Linear Algebra and Vectors

  • #1
vijay_singh
28
0
I am reading "Linear Algebra" by Strang. In the first lesson, he talks about how to solve equations with 2 unknowns and he shows 2 approaches i,e row approach and column approach. I understand the row approach because it makes sense. I understand the column approach too, but I don't understand how the "column" represents a vector?

BTW I do understand what a vector is and how the vector arithmetic works, but I some how cannot abstract the "column" into vector, in my mind. Can somebody please explain me?

example :

2x - y = 1
x + y = 5
 
Last edited:

Answers and Replies

  • #2
trambolin
341
0
How about

[itex]
\begin{pmatrix}2\\1\end{pmatrix}x+\begin{pmatrix}-1\\1\end{pmatrix}y = \begin{pmatrix}1\\5\end{pmatrix}
[/itex]
 
  • #3
jacobrhcp
169
0
or so:

2 -1 x 1
1 1 . y = 5

?

So you can just solve the equations by back substitution, or you can simplify the matrices using the rules of matrix sweeping, which is actually just the same thing.
 
  • #4
vijay_singh
28
0
How about

[itex]
\begin{pmatrix}2\\1\end{pmatrix}x+\begin{pmatrix}-1\\1\end{pmatrix}y = \begin{pmatrix}1\\5\end{pmatrix}
[/itex]

Thanks for responding. May be I didn't make it very clear in my question, it is not the solution of the problem but some part of approach which I can't understand. And following is what I don't understand

why is [itex]
\begin{pmatrix}2\\1\end{pmatrix}
[/itex] considered a "vector" (How is it abstracted as vector quantity?) in the equation you described in your earlier reply

[itex]
\begin{pmatrix}2\\1\end{pmatrix}x+\begin{pmatrix}-1\\1\end{pmatrix}y = \begin{pmatrix}1\\5\end{pmatrix}
[/itex]
 
  • #5
CaffeineJunky
26
0
Let F be a field. Then a vector space V over F is a set that satisfies the following properties:

1.) There exists a map + : V x V --> V such that for all u, v in V, +(u, v) is an element of V. For brevity, rewrite +(u, v) as u + v.
2.) For all u, v in V, u + v = v + u
3.) For all u, v, w in V, (u + v) + w = u + (v + w)
4.) There exists a map * : (F x V) --> V such that for r in F, v in V, *(r, v) is an element of V. We rewrite *(r,v) as r * v, or simply rv.
5.) For all r, s in F, v in V, (r + s) * v = r * v + s * v.
6.) For all r in F, u, v in V, r * (u + v) = r * u + r * v.
7.) For all r, s in F, v in V, r * (sv) = (rs) * v
8.) There exists an element 0 in V such that for all v in V, v + 0 = v.
9.) For all v in V, there exists an element w such that v + w = 0 = w + v. We denote w by -1.
10.) For all v in V, there 1 * v = v, where 1 is the identity element of F.

Any set and functions that satisfies these axioms is called a vector space over F. Any element of this set is then called a vector.

Your row vectors and column vectors both belong to vector spaces because they both belong to sets with operations that satisfy the properties above. You should begin to avoid the association of arrows with vectors, since not all vectors are equivalence classes of oriented segments. For example, the set of degree two polynomials is a vector space by letting F be the real or complex numbers.
 

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