# Finding the Jordan form of a particular degenerate matrix

Find the Jordan form of $$A = \begin{bmatrix} 1 & 1 & 1 \\ -1 & -1 & -1 \\ 1 & 1 & 1 \end{bmatrix}.$$

Characteristic polynomial: $p(\lambda) = (\lambda-1)(\lambda^2-1+1) + (\lambda-1+1) - (-1+\lambda+1) = \lambda^3-\lambda^2$

Eigenvalues: $\lambda_1 = 0, \lambda_2 = 0, \lambda_3 = 1$

Eigenvectors: $v_1 = (-1,1,0), v_2 = (-1,0,1), v_3 = (1,-1,1)$

I would think that the Jordan form is $$J = \begin{bmatrix} \lambda_1 & 0 & 0 \\ 1 & \lambda_1 & 0 \\ 0 & 0 & \lambda_2 \\ \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}.$$

If I query WolframAlpha with

jordan form of {{1,1,1},{-1,-1,-1},{1,1,1}}

I get:

$J = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}$

Which is correct? Please explain. Thank you!

• You have two (linearly independent) eigenvectors for the eigenvalue $0$. This means that there will be two Jordan blocks for this eigenvalue. – Martin Sleziak Mar 24 '15 at 6:48

## 1 Answer

If an $n \times n$ matrix has $n$ linearly independent eigenvectors, as is the case here, then the matrix is not defective, so all the Jordan blocks are $1 \times 1$, that is, the Jordan normal form is diagonal. In this case, the eigenvalues are $0, 0, 1$, so the Jordan form is $\text{diag}(0, 0, 1)$, in agreement with W.A.