# Invertible: A non-square matrix?

So I am doing a question were I have the set column matrix 1 = (3, -8, 1) and column matrix 2 = (6, 2, 5) and the question is asking if this is either a bases for R2 or R3. Can I just say that since the matrix is not a square matrix (nxn) it cannot be a bases for R3 since it is not invertible -- it's not invertible b/c it's not a square matrix. And that it cannot be a bases for R2, since that would be impossible. I unsure how the Invertible Matrix Theorem applies to non-square matrix, which is why I ask this. Thanks for the help and clarifications!

Hint:

1) A base of $\Bbb R^2$ is constituted of vectors belonging to $\Bbb R^2$

2) A base of $\Bbb R^3$ has at least $3$ vectors

• So a non-square matrix, say a 3x4 matrix, can be a base for R3? Or are you saying that is only one of the conditions to be a base for R3 and that there are more than just that one condition? Commented Oct 9, 2014 at 22:13
• No , a basis of $\Bbb R^3$ has exactly $3$ vectors[exactly $\Rightarrow$ at least.] Commented Oct 9, 2014 at 22:18
• Sorry for so many questions, so in that case would a 2x3 Matrix work. Does it matter that a free variable will be exist? Commented Oct 9, 2014 at 22:24
• no because the vectors of such matrix not belong to $\Bbb R^3$. [ask your questions without embarrassment] Commented Oct 9, 2014 at 22:28
• Thank you. So correct me if I'm wrong,but the matrix has to be a square matrix for this work at all -- so long as it is consistent? Commented Oct 9, 2014 at 23:13

A basis of $\mathbb{R}^3$ has exactly three elements in it. So a set of two vectors cannot be a basis for $\mathbb{R}^3$.

A basis of $\mathbb{R}^2$ has exactly two elements in it, and they are vectors in $\mathbb{R}^2$. So a set of two vectors from $\mathbb{R}^3$ can not be a basis for $\mathbb{R}^2$.

• Ye If a family $\mathcal B$ is a base of $\Bbb R^n$ the matrix of $\cal B$ in the canonical base of $\Bbb R^n$ is a $n \times n$ square matrix , conversely a square $n \times n$ matrix $M$ has coloumns of a base of $\Bbb R^n$ if and only if $\det(M)\neq 0$ Commented Oct 10, 2014 at 0:22