# For what $\lambda \in \mathbb{R}$ is this a spanning set (Erzeugendensystem) of the vector space $\mathbb{R}^3$ ? {(1, 3, 4), (1, λ, 5), (1, 4, 3)}

The set forms a spanning set (German: Erzeugendensystem) if the matrix is linearly independent, which is true if the determinant is non-zero.

We have $$1 \cdot (\lambda \cdot 3 - 4 \cdot 5) - 1 \cdot (3 \cdot 3 - 4 \cdot 4) + 1 \cdot (3 \cdot 5 - \lambda \cdot 4) = 3\lambda - 20 - 9 + 16 + 15 - 4\lambda = -\lambda + 2 \neq 0 \implies \lambda \neq 2$$

So we have $$\lambda \in \mathbb{R}$$ \ {$$2$$}

Is this correct ?

The course I take is in German, but it is not my mother tongue. Is "spanning set of vector space" an accurate translation for "Erzeugendensystem" ?

You have shown that the three given vectors in $$\mathbb R^3$$ are linearly independent if $$\lambda \neq 2$$.
In my point of view, you have only shown that if $$\lambda\neq2$$, then $$(1,3,4),(1,\lambda,5),(1,4,3)$$ is linearly independent. But, you need to show is if $$(x,y,z)\in\mathbb{R}^3$$, then $$(x,y,z)\in\text{span}\{(1,3,4),(1,\lambda,5),(1,4,3)\}$$. In other words, you need to prove that if $$(x,y,z)\in\mathbb{R}^3$$, then there exist $$\alpha,\beta,\gamma\in\mathbb{R}$$ such that $$$$(x,y,z)=\alpha(1,3,4)+\beta(1,\lambda,5)+\gamma(1,4,3).$$$$ Thus, you need to solve this equation for $$\alpha,\beta,\gamma$$ and prove that if $$\lambda\neq2$$, then the solution is \begin{align} \alpha&=\frac{(20-3\lambda)x+(\lambda-4)z-2y}{\lambda-2}\\ \beta&=\frac{-7x+y+z}{\lambda-2}\\ \gamma&=\frac{(4\lambda-15)x-(\lambda-3)z+y}{\lambda-2}. \end{align} .
• If $S$ is linearly independent, then $\dim \operatorname{span} S = |S|$. No need to go through the hard way. Nov 1, 2023 at 16:06