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Not sure if the following statement is true or false:

For any $I_n$ and any $A \in \Bbb R^{n\times n}$, $(I_n + A)(I_n − A) = I_n − A^2$

I am thinking true because: $(I_n + A)(I_n - A) = I_n*I_n - I_n*A + A*I_n - A*A = I_n - A + A - A^2 = I_n - A^2$

Shaun
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Pranay Agrawal
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    Your analysis is correct. – Robert Shore Dec 19 '19 at 16:24
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    Nothing wrong there. As a tip, all factoring rules and algorithms such as difference of squares or binomial expansion apply if the matrices involved all commute with each other. – Ninad Munshi Dec 19 '19 at 16:26
  • And to add to @NinadMunshi 's comment, note that $A$ always commutes with all of its powers (including $A^0=I_n$). – MPW Dec 19 '19 at 16:28

1 Answers1

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Your working is correct. It is good that you thought about the non-commutativity of matrices in general.

Samuel Barkes
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