Is it possible for matrices to be equal but their columns not equal?

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  • Thread starter Thread starter yaganon
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Discussion Overview

The discussion revolves around the possibility of matrices being equal while their columns are not equal. Participants explore this concept through various mathematical scenarios, including cases involving singular matrices and specific examples of matrix multiplication.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants propose that if AB = AC, it does not necessarily imply B = C, particularly when A is a singular matrix.
  • Others argue that if A has an inverse, then B must equal C when AB = AC, suggesting that singularity of A is crucial to the discussion.
  • A participant provides a general case where A(B - C) = 0 without B - C being equal to 0, indicating that non-invertible matrices can lead to such scenarios.
  • Another participant presents a specific example involving matrices that yield equal products while demonstrating that the individual matrices involved are not equal.
  • Concerns are raised regarding the necessity of A being singular for the arguments to hold, with some participants questioning this assumption.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of A being singular for the claims to hold. While some support the idea that singularity is essential, others challenge this notion, leading to an unresolved debate on the conditions under which matrices can be equal without their columns being equal.

Contextual Notes

Limitations include the dependence on the definitions of matrix equality and singularity, as well as the implications of matrix dimensions in the examples provided.

yaganon
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If AB = AC, it's possible that B /= C

conversely:
if BA = CA, is it possible that C /= B?

/= means doesn't equal
 
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Certainly. In both cases let A = 0, B = Anything, C = Anything else.
 
More generally, if A is a matrix that does not have an inverse, then we can have A(B- C)= 0 without B- C being equal to 0. And, similarly, (B- C)A= 0 without B- C= 0.
 
hallsofivy

why does A have to be singular?
 
yaganon said:
hallsofivy

why does A have to be singular?

say
AB=AC
suppose A has an inverse D such that AD=DA=
DAB=DAC
B=C
thus if B/=C A must be singular

Also the matrices might not be square
[[1,1,1]]*[[1,1,1]]=[[1,1,1]]*[[3,0,0]]=[[3]]
where * is the adjoint
[[1,1,1]]/=[[3,0,0]]
 
A basic example for you...

\begin{pmatrix}0 &amp; 0 &amp; 0\\0 &amp;1 &amp; 0\\ 0 &amp; 0 &amp;0\end{pmatrix} = \begin{pmatrix}1 &amp; 0 &amp; 1\\0 &amp;1 &amp; 0\\ 1 &amp; 0 &amp;1\end{pmatrix} \begin{pmatrix}1 &amp; 0 &amp; -1\\0 &amp;1 &amp; 0\\ -1 &amp; 0 &amp;1\end{pmatrix} = \begin{pmatrix}1 &amp; 0 &amp; 1\\0 &amp;1 &amp; 0\\ 1 &amp; 0 &amp;1\end{pmatrix} \begin{pmatrix}2 &amp; 0 &amp; -5\\0 &amp;1 &amp; 0\\ -2 &amp; 0 &amp;5\end{pmatrix} = <br /> \begin{pmatrix}0 &amp; 0 &amp; 0\\0 &amp;1 &amp; 0\\ 0 &amp; 0 &amp;0\end{pmatrix}<br />

But

\begin{pmatrix}1 &amp; 0 &amp; -1\\0 &amp;1 &amp; 0\\ -1 &amp; 0 &amp;1\end{pmatrix} \neq \begin{pmatrix}2 &amp; 0 &amp; -5\\0 &amp;1 &amp; 0\\ -2 &amp; 0 &amp;5\end{pmatrix}

If you look at the rank of A and the vector that is in the nullspace of A, you can see why this holds with the columns of the B and C are chosen as such.
 
Last edited:

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