Eigenvectors for an eigenvalue

Question

For a $n\times n$ matrix is it possible to have only one eigenvalue which gives $n$ linearly independent eigenvectors?

Answer 1

Take $$A=I_n$$ The eigenvalues are all $1$'s. You can see that the eigenvectors are all linearly independent.

Answer 2

Any matrix of the form $aI_n$, $a \in \mathbb{R}$, (called a scalar matrix) has one eigenvalue and $n$ linearly independent eigenvectors.

Answer 3

I am conscious that I do not answer directly to your question, but it is not bad to know that there exists a completely opposite category of matrices having a unique eigenvalue and a unique eigenvector : the maximum size so-called Jordan blocks of the form :

For example, $5 \times 5$ matrix

$$\begin{pmatrix}a & 1 & 0 & 0 & 0\\ 0 & a & 1 & 0 & 0\\ 0 & 0 & a & 1 & 0\\ 0 & 0 & 0 & a & 1\\ 0 & 0 & 0 & 0 & a\end{pmatrix}$$

will have 5 times value $a$ as its (unique) eigenvalue and the unique eigenvector $e_1$ made of $1$ followed by 4 zeros.