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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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Most popular questions from this chapter

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

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