Find ALL value's of $h$ so that the vectors $$a_1= (-1,h,7)$$ $$a_2=(4,-2,5)$$ $$a_3=(1,-6,2)$$ span $\mathbb{R}^3$.
Previously, I posted asking how to find a single value of $h$ so that the given vectors span $\mathbb{R}^3$. I am now trying to figure out how to find ALL values of such $h$. A got a great answer: Trying to find the value of $h$ such that the set of vectors spans $\mathbb{R}^3$
Don Thousand restated the question as "for what values of $h$ will the vectors NOT span $\mathbb{R}^3$ and he demonstrated that the vectors don't span $\mathbb{R}^3$ $\iff$ $h=60$
His response was based on this particular problem, where he used that in this case, if one vector can be written as a linear combination of the others, than they all can; however, this will not hold in general. Does anyone have a systematic way to approach this problem? Thank you.
Also, I'm trying to solve this problem without resorting to arguments involving linear independence, because we haven't covered that in class yet. Thank you.
Best Answer
Well, if $v=\lambda_1 a_1 + \lambda_2 a_2 + \lambda_3 a_3$ for some $\lambda_1, \lambda_2, \lambda_3 \in \mathbb R$, then $v=A\left(w\right)$ where $w=(\lambda_1, \lambda_2, \lambda_3)\in\mathbb R^3$ and $A$ is linear transformation with matrix $$\left[\begin{matrix} -1& 4 & 1 \\ h & -2 & 6 \\ 7 & 5 & 2\end{matrix}\right].$$ So, every $v\in \mathbb R^3$ can be written as linear combination of $a_1,a_2,a_3$ if and only if $A$ is surjective. And $A$ is surjective if and only if $\det A\ne 0$ (note that $A$ is injective iff it is surjective). This give us linear equation in terms of $h$. Zero of this (linear) polynomial is precisely value of $h$ for which $A$ is not surjective.
This is systematic way for solving this.
More elementary way for solving would be to write $v=Aw$ as system of linear equation and then with elementary transformations reduce this system to simpler one from which value of $h$ can be easily deduced.