We were given this identity during the lecture, but the proof was omitted:

$$\langle v, u \rangle u = (uu^T)v \text{ with } ||u||=1$$ If I write out the intermediate matrices, I see that the equality holds, but I wanted to prove it more analytically.

I tried starting with $\langle v, u \rangle u = (v^Tu)u$ using the definition of dot product, then the professor suggested we use associativity of matrix/vector multiplication to rewrite the operations as an outer product.

I'm not very familiar with the outer product and I do not get how to do use it for the last part. Any insight appreciated.


The inner product is scalar valued, so can be placed on the right of $u$ instead of on the left. We can also write it as $u^Tv$ instead, because the product is symmetric. The associativity of matrix multiplication completes the proof.

  • $\begingroup$ This is rather straightforward, thanks! I was not sure that writing the scalar on the right was considered formally correct. $\endgroup$ – Generalbrus Jul 12 at 13:56
  • $\begingroup$ @Generalbrus Technically we're really multiplying by that scalar times an identity matrix. See also here. $\endgroup$ – J.G. Jul 12 at 14:00

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