Linearly Independent?

• June 6th 2008, 01:49 PM
Jim Newt
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
• June 6th 2008, 02:20 PM
Isomorphism
Quote:

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.
• June 6th 2008, 02:39 PM
Jim Newt
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?
• June 6th 2008, 02:43 PM
arbolis
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.
• June 6th 2008, 03:40 PM
Reckoner
Quote:

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 $\left\{
\left[\begin{matrix}
2-t\\t
\end{matrix}\right],\;
\left[\begin{matrix}
t+1\\-2
\end{matrix}\right],\;
\left[\begin{matrix}
t\\t+2
\end{matrix}\right]
\right\},\text{ where }t\in\mathbb{R}$
is linearly dependent, because you have three vectors from a two-dimensional vector space.

For example:

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

(Try checking it yourself)