So, I am given the matrix $$A =\begin{pmatrix} 0 & 1 & 0 & 0 & \\ -1 & 2 & 0 & 0 & \\ -2 & 2 & 1 & 0 & \\ 0 & 1 & 0 & -1 \end{pmatrix}$$ which I need to put into Jordan canonical form (something like $A = P^{-1} J P$, where $J$ is the Jordan matrix). I have calculated the characteristic polynomial to be $(\lambda-1)^3(\lambda+1) = 0$, which implies there are two eigenvalues, $\lambda_1 = -1$, with multiplicity of one, and $\lambda_2 = 1$, with multiplicity of three.

Calculating the eigenvectors, I get that there is one eigenvector for $\lambda_1$ (call it $\mathbf{v_1} = (0, 0, 0, 1)^t$) and that $\lambda_2$ has two eigenvectors (call them $\mathbf{v_2} = (0, 0, 1, 0)^t$ and $\mathbf{v_3} = (2, 2, 0, 1)^t$).

So far I understand that I need to find one additional eigenvector since the dimension of $A$ is four, but I have run into trouble time and again while trying to compute it.

Since $\lambda_1$ has one eigenvector, I understand it will have a Jordan block of size one and that $\lambda_2$ will have a Jordan block of size two.

So, $$J = \begin{pmatrix} -1 & 0 & 0 & 0 & \\ 0 & 1 & 1 & 0 & \\ 0 & 0 & 1 & 0 & \\ 0 & 0 & 0 & 1 \end{pmatrix}$$ How do I complete $P$ using generalized eigenvectors, though? Thanks in advance :)

EDIT: I looked up the definition of generalized eigenvectors and got that $(A-I)^2 \mathbf{v_4} = 0$ is spanned by the set $\{(-1, 1, 0, 0)^t, (4, 0, 0, 1)^t\}$. Of these two, I concluded that since only $(4, 0, 0, 1)^t$ is linearly independent relative to $\mathbf{v_1},\mathbf{v_2}$, I should use it as a seed to generate $\mathbf{v_5} = (A-I)\mathbf{v_4} = (-4,-4,-8,-2)^t$. Why is it not possible to use $\mathbf{v_3}$ as a column in $P$, though? What is the rationale behind having to compute $\mathbf{v_5}$?

  • $\begingroup$ There is no fourth linearly independent vector since otherwise $J$ would be diagonalizable. $\endgroup$
    – Levi
    Commented Oct 29, 2019 at 6:24
  • $\begingroup$ Oh, right, sorry. Let me clarify the question. $\endgroup$
    – Dazed
    Commented Oct 29, 2019 at 6:30
  • $\begingroup$ The rank of $(A-I)^2$ is just $1$ (as you can see from $J$ too), so its kernel has dimension $3$, not $2$ as you imply (by giving just two spanning vectors). Indeed, any eigenvector for $\lambda_2=1$ is also in the kernel of $(A-I)^2$, so your $\mathbf v_2$ is in that kernel as well. $\endgroup$ Commented Oct 29, 2019 at 9:03

1 Answer 1


The first important thing to remember is that the Jordan form is only unique up to rearranging the blocks which appear in it. You have found the eigenvalues and the sizes of the blocks and written them in a specific order given by your $J$. Now, you want to find $P$ such that $P^{-1}AP = J$. Let's call the columns of $J$ by the names $u_1,u_2,u_3,u_4$. Then the following must be true:

  1. The first column $u_1$ must be an eigenvector of $A$ corresponding to the eigenvalue $-1$.
  2. The second the and the fourth columns $u_2,u_4$ must be (linearly independent) eigenvectors which correspond to the eigenvalue $1$. In fact, they will be a basis for the corresponding eigenspace.
  3. The third column $u_3$ must satisfy $(A - I)u_3 = u_2$ (and should be linearly independent from $u_2,u_4$).

Let's see what this means.

  1. You can take $v_1 = (0,0,0,1)^t$ to be the first column $u_1$ of $J$ but you an also take any other eigenvector corresponding to the eigenvalue $-1$ such as $(0,0,0,2)^t$. In this case, it is not really important which one you take because the coresponding eigenspace is one dimensional.
  2. The eigenspace corresponding to the eigenvalue $1$ is two dimensional and you have found that it is spanned by $v_2,v_3$. Now, you can try and take them as columns (say $u_2 = v_2, u_4 = v_3$) but you can also take as columns any other basis for the eigenspace.
  3. After you have chosen $u_2,u_4$, the third column $u_3$ must satisfy $(A - I)u_3 = u_2$. Here is where you run into a problem. It turns out that if you try and take $u_2 = v_2$ then the equation $(A - I)u_3 = u_2$ has no solution! However, if you take $u_2 = 8v_2 - 2v_3$ (or any other multiple of this) then the equation has a solution (in fact, infinitely many of them) and one of them is given by $u_3 = v_5$.

Since you don't want to "try" all possible vectors $v$ in the eigenspace and see if the equation $(A - I)u_3 = v$ has a solution, you work backwards. You find an element $u_3 \in \ker (A - I)^2$ (not all elements will work because you want at the end a basis) and then set $u_2 = (A - I)u_3$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .