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The method I'm using to solve this is to form an augmented matrix then use Row Reduction. However, I was wondering to what extent would I stop doing Row Reduction?

I know that the vector is in the span of vectors as long as the solution is consistent. I just don't know when to stop using Row Reduction.

I am also aware that there are other posts on this, but no one specifies when you should stop using Row Reduction.

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  • $\begingroup$ Once you determine the rank you can stop using row reduction. You don't need to solve completely, you just need to show that solution exists. $\endgroup$ – Sonal_sqrt Feb 28 '18 at 2:48
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We do row reducing operations in order to determine whether given vectors (spanning vectors and the vector we are checking) are linearly independent or not. To do that, we try to find Row Echelon Form of that matrix.

Since you are asking when to stop, I would say Row Echelon Form has some properties and when you get to a matrix satisfying those, then you are done. Those properties are:

  • All nonzero rows (rows with at least one nonzero element) are above any rows of all zeroes (all zero rows, if any, belong at the bottom of the matrix),

  • the leading coefficient (the first nonzero number from the left, also called the pivot) of a nonzero row is always strictly to the right of the leading coefficient of the row above it (some texts add the condition that the leading coefficient must be $1$).

(source: https://en.wikipedia.org/wiki/Row_echelon_form If you check here, there are examples of matrices that are in Row Echelon Form. I suggest you to see them in order to fully understand the properties)

When matrix is in this form, number of non-zero rows gives us the rank of that matrix and when rank is equal to number of vector i.e. number of columns, we can say that vectors are linearly independent. Otherwise, if we have a zero row, it means that at least one of the vectors can be written in linear combination of others therefore they are linearly dependent.

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  • $\begingroup$ Thank you. This is just the response that I needed! May I ask, what is meant when you say linearly independant? $\endgroup$ – Michael Johnson Feb 28 '18 at 3:09
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    $\begingroup$ Nevermind I just read the response again and I get it now. Thanks again! $\endgroup$ – Michael Johnson Feb 28 '18 at 3:10
  • $\begingroup$ Linearly independent means we can't write one of the vectors with respect to linear combination of other vectors. In other words, we can also say that if vectors $\vec{v_1}, \vec{v_2}, \vec{v_3}$ are linearly independent, then $a_1\vec{v_1} + a_2\vec{v_2} + a_3\vec{v_3} = 0$ if and only if $a_1 = a_2 = a_3 = 0$. And you are welcome :) $\endgroup$ – ArsenBerk Feb 28 '18 at 3:11

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