Two matrices A and B are said to be similar if there exists an invertible matrix P such that B = P^-1 * A * P.

Let's assume λ is an eigenvalue of A, with corresponding eigenvector x. Then we have:

A * x = λ * x

We can multiply both sides of this equation by P^-1 and P (on the left and right respectively) to get:

P^-1 * A * P * (P^-1 * x) = λ * (P^-1 * x)

This shows that λ is also an eigenvalue of B = P^-1 * A * P, with corresponding eigenvector P^-1 * x.

Therefore, if two matrices A and B are similar, they have the same eigenvalues.

Question

Two matrices A and B are said to be similar if there exists an invertible matrix P such that B = P^-1 * A * P.

Let's assume λ is an eigenvalue of A, with corresponding eigenvector x. Then we have:

A * x = λ * x

We can multiply both sides of this equation by P^-1 and P (on the left and right respectively) to get:

P^-1 * A * P * (P^-1 * x) = λ * (P^-1 * x)

This shows that λ is also an eigenvalue of B = P^-1 * A * P, with corresponding eigenvector P^-1 * x.

Therefore, if two matrices A and B are similar, they have the same eigenvalues.

Knowee AI · Accepted Answer

Two matrices A and B are said to be similar if there exists an invertible matrix P such that B = P^-1 * A * P.

Let's assume λ is an eigenvalue of A, with corresponding eigenvector x. Then we have:

A * x = λ * x

We can multiply both sides of this equation by P^-1 and P (on the left and right respectively) to get:

P^-1 * A * P * (P^-1 * x) = λ * (P^-1 * x)

This shows that λ is also an eigenvalue of B = P^-1 * A * P, with corresponding eigenvector P^-1 * x.

Therefore, if two matrices A and B are similar, they have the same eigenvalues.

Suppose two n × n matrices A and B are similar.(a) (8 pts) Show that A and B have the same eigenvalues