Im trying to figure out if I understand the conceptual basics.

All the time you see that vectors are scaled down/up for readability or for simplifying future calculations with that same vector.

As I've seen we scale vectors all the time, and we use equality signs between them, so that makes me conclude they are equal, right? But if you divide every element in a vector by the same value, what we refer to as scaling, you get a vector of a shorter length. A shorter length, but pointing in the same direction. Normalizing a vector we obtain by dividing every vector element by the norm/length of the entire vector, which returns a unit vector of 1 unit length, but pointing in the same direction as the initial vector before we normalized it. We can also say that normalizing vectors is the same as scaling them down to 1 unit length, as we mentioned before, dividing every element by the same value (the norm).

So far so good, but now comes my real question.

How do you know what "scaled" version of a vector you would apply, if using that vector further in a subtraction with another vector (one vector minus another). If you use 2 different scaled "versions" of the vector, it will return 2 different results/vectors if used in a subtraction with another vector. So am I right when concluding with that these end results will be different vectors pointing in different directions?

Lets say we have the following relations where $$ \vec{e} $$ is a scaled version of $$ \vec{c} $$: $$ \vec{a} - \vec{b} = \vec{c}$$ $$ \vec{a} - \vec{d} = \vec{e}$$

Now if Im right this relation is NOT valid as they are not equal: $$ \vec{c} = \vec{e}$$

So: $$ \vec{c} \neq \vec{e}$$

Now $$ \vec{c} $$ or $$ \vec{e}$$ is one of the column vectors in the Q-matrix in QR-factorization.

So the question is if these vectors are not equal, how do you know which is more "correct"/useful to use in QR-factorization or does it not matter at all? Any considerations to take into account?

Here are the specific details for the problem in question. Here are the specific details for the problem in question

$$ \vec{t}$$ is the projection vector of $$ \vec{c_2}$$ onto $$ \vec{c_1}$$

  • $\begingroup$ Are you asking about the Gram-Schmidt procedure? I am not sure about your question. Remember that the $Q$ needs to be orthogonal, that's why we need to normalize the vectors. $\endgroup$ – KittyL Apr 18 '15 at 18:53
  • $\begingroup$ No the vectors in question are simply provided in the question text/task, so no Gram Schmidt to obtain them. $\endgroup$ – jibo Apr 18 '15 at 19:15
  • $\begingroup$ Then I am not sure what they are. If you could provide some more details or examples, it would be easier to see. $\endgroup$ – KittyL Apr 18 '15 at 21:53
  • $\begingroup$ Added specific details in my initial post to the concrete problem that arised my question. $\endgroup$ – jibo Apr 19 '15 at 18:33

First of all, remember $Q$ is orthogonal, that means each column of it has to be a unit vector, that is essentially why you have to normalize the $q_1$ and $q_2$.

I am guessing that you wanted to ask whether it would change the subtraction $c_2-t_2$. The answer is no, the normalization does not change the subtraction. However it is because of the projection process. When you find the projection, you divide by the length of the vector $c_1$ or $q_1$. So if it is normalized, you will divide by $1$. If it is not normalized, you will divide by the norm. The final result doesn't change because of this. enter image description here

Look at the picture, it doesn't matter how long the base vector $\vec{c}$ is. The projection of $\vec{t}$ onto $\vec{c}$ stays the same. However, the scale of $\vec{t}$ would change the projection, hence change the subtraction.

  • $\begingroup$ If I understand you correctly: So we could in other words say that the final result has length of 1 unit in either case. The only difference in the methods is that the normalizing into unit length either happen in the end or in an earlier step. My question was regarding if using a t2 that is instead a scaled version of itself, will that change the results? $\endgroup$ – jibo Apr 20 '15 at 18:12
  • $\begingroup$ What I meant is the normalizing of the base vector does not affect the projection of the other vector onto it. I edited the answer. Sorry in the picture I used difference notations. $\endgroup$ – KittyL Apr 21 '15 at 9:43
  • $\begingroup$ Good point, thx $\endgroup$ – jibo Apr 21 '15 at 20:11

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