Some ridiculously tough true and false questions (Linear Algebra)

flyingpig
Messages
2,574
Reaction score
1

Homework Statement



http://img844.imageshack.us/img844/8878/truie.th.png

Uploaded with ImageShack.us




The Attempt at a Solution



a) I know that it contains the 0 vector, so one of the conditions for subspace. But how do I determine whether the plane contains the origin?

b) I am going to answer False, because Ax = b is still the columnspace (provided b is not 0). This is only true if b = 0 (which is the nullspace)

c) I honestly do not even know where to begin lol. The addition part is killing me

d) Is this too trivial? Does this fall from Diagonalization? I am going to say it is true. I don't have a formal proof though...
 
Last edited by a moderator:
Physics news on Phys.org
flyingpig said:

Homework Statement



http://img844.imageshack.us/img844/8878/truie.th.png

Uploaded with ImageShack.us




The Attempt at a Solution



a) I know that it contains the 0 vector, so one of the conditions for subspace. But how do I determine whether the plane contains the origin?
What are the coordinates of the origin? How would you tell if any given point is on the plane?
flyingpig said:
b) I am going to answer False, because Ax = b is still the columnspace (provided b is not 0). This is only true if b = 0 (which is the nullspace)

c) I honestly do not even know where to begin lol. The addition part is killing me
Isn't A2 + 5A = A(A + 5I)?
flyingpig said:
d) Is this too trivial? Does this fall from Diagonalization? I am going to say it is true. I don't have a formal proof though...
 
Last edited by a moderator:
a)What are the coordinates of the origin?

The coord of origin is <0,0,0>

How would you tell if any given point is on the plane?

How about 0 + 2(0) + 3(0) = 0 ≠ 4, so it doesn't contain the origin. Oh it is false then...
 
Isn't A2 + 5A = A(A + 5I)?

I swear, I did NOT see the 5A when I did the problem

So det(A(A + 5I)) = 0 = det(A)det(A + 5I) = 0 * det(A + 5I) = 0 = 0

So it is true!
 
flyingpig said:
How about 0 + 2(0) + 3(0) = 0 ≠ 4, so it doesn't contain the origin. Oh it is false then...
Yes.

flyingpig said:
I swear, I did NOT see the 5A when I did the problem

So det(A(A + 5I)) = 0 = det(A)det(A + 5I) = 0 * det(A + 5I) = 0 = 0

So it is true!
Yes.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top