Rank of a Matrix and whether the columns span R12

testme
Messages
68
Reaction score
0

Homework Statement


Let M be the 12 x 7 coefficient matrix of a homogeneous linear system, and suppose that this system has the unique solution 0 = (0, ..., 0) \in ℝ7.

1. What is the rank of M.
2. Do the columns of M, considered as vectors in ℝ12, span ℝ12.

Homework Equations





The Attempt at a Solution



1. Well since the matrix is a homogeneous matrix the rank of M can be from 0 \leq rankM \leq 7.

so then rank has a rank of 7 I believe

2. I'm not sure how to solve this but if the solution is in ℝ7 does that automatically mean the vectors can't span ℝ12 and aside from that a set of all 0s can't span ℝ7 can it?
 
Physics news on Phys.org
By 12 x 7 I believe you mean column x row in this case? I ask this because if you had it the other way around, your zero vector would be in ℝ12 not ℝ7.
 
No, my professor does it 12 x 7, row x column (I thought that's the norm way of doing it?)

Maybe the ℝ7 is a typo by the professor and it should be ℝ12

I'm not completely sure.
 
testme said:
2. Do the columns of M, considered as vectors in R12, span R12.

2. I'm not sure how to solve this but if the solution is in R7 does that automatically mean the vectors can't span R12 and aside from that a set of all 0s can't span R7 can it?
This question is asking about the columns of the matrix, not the solutions. The columns are in R12 and there are seven of them, so could the columns span R12. Hint: how many vectors does it take to span R3? R3?
 
Mark44 said:
This question is asking about the columns of the matrix, not the solutions. The columns are in R12 and there are seven of them, so could the columns span R12. Hint: how many vectors does it take to span R3? R3?
So then if I'm not mistaken since there are only 7 columns (or 7 vectors) and there must be a minimum of n vectors to span ℝn 7 vectors can't span all of ℝ12. So the answer is no.

As for the first one would I be correct to say that the rank is then 7?
 
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