Properties of Invertible Matrices: A2-AB+BA-B2 is Singular

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SUMMARY

If A and B are n x n matrices and A - B is singular, then the expression A² - AB + BA - B² is also singular. This conclusion is derived from the factorization of the expression into (A - B)(A + B). Since the determinant of A - B is zero (det(A - B) = 0), it follows that the determinant of the product (A - B)(A + B) is also zero, confirming that the matrix is singular.

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If A and B are n x n matrices such that A - B is singular then A2 - AB + BA - B2 is also singular.


I really have no clue how to solve this, but I am guessing that AB does not equal BA, I don't know how that can help or be relevant but just in case


Thanks a lot, any help is appreciated!
 
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A²-AB+AB-B²=(A-B)(A+B).

Try computing the determinant of this matrix.
 
alright I am not sure how to calculate the determinant, because we are not actually given any information besides to the matrices A and B are nxn matrices
 
Agreed, but the only thing you need to know is det(A-B)=0. Can you now calculate det((A-B)(A+B))?
 
well if det(A-B)=0. Then det((A-B)(A+B))=0 and the matrix is singular!

I am trying to understand though, how did you know that det(A-B)=0?

and thanks a lot for all the help!
 
Well, you said that A-B was singular. Thus that means that det(A-B)=0...
 
aha I see right, I am really lost with this whole matrices situation haha thanks a lot for the help!
 

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