Say we have matrix

$$M= \begin{pmatrix} 2 & 0 & 1 & -3\\ 0 & 2 & 4 & 8\\ 0 & 0 & 2 & 0\\ 0 & 0 & 0 & 3 \end{pmatrix}\DeclareMathOperator{\Id}{Id}$$

It follows that


I found that $\ker(M-2\Id)=\{(1,0,0,0)^{T},(0,1,0,0)^{T}\}$, so dimension 2.

Similarly, we get dimension of $\ker(M-3\Id)=1$.

I want to focus on $ker(M-2\Id)$:

Looking at $ker(M-2Id)^{2}$, we get a basis of $\{(1,0,0,0)^{T},(0,1,0,0)^{T}, (0,0,1,0)^{T}\}$, so dimension 3.

In our notes, we have written down:

Find a vector $v \in \ker(M-2\Id)^{2}$, such that $v \notin \ker(M-2\Id)$. It follows that $Mv \in \ker(M-2\Id)$ ( First question, should this not be $(M-2\Id)v \in \ker(M-2\Id)?).$ How does this help us in terms of the Jordan Form? I'm not sure how invariant subspaces fit into all of this either, other than the fact $\ker(M-2\Id)\subset \ker(M-2\Id)^{2}$.

An intuitive explanation would be of great assistance.

  • 2
    $\begingroup$ Yes, it should be $(M-2\operatorname{id})v\in\operatorname{ker}(M-2\operatorname{id})$ for $v\in\operatorname{ker}(M-2\operatorname{id})^2$. $\endgroup$
    – Christoph
    Jun 9, 2018 at 11:26

2 Answers 2


$\DeclareMathOperator\ker{ker}\DeclareMathOperator\id{id}$Here's the idea: Denote by $H_i=\ker(f-\lambda\id)^i$ the generalized eigenspaces and note that they form a chain $H_1\subseteq H_2 \subseteq \cdots$. Pick a vector $v_k\in H_k\setminus H_{k-1}$. Now define the vectors $v_{k-1},\dots,v_1$ by letting $v_i = (f-\lambda\id)(v_{i+1})$ for $i=k-1,\dots,1$. Note that $v_i\in H_i\setminus H_{i-1}$. Hence, the vectors $v_1,\dots,v_k$ will be linearly independent and by construction we have \begin{align*} f(v_k) &= \lambda v_k + v_{k-1}, \\ f(v_{k-1}) &= \lambda v_{k-1} + v_{k-2}, \\ &\,\,\,\vdots\\ f(v_2) &= \lambda v_2 + v_1, \\ f(v_1) &= \lambda v_1. \end{align*} Thus, the subspace $U=\langle v_1,\dots,v_k\rangle$ is $f$-invariant and the matrix of $f$ on $U$ with respect to the basis $v_1,\dots,v_k$ is $$ \begin{pmatrix} \lambda & 1\\ &\lambda & 1 \\ & & \lambda \\ & & & \ddots & 1 \\ & & & & \lambda \end{pmatrix}. $$


$\DeclareMathOperator{\Id}{Id}$In Your case You already found a vector $v\in ker (M-2\Id)^2\backslash ker(M-2\Id)$, namely $v=(0,0,1,0)^T$ and $(M-2\Id)v=(1,4,0,0)^T$. Furthermore $M(1,4,0,0)^T=2(1,4,0,0)^T$ and $Mv=2v+(1,4,0,0)^T$. This corresponds to the Jordan block $$\begin{pmatrix}2&1\\0&2\end{pmatrix}.$$ The subspace $\operatorname{span}\{(0,0,1,0)^T,(1,4,0,0)^T\}$ is thus an invariant subspace.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.