# Linear Math

• Dec 18th 2007, 04:21 PM
linearmath
Linear Math
questions 1: the three vectors x, y, z have real-valued components and form the sides of a triangle. Prove the Law of Cosines for these vectors, i.e...that

|z|^2 = |x|^2 + |y|^2 - 2*x*y = |x|^2 + |y|^2 - 2*x*y*cos(theta)

where theta is the angle between vectors x and y. (can be using scalar product to solve)

question 2: Let vector x = (1,1,0,0), y = (-1,1,0,0), z = ( 0,0, 1,1), t = (0,0,-1,1). Show that these vectors are linearly independent. Therefore, any subset of them is linearly independent. Discuss the space spanned by each pair of the vectors.

anyone can help ?
• Dec 18th 2007, 05:43 PM
1)
|z|^2 = |x|^2 + |y|^2 - 2|x||y|cos(theta)

Start from the following facts:
\$\displaystyle |z|^2 = z \bullet z\$ (definition)
\$\displaystyle z = x-y\$ (since the vectors make a triangle)

2) Show that ax+by+cz+dt = (0,0,0,0) if and only if a=b=c=d=0. This is done by showing that the system of simultaneous equations has only this solution. You will discover this if you solve the system in the normal way.
• Dec 18th 2007, 05:50 PM
linearmath
well, as you said I can read my notes. I just haven't seen any example of show vectors are linearly independent ( 4 vectors)(Handshake)
• Dec 20th 2007, 12:16 PM
topsquark
Quote:

Originally Posted by linearmath
question 2: Let vector x = (1,1,0,0), y = (-1,1,0,0), z = ( 0,0, 1,1), t = (0,0,-1,1). Show that these vectors are linearly independent. Therefore, any subset of them is linearly independent. Discuss the space spanned by each pair of the vectors.

A set of vectors is linearly dependent if you can construct at least one member of the subset by a linear combination of the remaining vectors.

So your question becomes four questions of the type:
\$\displaystyle x = ay + bz + ct\$

If no solution other than a = b = c = 0 exists then x is linearly independent from the rest. To prove that the whole set is linearly independent you have to do this for y, z, and t as well.

-Dan
• Dec 20th 2007, 03:09 PM