# Thread: Linearly Independent Vectors

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

My solution:
No, they are not. Since $\vec{v_{m}} = \vec{0}$, $rank[\vec{v_{1}} ... \vec{v_{m}}]$ 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!

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

My solution:
No, they are not. Since $\vec{v_{m}} = \vec{0}$, $rank[\vec{v_{1}} ... \vec{v_{m}}]$ 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 $0v_1+\cdots+v_m=0$?