Chapter 3: Q9P (page 159)

Show that $d e t (C^{- 1} MC) = d e t$ M. Hints: See (6.6). What is the product of $d e t (C^{- 1})$ and det $C$ ? Thus, show that the product of the eigenvalues of $M$ is equal to $d e t M$ .

Short Answer

Expert verified

The determinants of a matrix is equal to the product of its eigen values

Step by step solution

Given information from question

The matrix $C = (\begin{matrix} \frac{1}{\sqrt{5}} & \frac{- 2}{\sqrt{5}} \\ \frac{- 2}{\sqrt{5}} & \frac{1}{\sqrt{5}} \end{matrix})$

Eigen Values

Eigenvalues are a set of scalar values associated to a set of linear equations, and they are most commonly found in matrix equations. Characteristic roots are another name for eigenvectors. It's a non-zero vector that can only be altered by its scalar factor once linear transformations are applied.

Calculate the determinants of a matrix is equal to the product of its eigen values

$d e t D = d e t (C^{- 1} M C) = d e t C^{- 1} \times d e t M \times d e t C = \frac{1}{d e t C} \times d e t M \times d e t C = d e t M$

But since, we know D is a diagonal matrix, then its determinant equals the product of its diagonal elements, and also its diagonal elements are the eigenvalues for the matrix $M$ , therefore the determinants of a matrix is equal to the product of its eigen values.

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Show that $d e t (C^{- 1} MC) = d e t$ M. Hints: See (6.6). What is the product of $d e t (C^{- 1})$ and det $C$ ? Thus, show that the product of the eigenvalues of $M$ is equal to $d e t M$ .

Short Answer

Step by step solution

Given information from question

Eigen Values

Calculate the determinants of a matrix is equal to the product of its eigen values

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Show that det(C-1MC)=det M. Hints: See (6.6). What is the product of det(C-1) and det C ? Thus, show that the product of the eigenvalues of M is equal to detM.

Short Answer

Step by step solution

Given information from question

Eigen Values

Calculate the determinants of a matrix is equal to the product of its eigen values

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Fundamentals of Physics

Medical Physics

Electromagnetism

Dynamics

Torque and Rotational Motion

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Study anywhere. Anytime. Across all devices.

Show that $d e t (C^{- 1} MC) = d e t$ M. Hints: See (6.6). What is the product of $d e t (C^{- 1})$ and det $C$ ? Thus, show that the product of the eigenvalues of $M$ is equal to $d e t M$ .