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Chapter 5: Q5.3-24E (page 267)

Question: A is a \({\bf{3}} \times {\bf{3}}\) matrix with two eigenvalues. Each eigenspace is one-dimensional. Is A diagonalizable? Why?

Short Answer

Expert verified

According to diagonalize theorem, A cannot be diagonalizable.

Step by step solution

01

Write the given information

Matrix A is of the order \(3 \times 3\) has 2 eigenvalues and all the eigenspace for the matrix is one-dimensional.

02

Find that A is diagonalizable

Consider the matrix A has \({{\bf{v}}_1}\) and \({{\bf{v}}_2}\) as its eigenvectors that span two one-dimensional space. If \({\bf{v}}\) is any other eigenvector, then it belongs to one of the eigenspaces and \({\bf{v}}\) will be multiple of either \({{\bf{v}}_1}\) and \({{\bf{v}}_2}\).

So, three linearly independent eigenvectors cannot exist.

Therefore, according to diagonalize theorem, A cannot be diagonalizable.

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Let\(D = \left\{ {{{\bf{d}}_1},{{\bf{d}}_2}} \right\}\) and \(B = \left\{ {{{\bf{b}}_1},{{\bf{b}}_2}} \right\}\) be bases for vector space \(V\) and \(W\), respectively. Let \(T:V \to W\) be a linear transformation with the property that

\(T\left( {{{\bf{d}}_1}} \right) = 2{{\bf{b}}_1} - 3{{\bf{b}}_2}\), \(T\left( {{{\bf{d}}_2}} \right) = - 4{{\bf{b}}_1} + 5{{\bf{b}}_2}\)

Find the matrix for \(T\) relative to \(D\), and\(B\).
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