- #1
Ted123
- 446
- 0
Homework Statement
[PLAIN]http://img152.imageshack.us/img152/3162/linal.jpg [Broken]
Homework Equations
The Attempt at a Solution
How do I do part (i) and follow the hint?
Last edited by a moderator:
micromass said:What does the sentence "U is a subspace of V" mean? what do you need to check for this?
micromass said:Fine, so which of these three properties is bothering you? and why?
Mark44 said:For ii, you need to take two arbitrary matrices, say M and N, that are in U, and show that M + N is in U. For the two matrices M and N, the only entries that are important are the ones on the diagonal. I suspect that's where the hint will be successful.
You made some unwarranted assumptions in your matrix in part ii; namely, that the entries on the diagonal were 2a, -a, and -a. If the entries on the diagonal for M are a, b, and c, what makes it so that M is in U?
micromass said:Yes, this is a good proof.
micromass said:Well the trace is a linear map [tex]tr:M_{3\times 3}\rightarrow \mathbb{R}[/tex] and thus satisfies
[tex]\dim(ker(tr))+\dim(im(tr))=M_{3\times 3}[/tex]
You will need to find the dimension of ker(tr). So you just need to shom that the dimension of im(tr) is 1, and that the dimension of [tex]M_{3\times 3}[/tex] is 9.
If you didnt see the dimension theorem, then you could also proceed by explicitly finding a base of U...
micromass said:Well, take an equivalence class [A] (so all elements of this class have the same trace as A), to what number would you assign [A]?
micromass said:That's what the exercise is... You need to assign a real number to every equivalence class such that every real number corresponds to a unique equivalence class.
Now, every matrix in an equivalence class has the same trace. So it's obvious you need to do something with traces. What would then be the real number you assign to an equivalence class?
micromass said:Yes! To an equivalence class [A] just assign trace(A).
You'll need to check two things now:
- if [A]=, then the assigned number is also the same
- if [A] and are different classes, then the assigned number is different.
micromass said:Come on, that's not that difficult.
Take [A]=, then A and B are in the same equivalence class. It's easy to see that they have thesame trace.
The other thing is as easy...
micromass said:You've basically proven it youreself.
If A and B are in the same equivalence class, then (by definition) A and B are equivalent. And in page 1, you have shown that [tex]A\sim B[/tex] implies tr(A)=tr(B). So you've already proven it...
small b is a typo, it's supposed to be B, sorry
micromass said:Take a in R. You'll need to find a matrix A such that [A] is labelled by a. Remember that labelling is given by the trace...
micromass said:Yes, that matrix would be labelled by 0 (since the trace of the matrix is 0).
Can you find out what matrix is labelled by 2?
More generally, what matrix is labelled by a?
micromass said:Yes, can you now find in general what matrix is labelled by a?
micromass said:Correct! Thats the solution to this question.
micromass said:Yes, that seems to do it! It looks like you've got it!
The trace of a matrix is the sum of its diagonal elements. It is used to prove non-equivalence by showing that two matrices have different traces, indicating that they are not equivalent.
Yes, the trace can be used to prove non-equivalence for matrices of any size. It is a property that holds for all matrices.
Yes, there are other methods such as row reduction, determinant comparison, and rank comparison. However, using the trace is often a quicker and simpler method.
Yes, it is possible for two matrices to have the same trace and still be non-equivalent. The trace is just one aspect of a matrix and does not provide a complete picture of its properties.
No, there are other properties such as eigenvalues, determinants, and rank that can also be used to prove non-equivalence. It is important to consider multiple properties when determining the equivalence of matrices.