If $U$ is a unit vector, why is the matrix $UU^\top$ idempotent?

1k Views Asked by At

The title says it all: if $U$ is a unit vector, why is the matrix $UU^\top$ idempotent?

In other words, why is $UU^\top$ the matrix of a projection?

I am also interested in understanding the "physical" meaning of $UU^\top$. It is clear that $I-2UU^\top$ is an orthogonal symmetry with respect to the hyperplane $U^\perp$ so I could of course expand $(I-2UU^\top)^2=I$ but I would like another, more intuitive, explanation.

1

There are 1 best solutions below

5
On BEST ANSWER

Hint: By associativity, $$ (UU^T)(UU^T) = U(U^TU)U^T $$


The transformation $UU^T$ is the orthogonal projection onto the span of the unit vector $U$. As an illustrative example, you should consider $U = (1,0,\dots,0)^T$. All such maps look like this up to an orthogonal change of basis.