Proof of orthogonal and symmetric.

Question

Given $x$ is an $n$ dimensional vector, if $A = I_n- (2/x^Tx)xx^T$, show that it is orthogonal and symmetric.

I know that if $A$ is orthogonal and symmetric, $A = \operatorname{inverse}(A) = A^T$, however, how do we prove that from the statement above?

Answer 1

Let $x$ be an $(n \times 1)$-matrix and $A = I_n - \frac{2}{x^T x} xx^T$.

Claim 1: $A$ is symmetric.

Proof: $$\begin{align} A^T &= (I_n - \frac{2}{x^T x} xx^T)^T \\ &= I_n^T - \frac{2}{x^T x} (xx^T)^T\\ & = I_n - \frac{2}{x^T x} xx^T \\ &= A, \end{align}$$ so $A$ is symmetric.

Claim 2: $A$ is orthogonal.

Proof: $$\begin{align} AA^T &= A^2 \\ &= \left(I_n - \frac{2}{x^T x} xx^T\right)^2 \\ &= I_n^2 - 2 \frac{2}{x^Tx} xx^T + \frac{4}{(x^Tx)^2} (x x^T)^2 \\ &= I_n - \frac{4}{x^Tx} xx^T + \frac{4}{x^Tx} xx^T \\ &= I_n,\end{align}$$ so $A$ is orthogonal.

Note that for Claim 2:

• The first equation follows from $A^T = A$ which we have shown in Claim 1.
• $x^T x$ is a scalar which implies $(xx^T)^2 = x (x^T x) x^T = (x^T x) xx^T$, so the equation before last follows from cancelling $x^Tx$).

Answer 2

To prove that $A$ is symmetric, you can use the fact that sum of symmetric matrices is symmetric.

Lemma: Sum of symmetric matrices is symmetric.

Claim: $A$ is symmetric

Proof: To establish the claim, we just need to prove that $xx^T$ is symmetric. \begin{equation} (xx^T)^T = \left((x^T)^Tx^T\right) = xx^T \end{equation} Thus, $A = I_n + k xx^T$ is symmetric where $k = -2/x^Tx$ is a scalar.

Orthogonality requires that $A^TA = I_n$. Steps are shown in another answer.