1. ## Linearly Independent?

x1(t) =
[2-t]
[t ]

x2(t)=
[t+1]
[-2 ]

x3(t)=
[t]
[t+2]

These are supposed to be matrices, but I don't know how to format them correctly. Anyways, I want to determine whether or not this set of vector functions is linearly dependent. The way I know how to go about this would be to set up a matrix and then evaluate the determinant. I would then look at the coefficients of the determinant; if they add or subtract to zero, then it is linearly dependent. The problem is that these 3 functions would make for a 2x3 matrix and I can't evaluate the determinant of that one.

Any suggestions? Thanks,

Jim

2. Originally Posted by Jim Newt
x1(t) =
[2-t]
[t ]

x2(t)=
[t+1]
[-2 ]

x3(t)=
[t]
[t+2]

These are supposed to be matrices, but I don't know how to format them correctly. Anyways, I want to determine whether or not this set of vector functions is linearly dependent. The way I know how to go about this would be to set up a matrix and then evaluate the determinant. I would then look at the coefficients of the determinant; if they add or subtract to zero, then it is linearly dependent. The problem is that these 3 functions would make for a 2x3 matrix and I can't evaluate the determinant of that one.

Any suggestions? Thanks,

Jim
You should know that any three vectors in a vector space of dimension 2 are linearly dependent.

3. Thanks for the response, but the answer in the back of the book states that this set is linearly independent.

Here is the exact verbage of the question:

Determine whether or not the given set of vector functions is linearly dependent. The interval of definition is assumed to be the set of all real numbers.

x1(t) =
[2-t]
[t ]

x2(t)=
[t+1]
[-2 ]

x3(t)=
[t]
[t+2]

Any ideas?

4. Maybe you should learn how to produce matrix in Latex. This shouldn't be that hard. This way we'll be sure we're talking about the same thing.

5. Originally Posted by Jim Newt
Thanks for the response, but the answer in the back of the book states that this set is linearly independent.
Isomorphism is correct. The set $\displaystyle \left\{ \left[\begin{matrix} 2-t\\t \end{matrix}\right],\; \left[\begin{matrix} t+1\\-2 \end{matrix}\right],\; \left[\begin{matrix} t\\t+2 \end{matrix}\right] \right\},\text{ where }t\in\mathbb{R}$ is linearly dependent, because you have three vectors from a two-dimensional vector space.

For example:

$\displaystyle \forall t\in\mathbb{R},\; \left[\begin{matrix} t\\t+2 \end{matrix}\right] = \left(\frac{t^2+5t+2}{t^2-t+4}\right) \left[\begin{matrix} 2-t\\t \end{matrix}\right] + \left(\frac{2t^2-4}{t^2-t+4}\right) \left[\begin{matrix} t+1\\-2 \end{matrix}\right]$

(Try checking it yourself)