Does there exist a non-diagonalizable 3x3 matrix that has precisely 1 real eigenvalue and a multiplicity of 1? When it comes to multiplicity I'm trying to find a matrix that would give me something like $(\lambda-1)^3$ as the eigenvalue. This factors down to $\lambda^3 - 3\lambda^2+3\lambda-1$ so you could say the multiplicity is 3 but you can also say that it only has 1 real root. So could I use this to find a non-diagonalizable 3x3 matrix with only 1 eigenvalue. So would such a matrix exist?
3 Answers
Sure
\begin{bmatrix} 0&-1&0\\1&0&0\\0&0&1\end{bmatrix}
has one real eigenvalue of multiplicity 1.
For example
$$\begin{pmatrix}x&1&0\\0&x&1\\0&0&x\end{pmatrix}$$
has one unique eigenvalue $\;x\;$ of algebraic multiplicity $\;3\;$ and geometric multiplicity $\;1\;$ (if this is what you meant) , for any $\;x\in\Bbb R\;$ .
Assuming the matrix to be real, one real eigenvalue of multiplicity one leaves the only possibility for other two to be nonreal and complex conjugate. Thus all three eigenvalues are different, and the matrix must be diagonalizable.
If the matrix can be complex then it is possible to find a non-diagonalizable matrix with the only real eigenvalue of multiplicity one, for example $$ \begin{bmatrix} 1 & 0 & 0\\ 0 & i & 1\\ 0 & 0 & i \end{bmatrix} $$
