How does one calculate a ||f1||??? doing the Gram-Schmidt Algorithm
gram schmidt eh .... sounds orthogonal to me
if you are doing an algorithm, isnt that just a matter of following the instructions?
What do you mean? Like I have the equation and all, I just don't know how my prof calculated ||f1||
From what I can see you just multiple f1 by itself
can you post the problem, to give this a little more context? i know || is a magnitude of a vector but i see no convention of f1
||f1|| would be the square root of : f1 dot f1
Oh lol sorry. FInd the QR factorization of A = \[\left[\begin{matrix}1 & 3&2 \\ -1&1&0\\0&2&1\\1&1&4\end{matrix}\right]\]
so I followed along an example in my notes and did out to find f1, f2 and f3 with the equations f1=v1 f2=\[f1\times f1\left( v2 - \frac{ v2 \times f1 }{ f1 \times f1 } \times f1 \right)\]
and f3=\[f1 \times f2 \left( v3-\frac{ v3 \times f1 }{ f1 \times f1 } \times f1 - \frac{ v3 \times f2 }{ f2 \times f2 } \times f2\right)\]
the first vector is generally taken to be the first column vector \[u_1 =a_1 = [1~-1~0~ 1]\] \[e_1 =\frac{a_1}{|a_1|}\text{ the unit vector of }a_1\] \[u_2 =a_2-(a_2\cdot e_1)e_1~~...\]
so I solved for those three... then in my notes I seen that I need to find q1, q2 and q3 by q1 = \[\frac{ f1 }{ ||f1|| }\] and q2 =\[\frac{ f2 }{ ||f2|| }\] and q3= \[\frac{ f3 }{ ||f3|| }\]
But I'm not sure what ||f1|| means
Theses are all vectors btw lol
||v|| is notation for the length of a vector \(\vec v\). any vector, that is divided by its own length, become a unit vector of itself. you can use it as a blank slate to scale as needed
How do you like solve it though? Do you just times the vector by itself?
if youve already determined the vectors for the fs, make them into unit vector you can acheive the length of a vector by dotting it to itself and taking the sqrt of the results for example: let v = (3,1,2) v.v = 3,1,2 3,1,2 ------ 9+1+4 = 14 |v| = sqrt(14)
square add sqrt is also a good memory devies
Ohhhhh... So for example if f1=(1,-1,0,1) it'd be sqrt(3) ?
yep
Oh okay thank you :)
youre welcome, good luck :)
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