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Exercise 5.6.8 Suppose T R3 R? is a linear transformation given byTx=Axwhere A is a 3 x 3 matrix: Show that T iS an isomorphism if and only if A is invertible:...

Question

Exercise 5.6.8 Suppose T R3 R? is a linear transformation given byTx=Axwhere A is a 3 x 3 matrix: Show that T iS an isomorphism if and only if A is invertible:

Exercise 5.6.8 Suppose T R3 R? is a linear transformation given by Tx=Ax where A is a 3 x 3 matrix: Show that T iS an isomorphism if and only if A is invertible:



Answers

Determine whether the matrix transformation $T_{A}: R^{3} \rightarrow R^{3}$ is an isomorphism. $$A=\left[\begin{array}{rrr} 1 & -1 & 0 \\ 0 & 0 & 2 \\ -1 & 1 & 0 \end{array}\right]$$

Hello there. Okay. In this exercise we have here a matrix transformation. Uh and we need to determine if this is a nice organism. So in order to to show that these matrix indeed corresponds to a nice room of prism, we need to show that if we reduce this matrix to the relation forum, we have three linearly independent Rose. Okay. That means that we're mapping from from the three dimensional space To a three dimensional space. And also that these vectors are the veterans that we're going to mob arenal independent. And therefore we have a 1-1 relate. So basically, we just need to apply the Rose operations in order to let your mind if this is we're not a nice summer fizz. Um Okay, so we have these matrix. So the first operation that we're going to do is swab these two rules The first and the 2nd 1. So basically are one an art you're going to be swapped and we obtain the following matrix one, 01 -1. and here the 3rd 1 is the same group. Great. Now we need to put a zero in this position here. So in order to do that, we need to just some the third row With the 1st 1. After performing this operation, we have one, 02, 0 1, 9. It's one zero here, one and two. And the last thing is just put zero in disposition below the one in the taking position of the diagonal. So in order to do that, we just need to take the third road on. It's obstruct the 2nd 1. And in that sense we obtain the forming matrix 0, -1. 00 one. Rule three here. Yes. Three. Well, The last step is technically justified in the 3rd row by three. So we take our three and we divide by three, which is just one last step and went with the following matrix after all these reparations. And you can observe that we have we don't have any free variable here, Technically what we're saying is that these three rows and three columns are linearly independent and therefore we have a 1-1 relation. But also we have that we're mapping a three dimensional space to a three dimensional one. Therefore this transformation is onto and therefore this T. A. Is a nice rheumatism.

All right. So for this question, you want to find out if the linear transformation of the norm is going to be um a if the transformation of uh of a vector from our three to the norm is going to be a linear transformation. And we're going to look at homogeneity. And we're going to look at the second condition which is going to what you're going to compare the transformation of two vectors. The addition of two vectors. Yeah. Uh then transformed compared to to transform vectors, then added, I'm gonna be the same factor. So 101 and 010 Mhm. So for the first, for the first time you can see 10 one plus 010 is just gonna be 111 And the norm of that is just going to be all of the each individual vector squared and then added and square root. So that's going to be the root of three because one square dishes one. And then when we transform these vectors first into the norm, we then get one squared plus zero plus one. So it's going to be a roof two plus zero plus one squared plus one, which is gonna be one. Just gonna give us the root of two plus one. And since these are not the same thing, I mean they're not equal. And so this is not a linear transformation. Mm. Yeah.


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