## Saturday, November 11, 2006

### Simon Says Factor. Topic: Linear Algebra.

Problem: (2003 IMC Day 2 - #1) Let $A$ and $B$ be $n \times n$ real matrices such that $AB+B+A = 0$. Prove that $AB = BA$.

Solution: Well, seeing the $AB+B+A$, we think Simon's Favorite Factoring Trick. But, we have to remember that we're working with matrices, not numbers. So instead of $(x+1)(y+1) = xy+x+y+1$, we have

$(A+I_n)(B+I_n) = AB + I_nB+AI_n+I_n^2$,

where $I_n$ is the $n \times n$ identity matrix (the equivalent of $1$ in the matrix world). Recall that for any square matrix $M$ we have $MI = IM = M$ (where $I$ is the identity matrix of corresponding size), so

$AB + I_nB+AI_n+I_n^2 = AB+B+A+I_n = 0+I_n = I_n$

from the given condition. But the equation

$(A+I_n)(B+I_n) = I_n$

means that $A+I_n$ and $B+I_n$ are inverses of each other. Which consequently means we can multiply them in either order. So

$(A+I_n)(B+I_n) = (B+I_n)(A+I_n)$

$AB+B+A+I_n = BA+A+B+I_n \Rightarrow AB = BA$

as desired. QED.

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Comment: Just basic algebraic manipulations if you knew several key properties of matrices and recognized the factoring. Always good to keep these things handy!

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Practice Problem: (1999 IMC Day 1 - #1) Show that for all natural numbers $n$ there exists a real $n \times n$ matrix $A$ such that $A^3 = A+I$. Furthermore, prove that $|A| > 0$ for all such $A$. [Reworded]