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Chapter 7: Problem 89

Suppose \(B\) is a real nonsingular matrix. Show that: (a) \(B^{T} B\) is symmetric and (b) \(B^{T} B\) is positive definite.

Short Answer

Expert verified
In summary, given a real nonsingular matrix B, we proved that:\(B^T B\) is symmetric by showing that \((B^T B)^T = B B^T\), and \(B^T B\) is positive definite by demonstrating that for any non-zero vector x, \(x^T(B^T B)x > 0\).

Step by step solution

01

Recall definitions and properties

Before attempting this proof, we must be familiar with the following definitions and properties: 1. Symmetric Matrix: A matrix A is said to be symmetric if its transpose A^T equals A. 2. Positive Definite Matrix: A matrix A is said to be positive definite if for any non-zero vector x, x^T A x is positive (strictly greater than zero). 3. Transpose Properties: For any matrices A and B with appropriate dimensions, we have: (a) (A + B)^T = A^T + B^T (b) (AB)^T = B^T A^T (c) (A^T)^T = A Now we proceed to prove parts (a) and (b).
02

Prove that \(B^T B\) is symmetric

To prove that \(B^T B\) is symmetric, we need to show that its transpose is equal to itself. That is, we need to verify the following: \((B^T B)^T = B^T B\) Using the transpose properties mentioned in Step 1: \((B^T B)^T = B^{TT} (B^T)^T\) Since the transpose of a transpose is the original matrix: \(B^{TT} (B^T)^T = B B^T\) So, we have established that: \((B^T B)^T = B B^T\) Hence, \(B^T B\) is symmetric.
03

Prove that \(B^T B\) is positive definite

To prove that \(B^T B\) is positive definite, we need to show that for any non-zero vector x, the expression \(x^T(B^T B)x > 0\). Let x be any non-zero vector. Then: \(x^T(B^T B)x = (x^T B^T)(Bx)\) Now, let y = Bx. Since B is nonsingular (invertible), it has full rank, and hence Bx 鈮� 0 when x 鈮� 0. So, y is also a non-zero vector. Now, the expression becomes: \((x^T B^T)(Bx) = y^T y\) The expression \(y^T y\) is the dot product of vector y with itself, which is equivalent to the sum of the squares of the components of y. Since y is a non-zero vector, the sum of the squares of its components will always be positive. Thus: \(y^T y > 0\) This implies: \(x^T(B^T B)x > 0\) Therefore, \(B^T B\) is positive definite. By completing the proof for both parts (a) and (b), we have shown that if B is a real nonsingular matrix, then \(B^T B\) is symmetric and positive definite.

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Most popular questions from this chapter

Find the matrix \(P\) that represents the usual inner product on \(\mathbf{C}^{3}\) relative to the basis \(\\{1, i, 1-i\\}\).

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Let \(A\) be an \(n \times n\) matrix whose characteristic polynomial splits, \(\gamma\) be a cycle of generalized eigenvectors corresponding to an eigenvalue \(\lambda\), and \(W\) be the subspace spanned by \(\gamma\). Define \(\gamma^{\prime}\) to be the ordered set obtained from \(\gamma\) by reversing the order of the vectors in \(\gamma\). (a) Prove that \(\left[\mathrm{T}_{\mathrm{W}}\right]_{\gamma^{\prime}}=\left(\left[\mathrm{T}_{\mathrm{W}}\right]_{\gamma}\right)^{t}\). (b) Let \(J\) be the Jordan canonical form of \(A\). Use (a) to prove that \(J\) and \(J^{t}\) are similar. (c) Use (b) to prove that \(A\) and \(A^{t}\) are similar.

Let \(\mathrm{S}\) consist of the following vectors in \(\mathbf{R}^{4}\) : \\[u_{1}=(1,1,0,-1), u_{2}=(1,2,1,3), u_{3}=(1,1,-9,2), u_{4}=(16,-13,1,3)\\] (a) Show that \(S\) is orthogonal and a basis of \(\mathbf{R}^{4}\). (b) Find the coordinates of an arbitrary vector \(v=(a, b, c, d)\) in \(\mathbf{R}^{4}\) relative to the basis \(S\) (a) Compute \\[\begin{array}{lll} u_{1} \cdot u_{2}=1+2+0-3=0, & u_{1} \cdot u_{3}=1+1+0-2=0, & u_{1} \cdot u_{4}=16-13+0-3=0 \\ u_{2} \cdot u_{3}=1+2-9+6=0, & u_{2} \cdot u_{4}=16-26+1+9=0, & u_{3} \cdot u_{4}=16-13-9+6=0 \end{array}\\] Thus, \(S\) is orthogonal, and \(S\) is linearly independent. Accordingly, \(S\) is a basis for \(\mathbf{R}^{4}\) because any four linearly independent vectors form a basis of \(\mathbf{R}^{4}\) (b) Because \(S\) is orthogonal, we need only find the Fourier coefficients of \(v\) with respect to the basis vectors, as in Theorem \(7.7 .\) Thus, \\[\begin{array}{ll}k_{1}=\frac{\left\langle v, u_{1}\right\rangle}{\left\langle u_{1}, u_{1}\right\rangle}=\frac{a+b-d}{3}, & k_{3}=\frac{\left\langle v, u_{3}\right\rangle}{\left\langle u_{3}, u_{3}\right\rangle}=\frac{a+b-9 c+2 d}{87} \\ k_{2}=\frac{\left\langle v, u_{2}\right\rangle}{\left\langle u_{2}, u_{2}\right\rangle}=\frac{a+2 b+c+3 d}{15}, & k_{4}=\frac{\left\langle v, u_{4}\right\rangle}{\left\langle u_{4}, u_{4}\right\rangle}=\frac{16 a-13 b+c+3 d}{435}\end{array}\\] are the coordinates of \(v\) with respect to the basis \(S\).
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