Question

question 3

if a and b are similar n by n matrices, then they have the same eigenvalues.

true
false

Explanation:

Define similar matrices

Two \(n \times n\) matrices \(A\) and \(B\) are similar if there exists an invertible \(n \times n\) matrix \(P\) such that:

$$B = P^{-1}AP$$

Relate characteristic polynomials

We examine the characteristic polynomial of \(B\), which determines its eigenvalues:

$$\det(B - \lambda I) = \det(P^{-1}AP - \lambda P^{-1}IP)$$

Factor out the invertible matrix

Using the distributive property of matrix multiplication, we factor out \(P^{-1}\) and \(P\):

$$\det(P^{-1}AP - \lambda P^{-1}IP) = \det(P^{-1}(A - \lambda I)P)$$

Apply determinant properties

Using the multiplicative property of the determinant, \(\det(XY) = \det(X)\det(Y)\):

$$\det(P^{-1}(A - \lambda I)P) = \det(P^{-1})\det(A - \lambda I)\det(P)$$

Since \(\det(P^{-1}) = \frac{1}{\det(P)}\), the terms cancel:

$$\det(B - \lambda I) = \det(A - \lambda I)$$

Conclude eigenvalue equality

Because \(A\) and \(B\) share the exact same characteristic equation, they must have the same eigenvalues. Thus, the statement is true.

Answer:

• True (Correct answer)
• False