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I'm trying to understand the solution to a particular problem but can't seem to figure it out.

What kind of geometric object is represented by the equation:

$$(x_1, x_2, x_3, x_4) = (2,-3,1,4)t_1 + (4,-6,2,8)t_2$$

The answer is: a line in (1 dimensional subspace) in $\mathbb{R}^4$ that passes through the origin and is parallel to $u = (2,-3,1,4) = \frac{1}{2} (4,-6,2,8)$

I'm thinking it has something to do with the fact that $x_1 = 2x_3$, $x_2 = -3x_3$, $x_4 = 4x_3$, $x_3 = t_1+2t_2$ so the solution is $x_3$, which is a line (I think) but why would this line be in a 1-D subspace? What does that actually mean?

There's also another example:

$$(x_1, x_2, x_3, x_4) = (3, -2, 2, 5)t_1 + (6, -4, 4, 0)t_2$$

The answer: a plane in (2 dimensional subspace) in $\mathbb{R}^4$ that passes through origin and is parallel to $u = (3,-2,2,5)$ and $v = (6,-4,4,0)$

For this one, I don't even know why this object is a plane..

Can someone help connect the dots for me? Thanks!

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1 Answer 1

up vote 4 down vote accepted

The first one is a line because the vector $(4,-6,2,8)$ is twice the vector $(2,-3,1,4)$. Thus your collection of points is just the collection of all points of the form $(t_1+2t_2)(2,-3,1,4)$. So it is the collection of all points of the form $t(2,-3,1,4)$. The multiples of a non-zero vector are just a line through the origin.

In the second example, you are getting the set of all linear combinations of $2$ linearly independent vectors in $4$-dimensional space. Whether to call this a plane is a matter of taste. If we think of a $2$-dimensional subspace of $\mathbb{R}^n$ as a plane, then it certainly qualifies.

Similarly, if you are given a set $3$ linearly independent points $\{v_1,v_2,v_3\}$ in $\mathbb{R}^4$, then the set of all points of the form $t_1v_1+t_2v_2+t_3v_3$ is a $3$-dimensional subspace of $\mathbb{R}^4$.

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