Suppose we have to construct a $3 \times 3$ matrix with only one eigenvalue which has only one linearly independent eigenvector, what should be our approach?
I was asked this in an interview, so first thing that came on my mind was to look for a matrix with $\lambda=2$ on the diagonal such that char poly comes out to be $(\lambda-2)^3=0$.
I was going for something like $A=\begin{bmatrix}2&0&0\\0&2&1\\1&0&2\end{bmatrix}$. It has only one eigenvalue i.e. $\lambda =2$ but when I find $A-\lambda I= \begin{bmatrix}0&0&0\\0&0&1\\1&0&0\end{bmatrix}$ it's only element in null space is zero. And I end up with nothing. I was interrupted in between and asked another question? Was I completely in a wrong direction?
Best Answer
The standard matrix to pick would be a Jordan block $A=\begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{bmatrix}$.
The non-standard matrix to pick would be any conjugate of this one, i.e. $A' = PAP^{-1}$ for some invertible matrix $P$. The matrix you chose in your guess is of this form, with $P$ a permutation matrix $$P = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}.$$