A \(3 \times 3\) real matrix need not have any real eigenvalues.

Question

True or False - and Why?
a) A \(3 \times 3\) real matrix need not have any real eigenvalues.
b) If an \(n \times n\) matrix \(A\) is invertible, then it is diagonalizable.
c) If \(A\) is a \(2 \times 2\) matrix both of whose eigenvalues are 1 , then \(A\) is the identity matrix.
d) If \(\vec{v}\) is an eigenvector of the matrix \(A\), then it is also an eigenvector of the matrix \(B:=A+7 I\).

Answer 1

a) True. A \(3 \times 3\) real matrix can have complex eigenvalues, which are not real numbers.

 

b) False. An \(n \times n\) matrix \(A\) can be invertible but not diagonalizable. For example, the Jordan form of a matrix can be invertible but not diagonalizable.

 

c) False. If an \(2 \times 2\) matrix has both eigenvalues equal to 1 , it does not necessarily mean that it is the identity matrix. For example, the matrix:
\[\begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix}\]

has both eigenvalues equal to 1 , but it is not the identity matrix.

 

d) True. If \(A\) is a square matrix and \(\vec{V}\) is an eigenvector of \(A\) with eigenvalue \(\lambda\), then: \[A\vec{v} = \lambda;\vec{v}\]

Adding a multiple of the identity matrix to both sides of the above equation yields:

\[(A + 7I)\vec{v} = (\lambda; + 7)\vec{v}\]
This shows that if \(A\) has an eigenvector with eigenvalue \(\lambda\), then \(A+7I \) has an eigenvector with eigenvalue \(\lambda+7\)