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Consider the following matrix: $$\begin{bmatrix} 1 & 0& 1 & 0 \\ -1 &-1 &0& 0 \\ 1 &-1& 0 &-1 \\ -1 & 0& 1 &-1 \\ \end{bmatrix}$$ And I want to know if the associated set of vectors spans $\Bbb R^4$.

I took $\det(A)$ and got $-4$. Does this mean because the matrices determinant is not non-zero, that the set of vectors spans $\mathbb R^4$?

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  • $\begingroup$ There is a theorem stating that if a linear function $f$ satisfies that its determinant is non zero, then for every basis $(b_\alpha)$ the family $(f(b_\alpha))$ is also a basis. Hence, yes. $\endgroup$ – Will M. May 17 at 17:07
  • $\begingroup$ Please type the question when feasible, as here. Nobody wants to keep flipping back between the image and the question to see the relation between the vectors and the matrix. $\endgroup$ – saulspatz May 17 at 17:09
  • $\begingroup$ Sorry, this was my first time ever posting $\endgroup$ – gooddayjay May 17 at 17:09
  • $\begingroup$ Can someone fix the dollar that I forgot please 😆 $\endgroup$ – Maximilian Janisch May 17 at 17:10
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The determinant is non-zero implies that the vectors are linearly independent. Thus we have a set consisting of 4 linearly independent vectors in a vector space of dimension 4, thus the collection is a basis and in particular spans $\Bbb R^4$

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Yes, the determinant is nonzero if and only if the columns of your matrix are linearly independent.

Suppose you could find a set of four linearly independent vectors that don't span $\mathbb{R}^4$. We know that a basis for $\mathbb{R}^4$ has four vectors (take the standard basis). Then by the replacement theorem, adding another vector which is not in the span of your vectors will make them linearly dependent, a contradiction since the new vector was not in the span of the others.

So any set of four linearly independent vectors will span $\mathbb{R}^4$.

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I tend to think from linear transformation point of view. The determinant of a matrix indicates how a unit volume will be scaled up or down after the linear transformation. If det is $0$, it means this transformation collapse this unit volume into a lower dimension. In this case, det is non-zero. So the unit volume does not collapse. Therefore this linear transformation spans $\Bbb R^4$

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