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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.

Eoin
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orb
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2 Answers2

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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 $\Bbb R^2$.

Ivo Terek
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  • Thank you. I completely understand what you are saying. How do I show my work for a problem like this? Should I create an augmented matrix with one of the vectors set to the right side of the equations (normally called b)? Should I find the rref of that augmented matrix? – orb Sep 16 '14 at 01:50
  • What you did with the matrix is correct. Notice that the matrix having a row entire made of zeros is equivalent to saying that the vectors are parallel. This can be generalized for $n$-dimensional space, and $n$-vectors, and it uses the concept of linear dependence/independence, which I believe you will learn soon. – Ivo Terek Sep 16 '14 at 01:55
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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 $\mathbb{R}^2$.

Eoin
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