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For any non zero vector $x$ the following inner product is zero meaning that these two matrices are orthogonal to each other. $$\langle I- \frac{xx^T}{\|x\|_2^2},\frac{xx^T}{\|x\|_2^2}\rangle=0$$

What is the intuition behind this?

  • Hint: $\dfrac{xx^T}{\left| x\right|_2^2}$ is the matrix for orthogonal projection onto $\mathrm{span}\left{x\right}$ (make sure you can explain why!). – Minus One-Twelfth Mar 04 '19 at 04:27
  • @Minus One-Twelfth: could you explain why? –  Mar 04 '19 at 04:47
  • There is in fact a general formula as follows: if $X$ is a real matrix with linearly independent columns, then the matrix for orthogonal projection onto the column space of $X$ is $\color{blue}{X\left( X^{T}X\right)^{-1}X^{T}}$. Some proofs of this fact may be found here for example: https://math.stackexchange.com/questions/91278/projections-onto-ranges-subspaces. You can use this formula with $X = x$ in your case (a single column vector), remembering that $x^{T}x = \left|x\right|^2_2$. – Minus One-Twelfth Mar 04 '19 at 04:51
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    You’ve got the inner product of two matrices there. What specific inner product do you mean? – amd Mar 04 '19 at 06:27

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$\frac{xx^T}{\|x\|_2^2}$ is an orthogonal projection on the space spanned by $x$, while $I - \frac{xx^T}{\|x\|_2^2}$ is projection on $x$'s orthogonal complement, hence by definition/construction their inner product have to be zero.

V. Vancak
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