# idempotent matrices

#### Ackbeet

MHF Hall of Honor
What have you done so far?

#### mcemang

I figured that xx' and x'x are the same thing (just x1^2+x2^2...xn^2) so xx'/x'x=1 ??

therefore A=xx'/x'x is indepotent as A=1 and 1*1=1. subbing A=1 into B=In - A gives us B=In-1....am i on the right track?

#### Ackbeet

MHF Hall of Honor
I'm afraid you're completely on the wrong track. $$\displaystyle xx'$$ is a matrix, $$\displaystyle x'x$$ is a number (it's the dot product). To convince yourself of this, look at the sizes of the vectors $$\displaystyle x$$ (n x 1) against $$\displaystyle x'$$ (1 x n), along with the definition of matrix multiplication and the size of the resulting multiplication.

So, $$\displaystyle A$$ is the matrix $$\displaystyle \frac{1}{x'x}\,xx'$$. What's in the denominator is a number: the dot product.

I would ask yourself what the definitions of "idempotent" and "symmetric" are. What are those?

#### anon123

Break the problem down into 2 parts. First, try to show that the matrix is symmetric (i.e. transpose(A) = A). Then, try to show that the matrix is idempotent (A^2 = A). Since you have shown both separately, it means both are true.

Hint: You do not need to expand the vector x into its components to do this problem.