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.
Best Answer
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 $\mathbb{R}^2$.