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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" ?

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2 Answers 2

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You have shown that the three given vectors in $\mathbb R^3$ are linearly independent if $\lambda \neq 2$.
Assuming you have learned about/can use the Steinitz exchange lemma or similar results, this is all that had to be proven.

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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 \begin{equation} (x,y,z)=\alpha(1,3,4)+\beta(1,\lambda,5)+\gamma(1,4,3). \end{equation} 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} .

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    $\begingroup$ If $S$ is linearly independent, then $\dim \operatorname{span} S = |S|$. No need to go through the hard way. $\endgroup$
    – aschepler
    Nov 1, 2023 at 16:06

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