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Chapter 5: Q5.3-27E (page 267)

Question: Show that if A is diagonalizable and invertible, then so is \({A^{ - {\bf{1}}}}\).

Short Answer

Expert verified

Since \({D^{ - 1}}\) is a diagonal matrix, therefore the matrix \({A^{ - 1}}\) also a diagonal matrix.

Step by step solution

01

Check if matrix A is diagonalizable

If A is diagonalizable, then \(A = PD{P^{ - 1}}\).

Where P is an invertible matrix and D is a diagonal matrix. As the matrix A is invertible, so 0 is not an eigenvalue of A. So, the diagonal entries in D are not zero, and D is invertible.

02

Check if a matrix \({A^{ - {\bf{1}}}}\) is diagonalizable

According to the theorem on the inverse of a product,

\(\begin{array}{c}{A^{ - 1}} = {\left( {PD{P^{ - 1}}} \right)^{ - 1}}\\ = {\left( {{P^{ - 1}}} \right)^{ - 1}}{D^{ - 1}}{P^{ - 1}}\\ = P{D^{ - 1}}{P^{ - 1}}\end{array}\)

The matrix \({D^{ - 1}}\) is a diagonal matrix, therefore the matrix \({A^{ - 1}}\) also a diagonal matrix.

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Most popular questions from this chapter

Question: Let \(A = \left( {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right)\). Use formula (1) for a determinant (given before Example 2) to show that \(\det A = ad - bc\). Consider two cases: \(a \ne 0\) and \(a = 0\).

Question: Find the characteristic polynomial and the eigenvalues of the matrices in Exercises 1-8.

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