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1. ### Linear Algebra: Spans and Dimensions

Ok, so assuming we reduce the set in U to a basis: then if v is in U the given generating set for W is linearly dependent. Then v would be removed from this set, making it linearly independent. This implies that U and V share a basis, and thus their dimensions are the same. If v is not in U...
2. ### Linear Algebra: Spans and Dimensions

Okay, so if I reduced the generating set for U to a basis, I would know its linearly independent, but then where would I go with it?
3. ### Linear Algebra: Spans and Dimensions

Homework Statement Given v1, v2 ... vk and v, let U = span {v1, v2 ... vk} and W = span {v1, v2 ... vk, v}. Show that either dim W = dim U or dim W = 1 + dim U. The Attempt at a Solution I'm not really sure where to start. If I knew that {v1, v2 ... vk} was linearly independent, then it would...
4. ### Linear Algebra: Linear indepency of a set of Polynomials

Thanks for the reply! Today I was able to prove (1) by contraposition but I thought that meant I had to prove (2) by contraposition as well. Can I just prove the positive statement for (2)? Edit: Wait, taking statement A as {p, q, pq} is linearly independent and statement B as deg p ≥ 1 and deg...
5. ### Linear Algebra: Linear indepency of a set of Polynomials

Homework Statement Let {p, q} be linearly independent polynomials. Show that {p, q, pq} is linearly independent if and only if deg p ≥ 1 and deg q ≥ 1. Homework Equations λ1p + λ2q = 0 ⇔ λ1 = λ2 = 0 The Attempt at a Solution λ1p + λ2q + λ3pq = 0 I know if λ3 = 0, then the coefficients of...
6. ### Linear Algebra: Prove that the set of invertible matrices is a Subspace

Homework Statement Is U = {A| A \in nℝn, A is invertible} a subspace of nℝn, the space of all nxn matrices? The Attempt at a Solution This is easy to prove if you assume the regular operations of vector addition and scalar multiplication. Then the Identity matrix is in the set but 0*I and...
7. ### Differentiating This Function

Do you know what I did? I simplified the original function to get the one I posted here, just checked and they're not the same. That's why the derivative didn't match up. :surprised I think I've about got it sorted out now.
8. ### Differentiating This Function

I'm sorry, I keep typing it wrong. There's only meant to be a single variable on the right-hand side. Its fixed now.
9. ### Differentiating This Function

Somebody want to help with this derivative: y = (7.75x3 + 2250) / 750x I differentiate it with the quotient rule and get: dy/dx = 31x / 1500 - 3 / x2 But that's wrong. It's got a zero around 5 - 6 and the original function has its minimum at (4.03, 1.12). Not sure what I did wrong.
10. ### Help with This Derivative

No, I multiplied that in at the very beginning. I scaled up the fraction by 1000 times so I had a 4 in the numerator rather than a decimal. After fixing what Reptilian mentioned, my derivative looks correct, and the zero is at the x-value of that local minimum, so I think it's right.
11. ### Help with This Derivative

Using the quotient rule on the 4∏r2 / 3000 term. The denominator of the differentiated expression is 16∏2r4. A pi and an r cancel from the numerator which was 8∏r(3000).
12. ### Help with This Derivative

I typed that wrong, I did get 4/3000 (1/750).
13. ### Help with This Derivative

Yeah I noticed. No worries. :tongue: Do you know what's wrong with the derivative. I found just with the power rule and the quotient rule on the last term.
14. ### Help with This Derivative

After simplifying I get: (0.003 + 0.004√2) ∏r2 + 3/r - (4∏r2) / 300 Which is equivalent now according to a graphing calculator. Differentiating, I get: (0.006 + 0.008√2) ∏r - 3r-2 - 1500/∏r3 But thats not right. i.e. it has positive values when the slope of the function is negative. :/
15. ### Help with This Derivative

Does somebody want to help me with differentiating this function: (0.003 + 0.004\sqrt{2})∏r2 + 0.004∏r [(2250 - ∏r3) / 3∏r2)] Originally I tried simplifying it by distributing the 0.004∏r into the fraction, but whenever I did that the two expressions were never equivalent when I checked them...