Hello guys I have attached the problem.. could someone help me solve it?
I have made "an attempt" on the problem
The matrix of of with respect to , is the matrix satisfying where and are respectively the coordinates of anf with respect to . By a well known theorem:
Besides, is bijective if and only if , and if is bijective, the matrix of with respect to is .
Stupid me - I didn't read the problem correctly. My comments beneath the "%%%%%..." were thinking that you were supposed to find , but you're actually asked to find .
1. Your approach was correct, but you didn't order your columns correctly.
(When a basis is *labelled* for the purpose of writing linear maps as matricies, it will always be assumed that that's the order being used when referencing a position in the matrix.)
Thus .
2. Since , it follows that is surjective (onto).
Thus, from dim(Null(F)) + dim(Image(F)) = 3, it follows that Null(F) = {0}. Therefore F is injective (one to one).
Therefore F is bijective.
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Given , a linear map of real vector spaces and a basis,
and that ,
then show that is a bijection and find its matrix representation with respect to .
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Regarding your work, what are the columns of the matrix you made? What is ? It's . Likewise, you 2nd column is and your 3rd is . So that's not going to be the matrix representation of .
Also, I don't understand why you thought volume was relevant.
It appears that you didn't really know what to do. See FernandoRevilla's post above. The first step is to get those equations for , and in terms of , and . Then, as his post says, you can write down the matrix and, by computing its determinant, show that it's bijective.
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The way you acheive that 1st step is to use that is linear and the equations,
,
to solve for , and in terms of , and .
To you show how you do that, I'll do one:
, so .
Since F is linear, you can simplify the left hand side by:
.
Thus , so .
solves for in terms of , and . Now do it for the others.
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One thing that might help is to label the *vectors* and by and .
Then, using linearity, the equations become:
, where .
You then solve that for and just as you would a regular system of simultaneous equations, *except* remembering that all those things are vectors, not numbers. You can't multiply two vectors or divide by a vector (though you can multiply and divide vectors by real numbers!). However, to solve that system of equations, you won't need to do any of those forbidden acts. Thus, what you write on your paper when you solve it will look identical to how you'd solve a typical system of equations with 3 equations and 3 unknowns.
Anyway, you solve that system for and , which are and , and then you've finished the 1st step, and can proceed to solve the problem as described in FernandoRevilla's post above.