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So I am self-learning Linear Algebra and I came across a proof and I need a little verification to see if I am on the right track because it varies significantly from the answer provided.

Question: Consider the linear systems Ax = b and Bx = c where both A and B are invertible. Show that (A + B)x = b + c may have an infinite number of solutions.

My proof:

  • A & B are invertible, this means that they are square matrices
  • A & B are of same size in order to carry out matrix addition
  • Sum of two invertible matrices is not necessarily invertible
  • In order to have a unique solution $\ x = A^{-1} b $ must hold in the form $\ x = A^{-1}b + B^{-1}c$

$$ (A+B)x = b + c $$ $$ (A^{-1} A + B^{-1}B) = A^{-1}b + B^{-1}c $$ $$ (I + I)x = A^{-1}b + B^{-1}c $$ $$ 2Ix = A^{-1}b + B^{-1}c $$ $$ 2Ix = A^{-1}b + B^{-1}c \neq x = A^{-1}b + B^{-1}c $$

Thefore, the system does not have a unique solution. It is possible that the system is either inconsistent or has infinite number solutions.

Book's proof:

Let the matrix $\\ A = -B \implies Ax = -Bx = -c $ $$ (A + B)x = (-B + B)x = Ox = -c + c = O$$ O matrix is not invertible so system has infinite number of solutions.

My question:

Is my proof a valid one or did I just come up with a bogus proof? If it is incorrect can someone clarify where I might have gone wrong? Help is much appreciated.

iamlearningmath
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  • Why is your first bullet point true? – The Count Feb 02 '17 at 02:05
  • @TheCount He probably means that "because we define determinants only for square matrices." Refer to this question: http://math.stackexchange.com/q/854180/6715 – Amir Parvardi Feb 02 '17 at 02:10
  • @TheCount Well there is a definition 1.23 in the book which states that there must be a multiplicative inverse. AB = BA = I. If you have a matrix of two different sizes then the columns on the left may not necessarily match the rows on the right hand side in order to carry out the multiplication. A non-square matrix may just have left inverse or right inverse but it is not really invertible if this makes any sense. – iamlearningmath Feb 02 '17 at 02:11
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    @AmirHossein I am actually fine with it, but when first learning Linear Algebra, it is important that each step by small, deliberate, and careful. So I am trying to help in that direction. Thanks, though. :) – The Count Feb 02 '17 at 02:11
  • @Iamlearningmath Ok, good. I just wanted to make sure you could argue from your definitions. Looks like you're on the right track then. – The Count Feb 02 '17 at 02:12
  • @TheCount really appreciate it. Since I have been teaching it to myself I haven't really had a discussion with anyone about it. This helps my learning. – iamlearningmath Feb 02 '17 at 02:13
  • @TheCount Oh, sorry for that. – Amir Parvardi Feb 02 '17 at 02:16
  • @Iamlearningmath I just took my master's exam in Linear Algebra this afternoon, so I am happy to help others as they begin to learn. A good rule in math and especially linear algebra is that if something takes more than a single phrase or small sentence to explain, you have probably jumped too far, too quickly. – The Count Feb 02 '17 at 02:16
  • @AmirHossein Oh, no worries. I appreciate the help! – The Count Feb 02 '17 at 02:16
  • I don't know why anyone would downvote this. The OP is clear, responsive, receptive, and put thought and effort into the original question. – The Count Feb 02 '17 at 02:17
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    @Iamlearningmath Can you explain the first line of your fourth bullet? I mean, why would you have $x = A^{-1}b + B^{-1}c$? – Amir Parvardi Feb 02 '17 at 02:17
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    You could have done with your third point. That is since A+B is not necessarily invertible, if you prove that B exists or can be constructed such that A+B is not invertible, proof is done. – jnyan Feb 02 '17 at 02:20
  • @AmirHossein Well the reason behind that is if $\ x = A^{-1}b $ by definition represents a unique solution and if a unique solution exists for the system in question then we should technically have $\ x = A^{-1}b + B^{-1}c $ – iamlearningmath Feb 02 '17 at 02:29

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