem, we used the property \((LL^T)^T = (L^T)^T L^T\). Knowing that \((L^T)^T = L\), it simplified to \(L L^T\), showing that the matrix \(A\) remains the same when transposed, proving symmetry.
positive definiteness criteria
A matrix \(A\) is said to be positive definite if for any non-zero vector \(x\), the quadratic form \(x^T A x > 0\). Positive definiteness is crucial in optimization problems and ensuring system stability. To demonstrate that a matrix is positive definite, one typically checks that:
  • All its eigenvalues are positive
  • The given quadratic form is always positive for any non-zero vector \(x\)
In the problem at hand, we prove \(A = LL^T\) is positive definite by substituting \(y = L^T x\), leading to the form \(y^T y\), which is always positive for non-zero \(y\). Thus, \(x^T (LL^T) x > 0\), showing \(A\) is positive definite.
nonsingular matrices
A matrix is nonsingular if it is invertible, meaning there exists another matrix that when multiplied with it, results in the identity matrix. Nonsingular matrices have several key properties:
  • The determinant of a nonsingular matrix is non-zero
  • They have full rank, meaning their rows and columns are linearly independent
  • They exhibit stable and predictable behavior in systems
In the problem's context, the matrix \(L\) is given as real and nonsingular. Since \(L\) is nonsingular, \(L^T\) is also nonsingular, ensuring that the transformation \(L^T x = y\) does not result in a zero vector unless \(x\) is zero. This property is fundamental in proving both the symmetry and positive definiteness of the matrix \(A\).

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Most popular questions from this chapter

The system \(A x=b\), $$ A=\left[\begin{array}{rrrrrr} 4 & -1 & 0 & -1 & 0 & 0 \\ -1 & 4 & -1 & 0 & -1 & 0 \\ 0 & -1 & 4 & 0 & 0 & -1 \\ -1 & 0 & 0 & 4 & -1 & 0 \\ 0 & -1 & 0 & -1 & 4 & -1 \\ 0 & 0 & -1 & 0 & -1 & 4 \end{array}\right] \quad b=\left[\begin{array}{l} 2 \\ 1 \\ 2 \\ 2 \\ 1 \\ 2 \end{array}\right] $$ has the solution \(x=[1,1,1,1,1,1]^{T} .\) Solve the system using the Gauss- Jacobi iteration method, and then solve it again using the Gauss-Seidel method. Use the initial guess \(x^{(0)}=0\). Note the rate at which the iteration error decreases. Find the answers with an accuracy of \(\epsilon=.0001\).

Define the order \(n\) tridiagonal matrix $$ A_{n}=\left[\begin{array}{rrrrrr} 2 & -1 & 0 & & \cdots & 0 \\ -1 & 2 & -1 & 0 & & \\ 0 & -1 & 2 & -1 & & \vdots \\ \vdots & & & \ddots & \cdot & \\ 0 & & \cdots & & -1 & 2 \end{array}\right] $$ Find a general formula for \(A_{n}=L U .\) Hint: Consider the cases \(n=3,4,5\), and then guess the general pattern and verify it.

Using the Choleski method, calculate the decomposition \(A=L L^{T}\) for (a) \(\left[\begin{array}{rrr}2.25 & -3.0 & 4.5 \\ -3.0 & 5.0 & -10.0 \\ 4.5 & -10.0 & 34.0\end{array}\right]\) (b) \(\left[\begin{array}{rrrr}15 & -18 & 15 & -3 \\ -18 & 24 & -18 & 4 \\ 15 & -18 & 18 & -3 \\ -3 & 4 & -3 & 1\end{array}\right]\)

Consider solving the integral equation $$ \lambda x(s)-\int_{0}^{1} \cos (\pi s t) x(t) d t=1 \quad 0 \leq s \leq 1 $$ by discretizing the integral with the midpoint numerical integration rule (5.2.18). More precisely, let \(n>0, h=1 / n, t_{i}=\left(i-\frac{1}{2}\right) h\) for \(i=1, \ldots, n\). We solve for approximate values of \(x\left(t_{1}\right), \ldots, x\left(t_{n}\right)\) by solving the linear system $$ \lambda z_{i}-\sum_{j=1}^{n} h \cos \left(\pi t_{i} t_{j}\right) z_{j}=1 \quad i=1, \ldots, n $$ Denote this linear system by \(\left(\lambda I-K_{n}\right) z=b\), with \(K_{n}\) of order \(n \times n\), $$ \left(K_{n}\right)_{i j}=h \cdot \cos \left(\pi t_{i} t_{j}\right) \quad b_{i}=1 \quad 1 \leq i, j \leq n $$ For sufficiently large \(n, z_{i} \doteq x\left(t_{i}\right), 1 \leq i \leq n .\) The value of \(\lambda\) is nonzero, and it is assumed here to not be an eigenvalue of \(K_{n}\). Solve \(\left(\lambda I-K_{n}\right) z=b\) for several values of \(n\), say \(n=2,4,8,16,32,64\) and print the vector solutions \(z\). If possible, also graph these solutions, to gain some idea of the solution function \(x(s)\) of the original integral equation. Use \(\lambda=4,2,1, .5\)

Let \(A\) be symmetric, positive definite, and order \(\cdot n \times n .\) Let \(U=\) \(\left\\{u_{1}, \ldots, u_{n}\right\\}\) be a set of nonzero vectors in \(\mathbf{R}^{n} .\) Then if \(U\) is both an orthogonal set and an \(A\) -orthogonal set, then \(A u_{i}=\lambda_{i} u_{i}, i=1, \ldots, n\) for suitable \(\lambda_{i}>0 .\) Conversely, one can always choose a set of eigenvectors \(\left\\{u_{1}, \ldots, u_{n}\right\\}\) of \(A\) to have them be both orthogonal and \(A\) -orthogonal.

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