I Existence and Uniqueness of Inverses

jolly_math
Messages
51
Reaction score
5
Existence: Ax = b has at least 1 solution x for every b if and only if the columns span Rm. I don't understand why then A has a right inverse C such that AC = I, and why this is only possible if m≤n.

Uniqueness: Ax = b has at most 1 solution x for every b if and only if the columns are linearly independent. I don't understand why then A has a n x m left inverse B such that BA = I, and why this is only possible if m≥n.

Could anyone explain the logic behind this? Thank you.
 
Physics news on Phys.org
Let's start with uniqueness. Suppose A has two columns that are linearly dependent. Can you find x that's not equal to 0 with Ax=0? Hint: it only has two nonzero elements
 
Office_Shredder said:
Let's start with uniqueness. Suppose A has two columns that are linearly dependent. Can you find x that's not equal to 0 with Ax=0? Hint: it only has two nonzero elements
No. A would be a column vector, and only x=0 would work. Why does this lead to the left inverse B such that BA = I, and why it is only possible if m≥n?

Thank you.
 
jolly_math said:
No. A would be a column vector

It's not. You might have misread my question, give it another look :)
 
I'm not sure what the answer is, could you explain the reasoning? Thank you.
 
Can you write out a 2x2 matrix which has two columns that are linearly dependent?
 
Thread 'Determine whether ##125## is a unit in ##\mathbb{Z_471}##'
This is the question, I understand the concept, in ##\mathbb{Z_n}## an element is a is a unit if and only if gcd( a,n) =1. My understanding of backwards substitution, ... i have using Euclidean algorithm, ##471 = 3⋅121 + 108## ##121 = 1⋅108 + 13## ##108 =8⋅13+4## ##13=3⋅4+1## ##4=4⋅1+0## using back-substitution, ##1=13-3⋅4## ##=(121-1⋅108)-3(108-8⋅13)## ... ##= 121-(471-3⋅121)-3⋅471+9⋅121+24⋅121-24(471-3⋅121## ##=121-471+3⋅121-3⋅471+9⋅121+24⋅121-24⋅471+72⋅121##...
Back
Top