Why are the columns of Q linearly independent?

victoranderson
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Why column 1 is M^2*v? How can we know?

Please see attached. Many thanks.
 

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hi victoranderson! :smile:

if Q-1MQ = that matrix,

then MQ = Q(that matrix), which is … ? :wink:
 
tiny-tim said:
hi victoranderson! :smile:

if Q-1MQ = that matrix,

then MQ = Q(that matrix), which is … ? :wink:

I know what you mean
Let that matrix be D
I can find MQ=QD by trail and error
Is there any other method I can use to find the columns of Q without trial and error?
 
hi victoranderson! :smile:

(just got up :zzz:)
victoranderson said:
I can find MQ=QD by trail and error …

it isn't trial and error

what are the columns of QD ? :wink:
 
tiny-tim said:
hi victoranderson! :smile:

(just got up :zzz:)it isn't trial and error

what are the columns of QD ? :wink:


it's been a long time..

The columns of QD is [0,0,0]^T, [1,1,-1]^T and [2,3,1]^T
which are ##M^3v , M^2v , Mv ## respectively

I am stupid and I still do not understand..
 
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hi victoranderson! :smile:

for any matrix P with columns A B C:

the columns of PD are 0 A B

the columns of MP are MA MB MC​

so MP = PD (ie P-1MP = D) if 0 = MA, A = MB, B = MC

(so A = MB = M2C and M3C = 0)

so we choose the columns of P to be M2 MC and C :smile:

(and then we call it Q instead of P)
 
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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...
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