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$T: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ defined as $T(x,y,z) = (2x,z,y)$ is a linear transformation.

I need to prove that the following matrices cannot represent $T$ in ANY basis:

$$\begin{bmatrix} 2 & 0 & 1 \\ 0 & 1 & 0\\ 2 & 0 & 1 \end{bmatrix}$$

$$\begin{bmatrix} 1 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 2 & 1\end{bmatrix}$$

My attempt for the first matrix was to assume (negatively) existence of a basis in which the given matrix is the representation, and left multiplying it by $(x,y,z)^{T}$ to get $(2x+z,y,2x+z)^{T}$.

Can I conclude that given it is not in the form $(2x,z,y)^{T}$, there is NO such basis?

I want to know if this is correct and, in addition (Or alternatively), learn other ways to solve this exercise.

Thank you.

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2 Answers 2

up vote 5 down vote accepted

In the canonical basis: $[T]_e= \left( \begin{matrix} 2 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{matrix} \right)$

Any other representation of $T$ would be similar to $[T]_e$.

Similar matrices have the same trace, and this rules out the first option.

Similar matrices have the same determinant, and $|[T]_e|=-2$. This rules out the second option (Has determinant $2$).

About the reasoning you tried and asked about: It is not correct. The form you were talking about is just the form of the coordinates (The numbers you multiply some set of basis vectors by) and would certainly not retain It's form, generally, in different basis.

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Determinant and trace of a linear transformation don't depend on the basis. The 1st matrix has wrong determinant, and the other has wrong trace.

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Perfect answer; you should consider to make it a bit clearer to the OP. –  Michael Hoppe Feb 20 at 19:56
2  
The second matrix has the right trace, but wrong determinant. The first has both trace and determinant wrong. –  Daniel Fischer Feb 20 at 20:01
    
This is indeed a remarkable answer. –  Dror Feb 20 at 20:28
    
@Dror: I agree, the answer was very brief. The reason: I tried to point to the right direction w/out spelling everything out. This looks like a homework question, and it would benefit OP to transform the hint into a detailed answer. –  Michael Feb 22 at 4:19
    
@Michael first of all, you get the order of the matrices mixed up in your answer. 2nd of all, Linear transformations don't have determinants nor trace. Their matrices representatives have determinant and trace. –  Dror Feb 22 at 4:23

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