Find values of h such that the vectors (2, 4) and (h, 6) span $\mathbb{R}^2$

Question

My homework is asking me to answer problems such as the one that follows:

Find all values of $h$ such that the vectors $\{a_1, a_2\}$ span $\mathbb{R}^2$, where $a_1 = (2, 4)$ and $a_2 = (h, 6)$.

I created an augmented matrix \begin{pmatrix} 2 & h\\ 4 & 6\\ \end{pmatrix} and used elementary row operations to put the matrix in the form \begin{pmatrix} 1 & h/2\\ 0 & 6-2h\\ \end{pmatrix} The answer to the problem is $h \neq 3$. Obviously, $0 = 6 - 2h$ is true when $h = 3$. However, I don't understand why h can be any value except for that value that makes that equation true.

Answer 1

Since $\dim \Bbb R^2 = 2$ and you have two vectors, the only way that they cannot span $\Bbb R^2$ is if they are parallel, in which case, they will span a single line. The vectors $(2,4)$ and $(h,6)$ will be parallel only if $2/h = 4/6$, which gives $h = 3$. Meaning that if $h = 3$, they're parallel and span a single line. Otherwise, they are not parallel and span all of $\Bbb R^2$.

Answer 2

To answer your question, if $6-2h=0$, then we the vectors $(1,0)^T$ and $(h/2,0)^T$ are linearly dependent.

Since the $dim(\mathbb{R}^2)=2$, a set of vectors, $V=\{v_1,v_2,...\}$, with $v_i\in \mathbb{R}^2$, will span the vector space if and only if there are at least 2 linearly independent vectors in $v$.

Since we do not have 2 linearly independent vectors when $h=3$, the set of vectors listed $V=\{(1,0)^T$, $(h/2,0)^T\}$ will not span $\mathbb{R}^2$.