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Chapter 6: Problem 8

Let \(A\) be a symmetric positive definite matrix and let \(Q\) be an orthogonal diagonalizing matrix. Use the factorization \(A=Q D Q^{T}\) to find a nonsingular matrix \(B\) such that \(B^{T} B=A\)

Short Answer

Expert verified
The nonsingular matrix \(B\) that satisfies \(B^TB = A\) can be found by defining \(B = Q\sqrt{D}\), where \(\sqrt{D}\) is a diagonal matrix with the square roots of the eigenvalues of \(A\) on its diagonal. This choice of \(B\) makes use of the given factorization \(A = QDQ^T\) and the properties of the orthogonal diagonalizing matrix \(Q\).

Step by step solution

01

We are given the factorization of \(A\) as \(A = QDQ^T\). Here, \(A\) is a symmetric positive definite matrix, \(Q\) is the orthogonal diagonalizing matrix, and \(D\) is a diagonal matrix containing the eigenvalues of \(A\). #Step 2: Understand the properties of orthogonal matrix Q#

An orthogonal matrix has the property that \(Q^TQ = QQ^T = I\), where \(I\) is the identity matrix. This means that the inverse of \(Q\) is equal to its transpose, i.e., \(Q^{-1} = Q^T\). #Step 3: Define B using the given factorization and properties of Q#
02

We want to find a matrix \(B\) such that \(B^TB = A\). Let us define \(B = Q\sqrt{D}\), where \(\sqrt{D}\) is a diagonal matrix with the square roots of the eigenvalues of \(A\) on its diagonal. To show that this choice of \(B\) satisfies the required condition, we will compute \(B^TB\). #Step 4: Compute B^T and B^TB#

First, compute the transpose of \(B\): \(B^T = (Q\sqrt{D})^T\). Using the property \((AB)^T = B^TA^T\), we get \(B^T = \sqrt{D}^TQ^T\). Since \(\sqrt{D}\) is a diagonal matrix, its transpose is itself: \(\sqrt{D}^T = \sqrt{D}\). Now, compute \(B^TB\): \(B^TB = (\sqrt{D}^TQ^T)(Q\sqrt{D})\). Using the associativity of matrix multiplication, we get \(B^TB = \sqrt{D}^T(Q^TQ)\sqrt{D}\). #Step 5: Use the properties of orthogonal matrix Q to simplify B^TB#
03

Since \(Q^TQ = I\), we can substitute it into the expression for \(B^TB\): \(B^TB = \sqrt{D}^T \cdot I \cdot \sqrt{D}\). Since the product of a matrix with the identity matrix is the matrix itself, we get \(B^TB = \sqrt{D}^T\sqrt{D}\). #Step 6: Relate B^TB to the given factorization of A#

Since \(\sqrt{D}^T\sqrt{D}\) is a diagonal matrix with the eigenvalues of \(A\) on its diagonal (the square roots of the eigenvalues squared), we get \(B^TB = D\). Now, substituting the given factorization of \(A\) into this equation, we get \(B^TB = QDQ^T = A\), as required. Thus, the nonsingular matrix \(B\) that we've found is given by \(B = Q\sqrt{D}\).

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Most popular questions from this chapter

Let \(A\) be a Hermitian matrix with eigenvalues \(\lambda_{1}, \ldots, \lambda_{n}\) and orthonormal eigenvectors \(\mathbf{u}_{1}, \ldots, \mathbf{u}_{n} .\) Show that \\[ A=\lambda_{1} \mathbf{u}_{1} \mathbf{u}_{1}^{H}+\lambda_{2} \mathbf{u}_{2} \mathbf{u}_{2}^{H}+\cdots+\lambda_{n} \mathbf{u}_{n} \mathbf{u}_{n}^{H} \\]

Show that a nonzero nilpotent matrix is defective.

Show that if \(A\) is a symmetric positive definite matrix, then \(A\) is nonsingular and \(A^{-1}\) is also positive definite.

Let \\[ A=\left(\begin{array}{rrrr} 2 & -1 & 0 & 0 \\ -1 & 2 & -1 & 0 \\ 0 & -1 & 2 & -1 \\ 0 & 0 & -1 & 2 \end{array}\right) \\] (a) Compute the \(L U\) factorization of \(A\) (b) Explain why \(A\) must be positive definite.

For each of the following matrices, compute the determinants of all the leading principal submatrices and use them to determine whether the matrix is positive definite: (a) \(\left(\begin{array}{rr}2 & -1 \\ -1 & 2\end{array}\right)\) (b) \(\left(\begin{array}{ll}3 & 4 \\ 4 & 2\end{array}\right)\) (c) \(\left(\begin{array}{rrr}6 & 4 & -2 \\ 4 & 5 & 3 \\ -2 & 3 & 6\end{array}\right)\) (d) \(\left(\begin{array}{rrr}4 & 2 & 1 \\ 2 & 3 & -2 \\ 1 & -2 & 5\end{array}\right)\)
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