Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am currently studying for my linear algebra exam and I got quite confused when trying to find the Jordan normal form for some matrix. Let $$A = \begin{pmatrix} 2 & 0 & 0 & 1\\ 0 & 0 & -1 & 0\\ -1 & 0 & 1 & -1\\ 0 & 1 & 1 & 1 \end{pmatrix}.$$ The eigenvalue is $\lambda = 1$ with multiplicity 4. Then, $$\ker(A-I)^2 = \langle \begin{pmatrix} 1\\ -1\\ 0\\ 0 \end{pmatrix}, \begin{pmatrix} 1\\ 0 \\ -1 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 \\ 0 \\ 0 \\ -1 \end{pmatrix} \rangle$$ and $\ker(A-i)^3 = \mathbb{R}^4$. Now I want to find $v \in \ker(A-I)^3$ such that $v \notin \ker(A-I)^2$. The solution now suggests me to take $$v = \begin{pmatrix} 1\\ 0 \\ 0 \\ 0 \end{pmatrix}$$ and I understand why this is a possible choice. However, if I am not mistaken, also $e_2, e_3$ and $e_4$ are a valid choice. How is this possible? Where is my mistake? Or are really all four vectors a valid choice?

EDIT: I think I misformulated the question - my mistake.

I know that $\ker(A-I)^2 \subset \ker(A-I)^3$ and I want to find $v \in \ker(A-I)^3$ such that $$\langle v \rangle \oplus \ker(A-I)^2 = \ker(A-I)^3.$$ Now if I think about this, $\ker(A-I)^2$ is a subset of $\ker(A-I)^3$ and I want to find a basis of the subset of $\ker(A-I)^3$ which - when added to $\ker(A-I)^2$ - gives $\ker(A-I)^3$. Then, if the basis vector $e_1$ spans this whole subset that means that the subset contains only vectors of the form $c \cdot e_1$. But if I can also take $e_2$, that means the subset contains only vectors of the form $c \cdot e_2$ which is impossible. Where is my mistake?

Sorry again for the incorrectly formulated original question.

share|improve this question
    
Every vector not in the hyperplane H=ker(A-I)^2 is a solution. The analogue in two dimensions would be that if the kernel is H=Vect(e_1-e_2) every vector not in the line H=Vect(e_1-e_2) is a solution, for example e_1 and e_2 are. You see the two basis vectors are solutions. –  Did Aug 14 '11 at 16:49
1  
@Didier Piau - I think the question isn't carefully worded. It seems that Huy understands that there are lots of choices to complete a basis. The question is probably more along the lines of: Since I can use different bases to put a matrix into Jordan form doesn't this give different answers? How is this possible? Or something like that. –  Matt Aug 14 '11 at 16:58
    
When I asked the question, I was in a hurry and totally misformulated my question. Sorry about that. I hope the added part makes it clear now. –  Huy Aug 14 '11 at 17:12
    
I think the confusion stems from the fact that if I have a subspace $W \subset V$ then there are usually a lot of choices for a $W' \subset V$ such that $V = W \oplus W'$. –  Dylan Moreland Aug 14 '11 at 17:29
2  
...I want to find a basis of the subset of ker(A−I)^3 which - when added to ker(A−I)^2 - gives ker(A−I)^3... As I said in my first comment, the error (if any) is to think that for a given (strict) vector subspace W of the vector space V there is a unique subset S of V such that the direct sum of Vect(S) and W is V. There are plenty. And there are even plenty of subspaces U such that the direct sum of U and W is V. I can only recommend to study the 2D example I gave. –  Did Aug 14 '11 at 17:34

2 Answers 2

up vote 2 down vote accepted

Every vector not in the hyperplane $H=\ker(A-I)^2$ is a solution. The analogue in two dimensions would be that if the kernel is $H=\text{Vect}(e_1-e_2)$, then every vector not in the line $H$ is a solution, for example both $e_1$ and $e_2$ are. One sees the two basis vectors are solutions.

Thus, the error (if any) is to think that for a given (strict) vector subspace $W$ of the vector space $V$ there is a unique subset $S$ of $V$ such that $\text{Vect}(S)\oplus W=V$. There are plenty. And there are even plenty of subspaces $U$ such that $U\oplus W=V$, as the 2D example I gave shows.

Maybe the origin of the trouble is to confuse the Euclidean space $E=\mathbb R^4$ and the (simple) vector space $V=\mathbb R^4$. The former structure is the latter plus the usual scalar product. The orthogonal supplement of a subspace $W$ of the Euclidean vector space $E=\mathbb R^4$ is indeed unique, in the case at hand it is $\text{Vect}(u)$ with $u=e_1+e_2+e_3+e_4$. But the supplements of subspace $W$ of the vector space $V=\mathbb R^4$ are many.

share|improve this answer

I see no mistake. Denote your three basis vectors of $\ker(A-I)^2$ as $v_1,v_2,v_3$, with $v_1=(1,-1,0,0)^T$. You can always extend $\{v_1,v_2,v_3\}$ to a basis of $\mathbb{R}^4$ by adding a fourth vector $u\notin{\rm span}\{v_1,v_2,v_3\}$, but then any fifth vector would be a linear combination of $v_1,v_2,v_3$ and $u$. For example, you may add $e_1$ or $e_2$ to $\{v_1,v_2,v_3\}$ to form a basis of $\mathbb{R}^4$, but you cannot add both, because five vectors in $\mathbb{R}^4$ would make a linearly dependent set. (In particular, $e_1-e_2=v_1$.)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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