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 have a linear map $F: R^3 \longrightarrow R^3$ and the vectors

$F(v_1)=(0,1,0), F(v_2)=(1,0,0), F(v_3)=(0,0,1)$

$v_1=(1,0,0), v_2=(1,1,0), v_3=(1,1,1)$

I already found the matrix for the map with respect to the standard basis

$A=\begin{matrix} 1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{matrix}$

$a) $Now I have to determine if the map is invertible and isometry. About if it is invertible, is enough to see if $A^{-1}$ exists?

And about isometry I know that I have to prove that the length preserve, but I don't really know how to do it.

$b)$ I have to define the vector $v_1, v_2, v_3$ as an alternative (new) basis in $R^3$ and find the matrix F for this new basis. To find the matrix is the same as before once I have the new basis.

share|improve this question

1 Answer 1

Yes, it is sufficient to check the invertibility of the matrix to check the if the operator is invertible. Your operator is not an isometry as you can easily see by comparing the lengths of $v_{2}$ and its image. Now, for the matrix in new basis, observe that the first column is the coordinates of image of first basis vectors and so on.

share|improve this answer
    
Thank you very much. So the new basis can be: {(1,1,0),(1,0,0),(1,1,1)})? –  louis Jun 20 '13 at 12:15
    
Yes, that can be the new basis. And also, if you like an answer, you can upvote it, and even accept it. –  Vishal Jun 21 '13 at 6:31
1  
@louis It is a good practice to accept answers to acknowledge the users who help you and also to prevent bumping the question to the top later. –  Vishal Jul 14 '13 at 13:03

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.