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Chapter 7: Problem 6

Let \(A\) be an \(n \times n\) matrix with triangular factorization \(L U .\) Show that \\[ \operatorname{det}(A)=u_{11} u_{22} \cdots u_{n n} \\]

Short Answer

Expert verified
Using the properties of determinants, we have \(\operatorname{det}(A) = \operatorname{det}(L) \times \operatorname{det}(U)\). Since L has ones on the diagonal, its determinant equals 1. Thus, \(\operatorname{det}(A) = u_{11}u_{22} \cdots u_{nn}\), which is the product of the diagonal elements of U.

Step by step solution

01

Understand triangular factorization

In the given problem, we have a square matrix A with triangular factorization LU, where L is a lower triangular matrix, and U is an upper triangular matrix. This means that A = LU where L has only elements below the diagonal and ones on the diagonal, and U has only elements above the diagonal and non-zero elements on the diagonal.
02

Apply properties of determinants

Applying the property det(A) * det(B) = det(AB) on the LU factorization of A, we can write: \[ \operatorname{det}(A) = \operatorname{det}(L) \times \operatorname{det}(U) \] Now, using the property that the determinant of a triangular matrix is the product of its diagonal elements, we can write: \[ \operatorname{det}(A) = (\underbrace{1\cdot1\cdots1}_{n \text{ times}}) \times (u_{11}u_{22} \cdots u_{nn}) \]
03

Simplify the expression

As we have n ones in the product of diagonal elements of L, it becomes: \[ \operatorname{det}(A) = 1 \times (u_{11}u_{22} \cdots u_{nn}) \] Finally, we obtain: \[ \operatorname{det}(A) = u_{11}u_{22} \cdots u_{nn} \] which is the required proof.

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Most popular questions from this chapter

If \(\mathbf{x} \in \mathbb{R}^{m},\) we can think of \(\mathbf{x}\) as an \(m \times 1\) matrix. If \(\mathbf{x} \neq \mathbf{0},\) we can then define a \(1 \times m\) matrix \(X\) by \\[ X=\frac{1}{\|\mathbf{x}\|_{2}^{2}} \mathbf{x}^{T} \\] Show that \(X\) and \(\mathbf{x}\) satisfy the four Penrose conditions and, consequently, that \\[ \mathbf{x}^{+}=X=\frac{1}{\|\mathbf{x}\|_{2}^{2}} \mathbf{x}^{T} \\]

Let \\[ A=\left(\begin{array}{lll} 2 & 1 & 0 \\ 1 & 3 & 1 \\ 0 & 1 & 2 \end{array}\right) \quad \text { and } \quad \mathbf{u}_{0}=\left(\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right) \\] (a) Apply the power method to \(A\) to compute \(\mathbf{v}_{1}\) \(\mathbf{u}_{1}, \mathbf{v}_{2}, \mathbf{u}_{2},\) and \(\mathbf{v}_{3} .\) (Round off to two decimal places.) (b) Determine an approximation \(\lambda_{1}^{\prime}\) to the largest eigenvalue of \(A\) from the coordinates of \(\mathbf{v}_{3}\) Determine the exact value of \(\lambda_{1}\) and compare it with \(\lambda_{1}^{\prime} .\) What is the relative error?

Let \(A=X Y^{T}\), where \(X\) is an \(m \times r\) matrix, \(Y^{T}\) is an \(r \times n\) matrix, and \(X^{T} X\) and \(Y^{T} Y\) are both nonsingular. Show that the matrix \\[ B=Y\left(Y^{T} Y\right)^{-1}\left(X^{T} X\right)^{-1} X^{T} \\] satisfies the Penrose conditions and hence must equal \(A^{+} .\) Thus, \(A^{+}\) can be determined from any factorization of this form.

Let \(A \in \mathbb{R}^{m \times n}, B \in \mathbb{R}^{n \times r},\) and \(\mathbf{x}, \mathbf{y} \in \mathbb{R}^{n} .\) Suppose that the product \(A \mathbf{x y}^{T} B\) is computed in the following ways: (i) \(\left(A\left(\mathbf{x y}^{T}\right)\right) B\) (ii) \(\quad(A \mathbf{x})\left(\mathbf{y}^{T} B\right)\) (iii) \(\left((A \mathbf{x}) \mathbf{y}^{T}\right) B\) (a) How many scalar additions and multiplications are necessary for each of these computations? (b) Compare the number of scalar additions and multiplications for each of the three methods when \(m=5, n=4,\) and \(r=3 .\) Which method is most efficient in this case?

Let \(\mathbf{x}\) and \(\mathbf{y}\) be distinct vectors in \(\mathbb{R}^{n}\) with \(\|\mathbf{x}\|_{2}=\) \(\|\mathbf{y}\|_{2} .\) Define \\[ \mathbf{u}=\frac{1}{\|\mathbf{x}-\mathbf{y}\|_{2}}(\mathbf{x}-\mathbf{y}) \quad \text { and } \quad Q=I-2 \mathbf{u u}^{T} \\] Show that (a) \(\|\mathbf{x}-\mathbf{y}\|_{2}^{2}=2(\mathbf{x}-\mathbf{y})^{T} \mathbf{x}\) (b) \(Q \mathbf{x}=\mathbf{y}\)
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