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Chapter 4: Problem 14

Let \(A\) and \(B\) be similar matrices and let \(\lambda\) be any scalar. Show that (a) \(A-\lambda I\) and \(B-\lambda I\) are similar (b) \(\operatorname{det}(A-\lambda I)=\operatorname{det}(B-\lambda I)\)

Short Answer

Expert verified
In summary, since A and B are similar matrices, there exists a non-singular matrix P such that B = P^(-1)AP. Using this fact, we have shown that A - 位I and B - 位I are similar matrices as well, as we can rewrite B - 位I = P^(-1)(A - 位I)P, and by using the properties of determinants, we found that det(A - 位I) = det(B - 位I).

Step by step solution

01

Recall the definition of similar matrices

Two matrices A and B are similar if there exists a non-singular matrix P such that B = P^(-1)AP.
02

Part (a): Prove A - 位I and B - 位I are similar matrices

We are given that A and B are similar matrices, so there exists a non-singular matrix P such that: B = P^(-1)AP Now, we can try to express B - 位I in terms of A - 位I: B - 位I = P^(-1)AP - 位I To show that A - 位I and B - 位I are similar, we need to find a matrix P such that B - 位I = P^(-1)(A - 位I)P. Notice that we can rewrite B - 位I: B - 位I = P^(-1)AP - P^(-1)(位I)P B - 位I = P^(-1)(A - 位I)P As we can write B - 位I in the desired form, we conclude that A - 位I and B - 位I are similar matrices.
03

Part (b): Prove det(A - 位I) = det(B - 位I)

Now, we will show that the determinants of A - 位I and B - 位I are equal. We can use the fact that the determinant of a product of square matrices is the product of their determinants: det(A - 位I) = det(P^(-1)(A - 位I)P) det(A - 位I) = det(P^(-1))det(A - 位I)det(P) We will now use the property that det(P^(-1)) = 1/det(P): det(A - 位I) = (1/det(P))det(A - 位I)det(P) det(A - 位I) = det(B - 位I) Therefore, the determinants of A - 位I and B - 位I are equal.

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