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Chapter 6: Problem 5

Let \(A\) be a nondefective \(n \times n\) matrix with diagonalizing matrix \(X .\) Show that the matrix \(Y=\left(X^{-1}\right)^{T}\) diagonalizes \(A^{T}\)

Short Answer

Expert verified
The short answer is: Since A is diagonalizable by X, there exists a diagonal matrix D, such that \(A = XDX^{-1}\). By finding the transpose of A, we get \(A^T = YDY^T\), where \(Y = (X^{-1})^T\). Thus, by performing the required matrix multiplications and simplifications, we show that \(Y^{-1}A^T Y = DY\), which confirms that Y diagonalizes \(A^T\).

Step by step solution

01

The first step is to determine the transpose of the matrix A. The transpose of a matrix is obtained by interchanging its rows with its columns, denoted by \(A^T\). Since A is diagonalizable by X, there exists a diagonal matrix D, such that \(A = XDX^{-1}\). We must find the transpose of A using this relationship. #Step 2: Find the transpose of the product of matrices in A#

To find the transpose of \(A = XDX^{-1}\), we utilize the property of the transpose of the product of matrices: \((AB)^T = B^TA^T\). We apply this property to our equation, making sure to reverse the order of the matrices: \((XDX^{-1})^T = (X^{-1})^T D^T X^T\) We are also given the matrix Y which is equal to \((X^{-1})^T\), so we rewrite our equation as: \(A^T = YDY^T\) #Step 3: Show that Y diagonalizes A^T#
02

Now that we have established the relationship between A and A^T, we can show that the matrix Y diagonalizes A^T. We'll assume that Y is invertible (since we are given that X is invertible). We need to show that \(Y^{-1}A^{T}Y\) is a diagonal matrix. Begin by performing the left matrix multiplication by \(Y^{-1}\) and then performing the right matrix multiplication by Y. This gives us: \(Y^{-1}A^{T}Y = Y^{-1}(YDY^T)Y\) Since \(Y^{-1}Y = I\), where I is the identity matrix, we simplify the equation: \(Y^{-1}A^{T}Y = Y^{-1}YDY^TIY = IDY= DY\) #Step 4: Prove that Y diagonalizes A^T#

We found that \(Y^{-1}A^T Y = DY\), where D is a diagonal matrix. This shows that the matrix Y (which is equal to \((X^{-1})^T\)) diagonalizes the transpose of the matrix A, written as \(A^T\). Hence, the matrix Y = \((X^{-1})^T\) diagonalizes \(A^T\).

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Most popular questions from this chapter

Let \(A\) and \(C\) be matrices in \(\mathbb{C}^{m \times n}\) and let \(B \in \mathbb{C}^{n \times r}\) Prove each of the following rules: (a) \(\left(A^{H}\right)^{H}=A\) (b) \((\alpha A+\beta C)^{H}=\bar{\alpha} A^{H}+\bar{\beta} C^{H}\) (c) \(\quad(A B)^{H}=B^{H} A^{H}\)

Let \(A\) be a symmetric \(2 \times 2\) matrix and let \(\alpha\) be a nonzero scalar for which the equation \(\mathbf{x}^{T} A \mathbf{x}=\alpha\) is consistent. Show that the corresponding conic section will be nondegenerate if and only if \(A\) is nonsingular

Let \\[ A=\left(\begin{array}{rr} 1 & -\frac{1}{2} \\ -\frac{1}{2} & 1 \end{array}\right) \quad \text { and } \quad B=\left(\begin{array}{rr} 1 & -1 \\ 0 & 1 \end{array}\right) \\] (a) Show that \(A\) is positive definite and that \(\mathbf{x}^{T} A \mathbf{x}=\mathbf{x}^{T} B \mathbf{x}\) for all \(\mathbf{x} \in \mathbb{R}^{2}\) (b) Show that \(B\) is positive definite, but \(B^{2}\) is not positive definite.

Let \(\lambda\) be an eigenvalue of an \(n \times n\) matrix \(A\) and let \(\mathbf{x}\) be an eigenvector belonging to \(\lambda .\) Show that \(e^{\lambda}\) is an eigenvalue of \(e^{A}\) and \(\mathbf{x}\) is an eigenvector of \(e^{A}\) belonging to \(e^{\lambda}\)

Given \\[ \mathbf{Y}=c_{1} e^{\lambda_{1} t} \mathbf{x}_{1}+c_{2} e^{\lambda_{2} t} \mathbf{x}_{2}+\cdots+c_{n} e^{\lambda_{n} t} \mathbf{x}_{n} \\] is the solution to the initial value problem: \\[ \mathbf{Y}^{\prime}=A \mathbf{Y}, \quad \mathbf{Y}(0)=\mathbf{Y}_{0} \\] (a) Show that \\[ \mathbf{Y}_{0}=c_{1} \mathbf{x}_{1}+c_{2} \mathbf{x}_{2}+\cdots+c_{n} \mathbf{x}_{n} \\] (b) Let \(X=\left(\mathbf{x}_{1}, \ldots, \mathbf{x}_{n}\right)\) and \(\mathbf{c}=\left(c_{1}, \ldots, c_{n}\right)^{T}\) Assuming that the vectors \(\mathbf{x}_{1}, \ldots, \mathbf{x}_{n}\) are linearly independent, show that \(\mathbf{c}=X^{-1} \mathbf{Y}_{0}\)
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