# linear algebra

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• September 12th 2007, 08:59 PM
lord12
linear algebra
if u v and w are vectors that are linearly independent will u-v v-w and u-w be linearly independent?
• September 12th 2007, 09:22 PM
tukeywilliams
So you are given $a \vec{u} + b \vec{v} + c \vec{w} = 0 \Rightarrow a = b = c = 0$.

Then you want to show that $x(\vec{u} - \vec{v}) + y(\vec{v}- \vec{w}) + z(\vec{u} - \vec{w}) = 0 \Rightarrow x = y = z = 0$ is true or false.

Expand out: $x \vec{u} - x \vec{v} + y \vec{v} - y \vec{w} + z \vec{u} - z \vec{w} = 0$.

Then $(x+z) \vec{u} + (y-x) \vec{v} + (-y-z) \vec{w} = 0$
But that doesn't imply that $x = y = z = 0$. Only that $a = b = c = 0$, where $a = x+z, \ b = y-x, \ c = -y-z$. So they are linearly dependent.
• September 12th 2007, 09:38 PM
tukeywilliams
$A = \begin{bmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 2 & 1 \\ 0 & 3 & 2 & 1 \\ 0 & 1 & -1 & 1 \end{bmatrix}$

$\det A = -1 \begin{bmatrix} 0 & 0 & 2 \\ 0 & 3 & 2 \\ 0 & 1 & -1 \end{bmatrix} = -1(-2)(0) =0$

Linearly dependent.
• September 12th 2007, 09:58 PM
Jhevon
Quote:

Originally Posted by tukeywilliams
$A = \begin{bmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 2 & 1 \\ 0 & 3 & 2 & 1 \\ 0 & 1 & -1 & 1 \end{bmatrix}$

$\det A = -1 \begin{bmatrix} 0 & 0 & 2 \\ 0 & 3 & 2 \\ 0 & 1 & -1 \end{bmatrix} = -1(-2)(0) = 0$

Linearly independent.

i believe they are dependent if the Wronskian is zero.

recall that a set of two or more vectors is linearly dependent if and only if at least one of the vectors in the set can be expressed as the linear combination of finitely many other vectors in the set.

now, note that:

$u - v = u - v + w - w = (u - w) - (v - w)$

$v - w = v - w + u - u = (u - w) - (u - v)$

and

$u - w = u - w + v - v = (u - v) + (v - w)$

so we see that each of the vectors can be expressed as a linear combination of the other two, thus the vectors u - v, v - w, and u - w form a linearly dependent set of vectors
• September 12th 2007, 10:04 PM
tukeywilliams
yeah I did it a different way and got that they were linearly dependent also (my edit). You are correct, I forgot to edit the last sentence.
• September 12th 2007, 10:07 PM
Jhevon
Quote:

Originally Posted by tukeywilliams
yeah I did it a different way and got that they were linearly dependent also (my edit). You are correct, I forgot to edit the last sentence.

how did you set up those matrices?
• September 12th 2007, 10:10 PM
tukeywilliams
that was a different problem. My solution to the first problem is post #2.
• September 12th 2007, 10:12 PM
Jhevon
Quote:

Originally Posted by tukeywilliams
that was a different problem. My solution to the first problem is post #2.

o ok. yes, post #2 is a nice solution. did the poster delete the problem you responded to in post #3?
• September 12th 2007, 10:13 PM
tukeywilliams
yeah I think he did.
• September 12th 2007, 10:17 PM
lord12
FIND the coefficients of the linear combination.
• September 12th 2007, 10:40 PM
Jhevon
Quote:

Originally Posted by lord12
FIND the coefficients of the linear combination.

what are you talking about? what linear combination?
• September 12th 2007, 10:55 PM
tukeywilliams
in post #3, the determinant is 0, which implies that those vectors are linearly dependent.

Hes asking how you would find the coefficients.
• September 12th 2007, 10:56 PM
Jhevon
Quote:

Originally Posted by tukeywilliams
in post #3, the determinant is 0, which implies that those vectors are linearly dependent.

Hes asking how you would find the coefficients.

well, that question is for you then, i don't even know what the original question was
• September 12th 2007, 10:58 PM
lord12
the linear combination of the values in the matrix
• September 12th 2007, 11:02 PM
Jhevon
Quote:

Originally Posted by lord12
the linear combination of the values in the matrix

put the matrix in reduced row echelon form and you can read off the solutions from there. as the matrix looks now, it seems there will be a parameter for the first column. you really shouldn't delete your question
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