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.

Is the set $U(n,\mathbb R)$ of all upper triangular $n\times n$ matrices over $\mathbb R$ a connected set in $M(n,\mathbb R)$ (with its usual topology after identification with $R^{n^2})?$

I think the answer is yes since connectedness is a productive property, $\mathbb R,\{0\}$ are connected and $$U(n,\mathbb R)=\\\mathbb R\times\mathbb R\times...\times\mathbb R\\\times\{0\}\times \mathbb R\times...\times\mathbb R\\...\\\times\{0\}\times\{0\}\times...\times\mathbb R$$

Please tell me whether the attempt is right or wrong!

share|improve this question
    
Yes, it is right. Also, you could have constructed by hand the path from one upper triangular matrix to another upper triangular matrix. –  Evgeny Aug 9 '13 at 11:04
4  
Also: it is a subspace, so it's even convex. –  Daniel Fischer Aug 9 '13 at 11:06
add comment

2 Answers 2

If by "Productive" you mean "stable under product" then your reasoning is right.

You can also show it "by hand": for any two matrices in $U(n,\mathbb R)$, it is not hard to build a continuous path from one to the other.

share|improve this answer
2  
There is even a linear segment joining the two matrices, which proves that $U(n,\mathbb R)$ is convex. –  lhf Aug 9 '13 at 11:07
add comment

It is more, They are path connected too!

According to your notation, $A,B\in U(n,\mathbb{R})$, The continous path from $[0,1]$ is $f(t)=tA+(1-t)B$

share|improve this answer
add comment

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.