My textbook derives the Gram-Schmidt process in many steps, and the final step being:

Suppose we have already constructed orthogonal vectors $W = \{v_1, ...., v_k\}$, and these vectors can be made orthonormal by dividing them by their norm to result in: $W = \left\{\frac{v_1}{\|v_1\|}, ...., \frac{v_k}{\|v_k\|}\right\}$, W is orthonormal.

So... shouldn't the projection mapping of a vector $v$ on subspace $W$ then be the orthonormal vector representation as in:

$$P_W(v) = \sum_{i=1}^k\frac{1}{\|v_i\|}\langle v_i, v \rangle v_i$$

Should it not? But my textbook says that the projection mapping of some vector $v$ on the orthonormal set we have should therefore be:

$$P_W(v) = \sum_{i=1}^k\frac{\langle v_i, v\rangle }{\langle v_i, v_i\rangle}v_i$$

But aren't we missing the required coefficient to make the set orthonormal?


1 Answer 1


So actually $P_W(v) = \sum_{i=1}^k\langle\frac{v_i}{||v_i||}, v\rangle\frac{v_i}{||v_i||}$ right? Because the elements of $W$ are $\frac{v_i}{||v_i||}$.

So $P_W(v) = \sum_{i=1}^k\frac{1}{||v_i||^2}\langle v_i, v\rangle v_i=\sum_{i=1}^k\frac{1}{\langle v_i, v_i\rangle}\langle v_i, v\rangle v_i$ as your textbook says.

  • $\begingroup$ ...nevermind.., thank you so much $\endgroup$
    – q.Then
    Commented Apr 18, 2016 at 16:35

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