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Edited: Suppose $M$ represents a linear transformation from $\mathbb R^3\to \mathbb R^3$, where $M = \left( \begin{smallmatrix} a&1&0\\ 0&a&1 \\ 0&0&a \end{smallmatrix} \right)$. What are the bases for the domain and range s.t. $M$ is the identity?

(Sorry about the previous edition.)

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You mean existence of $S$ such that $S^{-1}M S = I$? –  user13838 Sep 13 '11 at 9:53
    
You can't. An identity matrix remains the same under any conjugation. Rewrite the question. –  KCd Sep 13 '11 at 9:55
    
@KCd: Edited. Thanks for pointing that out. –  userN Sep 13 '11 at 9:59
    
The question doesn't makes sense. You speak of "the basis" but then of two different bases for two different spaces. To which space does "the basis" refer? Also, $M$ is a matrix and not a basis-dependent entity. Do you mean something like "in which basis is the matrix for the linear transform whose matrix in the canonical basis is $M$ the identity matrix?"? –  joriki Sep 13 '11 at 10:03
    
Edited. :-) Sorry about the confusion. –  userN Sep 13 '11 at 10:13

1 Answer 1

up vote 2 down vote accepted

Thanks for improving the formulation of the question. It's still not really clear, since $M$ is a given matrix and it makes no sense to ask when it is the identity. A clearer formulation of what I think you mean would be:

Suppose a linear transformation from $\mathbb R^3$ to $\mathbb R^3$ is represented by the matrix $M = \left( \begin{smallmatrix} a&1&0\\ 0&a&1 \\ 0&0&a \end{smallmatrix} \right)$ if the canonical basis is used for both the domain and the codomain. What are bases of the domain and the codomain such that the matrix representing the transformation with respect to these bases is the identity matrix? (Note the lack of a definite article on "bases", which in your version seems to imply that there is a unique solution, which is not the case.)

The are infinitely many pairs of bases with this property. Two particularly simple ones are obtained if you use the canonical basis for one of the spaces and only look for a new one for the second one. This is particularly straightforward if you only change the basis of the codomain: Since the columns of the matrix are the images of the basis vectors of the domain, represented in the basis of the codomain, the matrix will be the identity matrix if you choose the column vectors of $M$ as the basis vectors of the codomain.

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Thanks a lot, joriki! :-) –  userN Sep 13 '11 at 11:12

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