I have a pivot in every row, but it is still not linearly independent

[flyingpig]

Homework Statement

I need to argue this properly

Let's say I have a matrix A and rref(A) is given as

\begin{bmatrix}<br /> 1 &amp; 0&amp;-1 \\ <br /> 0&amp; 1 &amp; -1<br /> \end{bmatrix}

Since I have a pivot in every row, why isn't this linearly independent? Don't give me other arguments like "because there are more vectors than entires", "it's not a good basis for R^2".

 

[lanedance]

the column vectors are clearly linearly dependent as the 3rd can be written as a linear combo of the first 2
c3 = -c1 -c2

 

[flyingpig]

I don't need other counterarguments. I want to know why pivots in every row does not mean linear indepedence

 

[Mark44]

Why isn't what linearly independent? You need to be more specific about what you're describing as linearly independent/dependent. The rows, taken as vectors in R³, are clearly linearly independent (neither one is a multiple of the other). The column vectors, taken as vectors in R², are clearly linearly dependent, since there are three of them.

Let's call your matrix A. If your matrix represents the equation Ax = 0, where x is a vector in R³, then there are nontrivial solutions for x. This is the same as saying that the nullspace of A is not just the zero vector.

 

[lanedance]

I don't need other counterarguments. I want to know why pivots in every row does not mean linear indepedence

it wasn't a counter-argument, it was an observation of the linear dependence of the column vectors

As Marks says, you need to be clear about what you are actually asking, as your question does not seem well formed.

The row vectors are clearly linearly independent.

If you are trying to solve for a point in R^3, as in the solution Ax=0 , then each row acts as a constraint (geometrically it represents a plane of allowable solutions in R^3)

as you have two non-parallel planes, their intersection is a line of points that satisfy both constraints - giving infinite solutions...

 

[flyingpig]

I got mixed up with row vectors and column vectors, that's the answer I was looking for.

I still don't understand what is a row vector, I still think the columns are defined as x, y, z, and so on continuously on the vertical

 

[HallsofIvy]

If you don't know the difference between rows and columns in a matrix you need to talk to your teacher!

You start of by asking "Why isn't this independent" without any clear statement of what "this" is. Grammatically, you would be referring to the matrix, but "independent" applies to a set of vectors, not a matrix.

If you are referring to the columns of the matrix as vectors,
\begin{bmatrix}1 \\ 0\end{bmatrix}\begin{bmatrix}0 \\ 1\end{bmatrix}\begin{bmatrix}-1 \\ -1 \end{bmatrix}
They are dependent because any set of 3 2-dimensional vectors (any set of more than n n-dimensional vectors) must be dependent. The pivots have nothing to do with that!

If you are talking about the rows of the matrix as vectors,
\begin{bmatrix}1 &amp; 0 &amp; -1\end{bmatrix}, \begin{bmatrix}0 &amp; 1 &amp; -1\end{bmatrix}
then they are independent.