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.

Possible Duplicate:
Questions about rank and eigenvalues of a matrix

Let $$A=\begin{pmatrix}1&w&w^2\\w&w^2&1\\w^2&w&1\end{pmatrix}$$ Where $w$ is a complex no. s.t. $w^3=1$. Its clear by adding columns of matrix that $0$ is an eigen value of $A$.

Do there exist linearly independent vectors $u,v\in\mathbb{C}^3$ s.t. $Au=Av=0$?

can anyone help me please....

share|improve this question

marked as duplicate by DonAntonio, joriki, draks ..., Amr, QiL Dec 17 '12 at 12:46

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

2  
You may learn how to type mathematical formulae here. It is a basic etiquette to format your question in a readable form. –  user1551 Dec 17 '12 at 10:24
    
Also, if you read a post and want to see how a mathematical expression in it was typed, you may right-click on it and choose "Show Math As > TeX Commands" from the context menu. –  user1551 Dec 17 '12 at 10:30
    
So you want to find the rank of A? What do eigenvalues have to do with it? –  Karolis Juodelė Dec 17 '12 at 10:48
    
I want to find if there exist linearly independent vectors u,v∈C3 s.t. Au=Av=0? –  prakash Dec 17 '12 at 10:50

1 Answer 1

You just want to what is the rank of $A$. Notice that the second and third column are just the first column scaled by $w$ and $w^2$ respectively. Thus, the rank is at most $1$ (so it's precisely $1$, since $A$ is not the zero matrix). In other words, the dimension of the kernel of $A$ is $2$, i.e. there are two linearly independent vectors $u,v$ with $Au = Av = 0$. You can even find them explicitly: one is $(1,1,1)$, another is $(1,w,w^2)$ (other choices are of course possible).

share|improve this answer
    
rank of A is 2 not 1 –  prakash Dec 17 '12 at 11:03
1  
third column aren't the first column scaled by w2 so rank of A is 2 not –  prakash Dec 17 '12 at 11:05
    
third column aren't the first column scaled by w2 so rank of A is 2 not 1 –  prakash Dec 17 '12 at 11:08
    
@prakash: are you sure? to me it looks as if $w^2 \cdot (1,w,w^2) = (w^2,w^3,w^4) = (w^2,1,w)$. –  Feanor Dec 17 '12 at 11:08
    
@Feanor My fault: when I TeX-ified the OP's question, I wrongly typed the last row of the matrix as $(w^2,1,w)$. It should be $(w^2,w,1)$. –  user1551 Dec 17 '12 at 11:42

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