Matrix Operation

Question

Let $x$ be a $n \times 1$ vector whose jth element is $x_j$. Show that $A = xx^{T}/x^{T}x$ and $B = I_n - A$ are symmetric idempotent matrices.

Note that $x^Tx$ is a scalar (real number)

Answer 1

Notice that if $x\neq 0$ then $x^Tx\neq 0$ and then $$A^T=\left(xx^T/x^Tx\right)^T=\frac{1}{x^Tx}\left(xx^T\right)^T=xx^T/x^Tx=A$$ so $A$ is symmetric and then $B$ is also symmetric, moreover we have: $$A^2=\frac{1}{(x^Tx)^2}x(x^T x)x^T=\frac{1}{x^Tx}xx^T=A$$ so $$B^2=(I-A)(I-A)=I-2A+A^2=I-A=B$$ and then $A$ and $B$ are idempotent matrices.

Answer 2

Since it is a little bit confusing to use $u v^T$ to represent the outer product of a $n\times 1$ column vector $u$ with a $1\times n$ row vector $v^T$, let us use an alternate notation $u \otimes v^T$ for such outer products.

Notice for any column vectors $a, b, c, d$, we have:

$$(a \otimes b^T)^T = (b \otimes a^T) \quad\text{ and }\quad (a \otimes b^T) (c \otimes d^T) = (b^T c) (a \otimes d^T)$$

These two equalities can be proved by expanding everything out and one can check the matrix entries on both sides do match with each other. Substitute $a,b,c,d$ by $x$ gives us: $$ (x \otimes x^T)^T = (x \otimes x^T) \quad\text{ and }\quad (x \otimes x^T)^2 = (x^T x) (x \otimes x^T)$$ Divide the first equality by $(x^T x)$ and second by $(x^T x)^2$, we get $A^T = A$ and $A^2 = A$, i.e. $A$ is a symmetric idempotent matrix. As a consequence,

$$\begin{align} & B^T = (I_n - A)^T = I_n - A^T = I_n - A = B\\ \text{ and }\quad & B^2 = (I_n - A)^2 = I_n - 2 A + A^2 = I_n - 2 A + A = I_n - A = B \end{align}$$

i.e. $B$ is also a symmetric idempotent matrix.