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frozenguy
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Homework Statement
He made some notes, but I'm still confused.
No, you didn't. When asked to find a basis for W, you need to provide a set of linearly independent vectors that spans W. The number of vectors in that set is the dimension.frozenguy said:Well like number one. he gave me 3 out of 10 points.
Now that I look at it, since its in R3, its dimension should be 3 ya? I put two because of the s and t but that doesn't really seem to make much sense to me now..
Did I find a basis in number one?
You could have explained what the bottom row of your augmented matrix meant; namely that there are no solutions for the equation 0x + 0y = 8, which means that <2, 4> is not in the column space of the matrix.frozenguy said:Or like in number two. I thought I did show b is not in the column space. How much farther do I have to go and what could I have done?
Find the vectors x such that Ax = 0. Your work after the matrixfrozenguy said:The only problem I got correct was number three.
What did I do wrong in problem four? I know I didn't find the nullspace, how would I have gone about that? Does the other work look ok up to finding the nullspace?
You found a basis, but the instructor wanted some explanation of why {1, x, x^2} is a basis for the given subspace.frozenguy said:And problem five, I thought I found the basis, but I guess I didn't.
vela said:No, you didn't. When asked to find a basis for W, you need to provide a set of linearly independent vectors that spans W. The number of vectors in that set is the dimension.
Hmm ok I see what you're saying.. And as for getting points off, the only reason I missed points was because he wanted me to say there were no solutions? Isn't that implied? I have 0+0=8 which doesn't make sense. Why do I have to use English when math says it? At least I wrote the conclusion I came to. It only asked to show. Am I being unreasonable here? He took off three points for that, out of ten.Mark44 said:You could have explained what the bottom row of your augmented matrix meant; namely that there are no solutions for the equation 0x + 0y = 8, which means that <2, 4> is not in the column space of the matrix.
A simpler approach that requires no computation is to recognize that the two columns in the matrix are multiples of each other, so the column space is the set of multiples of <-1, 2>. The vector <2, 4> is not a multiple of <-1, 2> so is not in the column space.
The question was posed in English, so the answer to it should also be in English. The math is the tool whereby you reach your conclusion. The notes your instructor wrote for this problem should give you a good idea about what he was looking for; namely that <2, 4> is not in the column space plus some justification that shows why you reached that conclusion.frozenguy said:Hmm ok I see what you're saying.. And as for getting points off, the only reason I missed points was because he wanted me to say there were no solutions? Isn't that implied? I have 0+0=8 which doesn't make sense. Why do I have to use English when math says it? At least I wrote the conclusion I came to. It only asked to show. Am I being unreasonable here? He took off three points for that, out of ten.
frozenguy said:I probably should of taken this class during the 18 week semester and not the six week :/
So I think I have a severe problem when it comes to understanding what a basis is. I guess I can understands its mathematical definitions, but what is a basis? Whats the point, what does it do.
When a set of vectors {v1, v2, ..., vn} spans a vector space W, it just means you can write any vector x from W as a linear combination of the vectors. In other words, you could solve the equationI get so confused sometimes. Maybe its the book or the instructor but I thought linear algebra was supposed to be easier then vector calculus! lol.. What does it mean in words, or english, when a set of vectors span a vector space W.
In problem 1, you were given what the elements of W look like. You could do what you said above or just use regular vector manipulations:So for problem one, should I have chosen t to be 0 and find the resulting coefficient matrix, then the same for s = 0? Then I have two (what are they? sets of vectors?), one multiplied by s, the other by t.
Like s(1,0,2) + t(-1,3,0)?
In addition to what Mark said, I'll add that you should understand that you need to show to the grader that you know what you're doing. That not only means you need to know the material, but you need to be able to express that fact clearly. You have to write down enough so the grader can see your train of thought.frozenguy said:Hmm ok I see what you're saying.. And as for getting points off, the only reason I missed points was because he wanted me to say there were no solutions? Isn't that implied? I have 0+0=8 which doesn't make sense. Why do I have to use English when math says it? At least I wrote the conclusion I came to. It only asked to show. Am I being unreasonable here? He took off three points for that, out of ten.
Basis refers to a set of linearly independent vectors that span a vector space, while dimension is the number of vectors in the basis set. In other words, the dimension of a vector space is the number of basis vectors needed to span the space.
To find the basis of a vector space, you need to find a set of linearly independent vectors that span the space. This can be done by solving a system of equations or using other methods such as row reduction or determinants.
The maximum number of vectors in a basis set is equal to the dimension of the vector space. This is because the basis set must span the entire vector space and any more vectors would result in linear dependence.
Yes, a basis set can have more than one solution. In fact, a vector space can have infinitely many basis sets, as long as they all satisfy the requirements of being linearly independent and spanning the space.
Some common mistakes include not checking for linear dependence, not verifying that the basis vectors span the entire space, and miscalculating the dimension. It is important to double check your work and make sure all the requirements are met to ensure an accurate answer.