Exercise 4.1.1

Question

True or false:
\begin{enumerate}[label = \alph*)]
    \addtocounter{enumi}{1}
    \item If a matrix has one eigenvector, it has infinitely many eigenvectors;
    \item There exists a square matrix with no real eigenvalues;
    \item There exists a square matrix with no (complex) eigenvectors;
    \addtocounter{enumi}{1}
    \item Similar matrices always have the same eigenvectors;
    \item The sum of two eigenvectors of a matrix $A$ is always an eigenvector;
\end{enumerate}

Jing Huang · Accepted Answer

(Part) True or false:
\begin{enumerate}[label = \alph*)]
    \addtocounter{enumi}{1}
    \item True, if $A \mathbf{x} = \lambda \mathbf{x}$, $A (\alpha \mathbf{x}) = \lambda(\alpha \mathbf{x})$, $\alpha$ is an arbitrary scalar.
    \item True, like the 2D rotation matrix $R_{\alpha}, \alpha 
eq n \pi$.
    \item False, when discussing in complex space, there are always eigenvalues and as a result, $A - \lambda I$ has nonempty null space.
    \addtocounter{enumi}{1}
    \item False, if $A, B$ are similar and $A = SBS^{-1}$. If $A \mathbf{x} = \lambda \mathbf{x}$, then $SBS^{-1} \mathbf{x} = \lambda \mathbf{x}$. i.e., $B(S^{-1} \mathbf{x}) = \lambda (S^{-1} \mathbf{x})$, $S^{-1} \mathbf{x}$ is an eigenvector of $B$, not $\mathbf{x}$.
    \item False
\end{enumerate}

Exercise 4.1.1

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Exercise 4.1.1

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