# A $3 \times 3$ matrix with one eigenvalue and one eigenvector?

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?

• The only element in its null space is not the zero vector! Look carefully at second column. It has all its entry zero making the second component of the vector free to take any value! Thus, the null space will have a vector other than the zero vector. – Aniruddha Deshmukh Jul 22 '17 at 8:48
• A 3D rotation that preserves the origin works. – Kaj Hansen Jul 22 '17 at 8:56
• Jordan Block is the key. – Surb Jul 22 '17 at 8:57

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 $$P = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}.$$

I would move the $1$ at $(3,1)$ position to $(1,2)$ position.

$$A=\begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2 \end{bmatrix}$$

$$A-2I = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$$

The nullity of $A-2I$ is $1$, hence there is one independent eigenvector.

Also, note that your working was right until you misjudged that there is only the zero solution. Your matrix actually work. For your matrix, the rank of your $A-\lambda I$ is $2$, the nullity is $1$ as well. My choice that differs from yours is simply due to I know Jordan block works.