# Decomposition of Vector through Dot Product with Basis Vector: Where does Necessity of Normalization Follow from?

If we have an orthonormal set $$E$$ of vectors $$e_1$$ to $$e_n$$ we can decompose any given vector $$v$$ from the same space as $$E$$ by dotting it with the $$e_i$$. Proof:

As $$v$$ is in the space of $$E$$ it must be the case that $$v = \sum_{i=1}^n \alpha_i e_i$$. If we wish to get the $$k$$th component of $$v$$ we can thus do

$$e_k \bullet v = e_k \sum_{i=1}^n \alpha_i e_i = \underbrace{\sum_{i=1}^n e_k \alpha_i e_i = a_k e_k \bullet e_k}_\text{as e_k \bullet e_i = 0 for all i \neq k, as all e_i are orthogonal.} = \alpha_k$$

If $$E$$ is not orthonormal, but merely orthogonal, all $$e_i$$ need to be normalised, so we end up with

$$\alpha_k = \frac{e_k \bullet v}{e_k \bullet e_k}$$ for the $$k$$th component of $$v$$.

At least that is what I was told. I don't see however, where the proof ever uses the fact that all $$e_i$$ are unit length. It works just as well if the $$e_i$$ are of different length. It still holds that $$e_k \bullet e_k = 1$$ and that $$e_k \bullet e_i = 0 ~ \forall ~i \neq k$$. Therefore the $$k$$th component of $$v$$ should still simply be $$v \bullet e_k$$, according to the logic of the proof.

However, I also realise that in general $$v \bullet e_k \neq v \bullet c e_k$$ for some factor $$c$$. So... is there a hole in the proof or am I missing the point where the property of the $$e_i$$ being unit length is being used?

• $e_{k}$ is of unit length means that $e_{k} \cdot e_{k}=1$. Dec 21 '18 at 14:48

"It still holds that $$e_k \bullet e_k = 1$$"

Under your assumptions (the $$e_k$$ are still orthogonal, but not necessarily unit length), this is in general false. You might want to look at this example:

$$e_1 = \pmatrix{2 \\ 0}\\ e_2 = \pmatrix{0 \\ 2}$$ in which $$e_k \cdot e_k = 4$$ for $$k = 1, 2$$.

If you try to decompose the vector $$v = \pmatrix{2\\2} = e_1 + e_2$$ using your proposed formula, you'll see what goes wrong.

The fact that $$a_k e_k\cdot e_k=a_k$$ is because $$|e_k|^2=1$$. If $$e_k$$'s are not orthonormal but only orthogonal then:

\begin{align} a_k e_k\cdot e_k=a_k\cdot|e_k|^2\\ \implies a_k=\frac{a_k e_k\cdot e_k}{|e_k\cdot e_k|}\\ \implies a_k=\frac{e_k\cdot v}{|e_k\cdot e_k|} \end{align}

• I know. That was not my question. My question was why the proof is valid... because the necessity to normalise does not follwo from it. The proof works with non-unit vectors too, but the formula obtained from the proof applied to non-uni tlength vectors yields an incorrect result. Thereofre either I am missing something or the proof has a flaw. Dec 21 '18 at 14:39
• Yes, thanks. I somehow confused the dot product with the cosine between the vectors, that is why I thought $e_k \bullet e_k = 1$, even when $|e_k| \neq 1$. Dec 21 '18 at 14:48
• Aah I see $e_i\cdot e_j=||e_i|||e_j||cos(\theta)$ is also true but not used in this proof you gave. Dec 21 '18 at 14:51