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paddy.dude

what are linearly independent vectors

amitp91

Linearly independent vectors

Definition Let $ S= \{ {\bf v}_1, {\bf v}_2, \dots {\bf v}_n \}$ be a set of vectors in the vector space $V$. The set $S$ is a linearly independent set if non of the vectors $ {\bf v}_1, {\bf v}_2, \dots {\bf v}_n $ is expressible as a linear combinatio n of the other vectors in $S$. This is equivalent to the following:

Vectors $ {\bf v}_1, {\bf v}_2, \dots {\bf v}_n $ are linearly independent if and only if the vector equation

\begin{displaymath}x_1 {\bf v}_1\; + \; x_2 {\bf v}_2\; + \;\dots \; + \; x_n {\bf v}_n \; = \; 0\end{displaymath}

has only trivial solution. That is $x_1 = x_2 = \dots = x_n = 0 .$

How could we use MATLAB to show that the vectors $ {\bf v}_1, {\bf v}_2, \dots {\bf v}_n $ are linearly independent?

To answer this question we need to show that $ AX =0$ has only trivial solution where $A= [ {\bf v}_1, {\bf v}_2, \dots {\bf v}_n ]$.

In general, if the rref of A looks like $\left[ \begin{array}{c} \rm I 0 \end{array} \right] $ then the

solution to $ AX =0$ is

\begin{displaymath}\begin{array}{*6{c@{\;} }} x_1&&&& =0 &x_2&&&=0 &&\ddots&&\vdots &&&x_n& =0 \end{array} \end{displaymath}

so the columns of A are linearly independent. If any column of the rref of A does not have a leading 1 then the corresponding variable is free and so $ AX =0$ would have nontrivial solutions making the columns of A linearly dependent.

The special case is if A is a square matrix. Then the fact that $ AX =0$ has only the trivial solution exactly when A is invertible,( i.e. when det(A) $\neq$ 0 ) is very useful.

So the columns of a square matrix A are linearly independent only if det(A) $\neq$ 0.

amitp91

rate if useful

paddy.dude

man what is this u have written can u explain it in short

amitp91

linearly independent vecotrs

amitp91

amitp91

The general definition of linear independency is:

sigma(a_i*v_i) for i = 0....n = 0 [the zero vector] ----> (a_1*v_1)+(a_2*v_2)..+(a_n*v_n) = 0
where a_1......a_n are arbitrary numbers and v_1.....v_n are vectors

If this is true, then these vectors are linearly independent. An example can help.

Let's take the 2-D case. If we want to prove that (1,2) and (3,4) are linearly independent, then:

a_1(1,2)+a_2(3,4) = (0,0)

(a_1+3a_2, 2a_1+4a_2) = (0,0). Thus you get:

a_1+3a_2 = 0
2a_1+4a_2 = 0

Solving for a_1 and a_2, the only unique solution is a_1 = a_2 = 0
Thus, this proves the definition, and these vectors are linearly independent.

Another way to prove this is by a matrix. Define the matrix, 'A' such that the columns are the vectors themselves. Using the example above,

A = [v_1 v_2] =

[1 3]
[2 4]

Using a matrix, these vectors are linearly independent if the determinant is not = 0.

For a 2x2 matrix, 'A':

[a b]
[c d]

det (A) = ad - bc. From our example, det(A) = (1)(4)-(2)(3) = -2

Thus, the determinant is not = 0, thus the vectors are linearly independent.

These derivations can expand to an N-dimensional vector case

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Hope this helps
rate if useful

akshay.khare91

simply the vector are called are linearly independent if the linear combination of vectors is 0 and there scalars are also

for ex. if a , b, c are three linearly independent vectors then for any scalars x ,y,z

xa+yb+zc = 0
then x= y =z 0 ..

paddy.dude

thnks akshay i dont what this user amit was copying and pasting here from other sites

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