Relationship between determinants and basis formation

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SUMMARY

The discussion clarifies that a set of three vectors in R3 forms a basis if and only if the determinant of the matrix formed by these vectors as columns is non-zero. This is established through the concept of linear independence, where the only solution to the equation involving these vectors equating to the zero vector is the trivial solution. The confusion arises from the interpretation of parameterization in relation to linear dependence, particularly when a row of zeros appears in the matrix after Gaussian elimination.

PREREQUISITES
  • Understanding of determinants in linear algebra
  • Familiarity with Gaussian elimination techniques
  • Knowledge of linear independence and dependence concepts
  • Basic proficiency in vector spaces, specifically R3
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  • Study the properties of determinants in linear algebra
  • Learn about Gaussian elimination and its applications in solving systems of equations
  • Explore the concept of linear independence and dependence in vector spaces
  • Investigate the implications of parameterization in the context of linear combinations
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Students and educators in mathematics, particularly those studying linear algebra, as well as anyone seeking to deepen their understanding of vector spaces and determinants.

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Hi,

I'm scratching my head over the statement from my textbook which states when determinant is non-zero, the set of vectors blah blah is a basis for r^3.

That does not make any sense to me because I know when a row of zeros in a matrix occur; the determinant is zero (through Gaussian Elimination), which means that a column has to be parameterized; then you are to set a arbitrary variables accordingly and from that one may create the needed basis.

What am I not understanding?
 
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I'm afraid what you are saying makes no sense to me. I assume that the determinant you are talking about is the determinant of a matrix having the given vectors as columns. Second, I assume you are talking about a set of three given vectors in R3.

Yes, it is true a set of three vectors in R3 will form a basis for R3 if and only if the determinant of the matrix having those vectors as columns (or rows) is non-zero.

That is true because three vectors will be a basis for R3 if and only if they are independent: that the equation
x\begin{bmatrix}a \\ b\\ c\end{bmatrix}+ y\begin{bmatrix}d \\ e\\ f\end{bmatrix}+ z\begin{bmatrix}g \\ h \\ i\end{bmatrix}= \begin{bmatrix} 0 \\ 0 \\ 0\end{bmatrix}
has only the "trivial" solution x= y= z= 0.

That is the same as saying that the three equations ax+ dy+ gz= 0, bx+ ey+ hz= 0, and cx+ fy+ iz= 0 have only the solution x= y= z= 0.

And that is true if and only if the "matrix of coefficients"
\begin{bmatrix}a & d & g \\ b & e & h \\ c & f & i\end{bmatrix}
has non-zero determinant.

But I have no idea what you mean by "which means that a column has to be parameterized". What do you mean by "parameterizing" a column?
 
Have a look at the http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/lindep/lindep.html" . Three vectors u,v and w are tested for linear dependence.
The corresponding matrix has a row of zeroes (after row reduction). The parameter is then z and you can show that the vector w can be expressed as a linear combination of two other vectors u and v.
 
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