Given the Matrix: $$A = \begin{bmatrix} 0.44 & 0.19 & 0.71 & 0.66 & 0.96 & 0.75\\ 0.38 & 0.49 & 0.75 & 0.16 & 0.34 & 0.26\\ 0.77 & 0.45 & 0.28 & 0.12 & 0.59 & 0.51\\ 0.80 & 0.65 & 0.68 & 0.50 & 0.22 & 0.70 \end{bmatrix} $$
Do the columns of $A$ form a basis for $\mathbb{R}^4$? Do they form a basis for $\mathbb{R}^6$?
The $\text{rank}(A) = 4$ and there are 6 columns, so the columns of $A$ are not a basis for $\mathbb{R}^4$
The columns of $A$ are also not a basis for $\mathbb{R}^6$ because none of the columns of $A$ belong to $\mathbb{R}^6$
Questions:
Would the rows of $A$ form a basis of $\mathbb{R}^6$? No, because the $\text{Rank}(A) = 4$?
What is a scenario (if any) is there that the dimension (number of basis vectors) of a basis set $\neq$ the number of elements in each basis vector?
My thought here would be something related to the null space. For example, $$ \begin{bmatrix} 1 & -4 & 3 & -1 \end{bmatrix} \begin{bmatrix} x_1\\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = [ 0 ] $$ where each null space basis vector would be in $\mathbb{R}^4$ (have 4 components) but the dimension of the set would be 3 (3 basis vectors in total). Is this correct?
Thanks!