Linear Algebra and Its Applications, Review Exercise 2.18

Posted on December 1, 2013 by hecker

Review exercise 2.18. Suppose that $A$ is an $n$ by $n$ matrix with rank $n$ . Show that if $A^2 = A$ then $A = I$ .

Answer: Since the rank of $A$ is $n$ we know that the columns of $A$ are linearly independent and that the inverse $A^{-1}$ exists. We then have

$I = AA^{-1} = A^2A^{-1} = A(AA^{-1}) = AI = A$

So if $A^2 = A$ and the rank $r = n$ then $A = I$ .

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

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