# Linearly Independent Vectors

• May 1st 2011, 08:48 PM
tangibleLime
Linearly Independent Vectors
Again, I think I have come up with the correct answer, but the answer is not provided as theoretical answers are omitted from the answer key (bad system if you ask me.)

Problem:
Consider the vectors $\displaystyle \vec{v_{1}},\vec{v_{2}},...,\vec{v_{m}}$ in $\displaystyle \mathbb{R}^{n}}$, with $\displaystyle \vec{v_{m}} = \vec{0}$. Are these vectors linearly independent?

My solution:
No, they are not. Since $\displaystyle \vec{v_{m}} = \vec{0}$, $\displaystyle rank[\vec{v_{1}} ... \vec{v_{m}}]$ will fail to equal m. There will not be a pivot in every column, which is a requirement for a set of linearly independent vectors.

Am I on the right track? Any input appreciated!
• May 1st 2011, 08:52 PM
Drexel28
Quote:

Originally Posted by tangibleLime
Again, I think I have come up with the correct answer, but the answer is not provided as theoretical answers are omitted from the answer key (bad system if you ask me.)

Problem:
Consider the vectors $\displaystyle \vec{v_{1}},\vec{v_{2}},...,\vec{v_{m}}$ in $\displaystyle \mathbb{R}^{n}}$, with $\displaystyle \vec{v_{m}} = \vec{0}$. Are these vectors linearly independent?

My solution:
No, they are not. Since $\displaystyle \vec{v_{m}} = \vec{0}$, $\displaystyle rank[\vec{v_{1}} ... \vec{v_{m}}]$ will fail to equal m. There will not be a pivot in every column, which is a requirement for a set of linearly independent vectors.

Am I on the right track? Any input appreciated!

I mean, if you want to do it in terms of matrices you're correct, but isn't it just simpler to note that $\displaystyle 0v_1+\cdots+v_m=0$?