Jordan Canonical Forms of an endomorphism in $\mathbb{R}^3$

If f is an endomorphism of $\mathbb{R}^3$ with only one eigenvalue $\lambda \in \mathbb{R}$ with algebraic multiplicity 3. The matrix of $f$ in standard basis is $A$.

a) What is the characteristic polynomial of $f$?

b) Show all possible forms of Jordan Canonical form for $A$

a) Since $\lambda$ is the only eigenvalue and it has algebraic multiplicity = 3, we get:

$$(n-\lambda)^3$$

b) \begin{pmatrix} \lambda & 0 & 0\\ 1 & \lambda & 0\\ 0 & 0 & \lambda\\ \end{pmatrix}

\begin{pmatrix} \lambda & 0 & 0\\ 0 & \lambda & 0\\ 0 & 1 & \lambda\\ \end{pmatrix}

\begin{pmatrix} \lambda & 0 & 0\\ 1 & \lambda & 0\\ 0 & 1 & \lambda\\ \end{pmatrix}

Is there anything wrong?! I'm new to this Jordan canonical form...

• I think only the third matrix is correct. Each Jordan block of a matrix must contain distinct eigenvalues. Hence, since there is only one distinct eigenvalue, there is only one Jordan block, which looks like the third matrix. Of course, the characteristic polynomial is correct. – астон вілла олоф мэллбэрг Oct 26 '16 at 0:08
• It's also possible that $A = \lambda I$. Moreover, the first and second Jordan canonical forms are effectively the same; those two matrices are similar. – Omnomnomnom Oct 26 '16 at 0:08
• so other than the third one, what would be the other forms for jordan canonical form? Thanks! – Bruno Reis Oct 26 '16 at 0:11
• @Omnomnomnom But wait... if $A=\lambda I$ it would be diagonalizable... hence, I wouldn't use anything related to Jordan form... – Bruno Reis Oct 26 '16 at 0:15
• Yep, that's right – Omnomnomnom Oct 26 '16 at 0:25

The three are correct, minus the fact that it is common to put the ones in the upper part. The one you are missing is $$\begin{bmatrix}\lambda &0&0\\0&\lambda&0\\0&0&\lambda \end{bmatrix}.$$
• Thanks for your answer Martin! But if $A=\lambda I$ it would be diagonalizable... hence, I wouldn't use anything related to Jordan form... – Bruno Reis Oct 26 '16 at 0:17