Determine whether the span of the column vectors of the given is in .?

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SUMMARY

The discussion focuses on determining whether the span of the column vectors of a given 4x4 matrix is equal to the entirety of Euclidean space R4. The matrix provided is:

1 0 1 -1
0 -1 -3 4
1 0 -1 2
-3 0 0 -1

To establish if the span is R4, the vectors must be independent, which is equivalent to the matrix being invertible. The properties of a basis for a vector space are outlined, emphasizing that if the vectors are independent and span the space, the matrix will be invertible.

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  • Understanding of vector spaces and their properties
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  • Familiarity with the concept of span in linear algebra
  • Basic comprehension of Euclidean spaces, specifically R4
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Determine whether the span of the column vectors of the given is in...?

Homework Statement



determine whether the span of the column vectors of the given matrix is in euclidean space R=4

1 0 1 -1
0 -1 -3 4
1 0 -1 2
-3 0 0 -1

this question is under the inverse of square matrix section of my textbook. Unfortunately it dosent show any examples for this kind of question so I am really clueless.

whats the connection between span and inverse matrices? How can i prove if span of some vectors is in some euclidean space?
 
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Have you given the exact wording of the problem? The phrase "in euclidean space R4" seems strange to me- each of the columns has four numbers and so is in R4 and so their span must be "in" R4 (a subspace of R4). I suspect the question is not whether the span is in R4 (which is trivially true) but whether or not the span is all of R4.

A "basis" for a vector space has three properties:
1) the vectors are independent
2) the vectors span the space
3) the number of vectors in the set is that same as the dimension of the space.

And if any two of those are true, the third is also,

Of course, R4 has dimension 4 and there are 4 columns in that matrix. The vectors will span R4 if and only if the vectors are independent. In that case, it will also be true that the matrix is invertible.
 

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