Alg1231W4T4 - Linear independence of orthogonal vectors
Friday, 17 August 2018
1:09 AM
Created with Microsoft OneNote 2016.
![Untitled picture.png Machine generated alternative text:
Recall that two vectors VI —
IS zero.
3
and v 2 ¯
in the plane are orthogonal, or perpendicular, precisely when their dot product
VI +YIY2
Thus for example
is orthogonal to [select all that apply]
5
-10
6
¯ 14
23
5
-8
35
-21
In general to get a simple vector orthogonal to v
we could just choose the vector
or any multiple of it.
Using this definition, the only vector which is orthogonal to itself in IR2 is the vector
1
Recall: the Maple notation for the vector
2
Untitled picture.png Machine generated alternative text:
In three dimensions we define orthogonality also in terms of the dot product. If VI —
and v'2
then the dot product is
VI + YlY2 + ZIZ2,
and again VI is defined to be orthogonal, or perpendicular, to v2 precisely when
Thus for example
4
is orthogonal to
23
Using this definition, the only vector which is orthogonal to itself in IR3 is the vector
Recall: the Maple notation for the vector 2 is < 1 , 2 , 3 >.](Alg1231W4T4%20-%20Linear%20independence%20of%20orthogonal%20vectors_files/image001.png)
![Untitled picture.png Machine generated alternative text:
In three dimensions we define orthogonality also in terms of the dot product. If VI —
and v'2
then the dot product is
VI + YlY2 + ZIZ2,
and again VI is defined to be orthogonal, or perpendicular, to v2 precisely when
Thus for example
4
is orthogonal to
23
Using this definition, the only vector which is orthogonal to itself in IR3 is the vector
Recall: the Maple notation for the vector 2 is < 1 , 2 , 3 >.
Untitled picture.png Machine generated alternative text:
A set of vectors S {VI v 2'
vectors are orthogonal, that is
in a vector space V is an orthogonal set if each v
for 'i j.
is a non-zero vector, and if any two vectors distinct
Which of the following sets of vectors are orthogonal? [select all that apply]
10
¯ 14
Untitled picture.png Machine generated alternative text:
A setof three vectors {VI, v2,v3} in a vector space is an orthogonal set if each is vector is a non-zero vector; and if every vector is orthogonal with the other two.
This means explicitly that
5
-2
So for example
Note:
3
4
1
1
—1
1
is an orthogonal set in R
Can you find a set of three orthogonal vectors in R
which includes the vector
2
-2
Such a set is
• express your set of vectors using the Maple notation
• don't forget to include< 2, -4, -2 > in your set
Hint: Once you have one vector orthogonal to
2
—4
—2
you can use the cross product to find a third.
](Alg1231W4T4%20-%20Linear%20independence%20of%20orthogonal%20vectors_files/image002.png)
![Untitled picture.png Machine generated alternative text:
We saw in the previous question that the set
3
4
1
5
is an orthogonal set of vectors Let's illustrate the lesson of the video, that an orthogonal sets of vectors is
Please complete the following argument
We suppose first that we have a combination of the form
for some constants r, s and t. To show that {v, u, w} is a linearly independent set; we want to show that
To show that r = 0, we take the dot product of both sides of (1) with the vector v. This gives
1
—1
1
always linearly independent
(1)
we must nave
(2)
Now, looking at the vectors in S above, we see that
26
Substituting these values into equation (2) gives
from which we easily deduce that r must equal (b
Employing exactly the same reasoning, we deduce that we must have s
So; since r = 0, s = = Ois tne only solution
O and v w
O and t
we conclude that {v,
IS a
linearly independent set
Untitled picture.png Machine generated alternative text:
It is also possible to define functions as being orthogonal if we can find the correct definition to replace the dot product
If we take two functions f and g , defined over the interval [0, 2%] , then we can define a fancy inner product
This will play the same role as the dot product, taking tvo vectors as input and spitting out a number Using this; we define two functions to be orthogonal if
The same way we do for geometric vectors.
Let's consider the functions
Using the inner product definition above we can determine
sin(6x) = o
f(x)
= sin(6x),
Showing that f@)
sin(6c) and g@)
sin(4x) are orthogonal with respect to this cool new product.
Note: you may need to remember that the Maple notation for is Pi.
Hint:
• you might find it easier to evaluate the integrals above if you expand the product
—i6x
sin(4c).
i4x
2
and (sin(4x))2
• this trick also works for (sin(6x))
i6x
sin(6x) sin(4x)
and should yield familiar relations.
—i4x
Untitled picture.png Machine generated alternative text:
Use the hint provided, or if you're running out of time, you can simply use
WolframAlpha!
Also, to integrate or sin2(3x), you can convert it to:
1
2
1
2
— cos(16x))](Alg1231W4T4%20-%20Linear%20independence%20of%20orthogonal%20vectors_files/image004.png)
