Can Vectors Form Equivalent Spans in Linear Algebra?

[Mathman23]

(a)

show that the vector (2,7,6) be we written as a linear combination of the vectors

(1,3,2) and (0,1,2)

(b) show that the vector (-1,0,4) can be written as a linear combination of the vectors (1,3,2) and (0,1,2)

(c)

show that Span((1,3,2),(0,1,2)) = Span( (2,7,6), (-1,0,4))

My solution (a).

I write vectors in equation form

$$x_1 \[ \[ \begin{array}{c} 1 \\ 3 \\ 2 \end{array} \] + x_2 \[ \begin{array}{dd} 0 \\ 1 \\ 2 \end{array} \] = <br /> \[ \begin{array}{c} 2 \\ 7 \\ 6 \end{array} \]$$

which can be rewritten to

$$\[ \begin{array}{ccc} x_{1} \\ 3x_{1} + x_{2} \\ 2x_{1} + 2x_{2} \end{array} \] = \[ \begin{array}{c} 2 \\ 7 \\ 6 \end{array} \]$$

There must exist x_1 and x_2 which makes the above set of equations true.

These are found be rewritten the system into its equivalent coefficient matrix.

$$\[ \begin{array}{ccc} 1 & 0 & 2 \\ 3 & 1 & 7 \\ 2 & 2 & 6 \end{array} \]$$

using row reduction I get

$$\[ \begin{array}{ccc} 1 & 0 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{array} \] \mathrm$$


which gives x_1 = 2 and x_2 = 1

If I insert into the original equation, then values of x_1 and x_2 make it true.

(b)

Following the same method used in (a) I get x1 = -1 and x_2 = 3.

(c)

How do I use these results to prove that Span((1,3,2),(0,1,2)) = Span((2,7,6),(-1,0,4)) ?

Can I claim the set of vectors are dependent, and therefore their spans equal each other?

Best Regards
Fred

 

[StatusX]

The span of a set of vectors is the set of all (finite) linear combinations of vectors in the set. So a vector is in span((1,3,2),(0,1,2)) iff it is of the form a(1,3,2)+b(0,1,2) for some real numbers a and b, and similarly for span((2,7,6),(-1,0,4)), with vectors in this space having the form c(2,7,6)+d(-1,0,4). See if you can always get a pair (c,d) from (a,b) and vice versa.