Linear independence of columns of a matrix

AI Thread Summary
The discussion centers on determining the linear independence of the columns of a given matrix, particularly in light of a row of zeros. It is noted that the presence of a zero row indicates that the dimension of the space spanned by the column vectors is less than the number of columns. However, the specific set of vectors presented does not include the zero vector, allowing for the possibility of linear independence despite the zero row. The conclusion drawn is that while a row of zeros can signal dependence in some cases, it does not automatically determine the independence of the columns in this scenario. Ultimately, the analysis confirms that the columns of the matrix can be considered linearly independent.
TrippingBilly
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Homework Statement


Are the columns of this matrix linearly independent?
1...3...-2
0...-8...11
0...0...1
0... 0... 0
(periods are just to make spacing clear)

The Attempt at a Solution


What is confusing me is the last row of zeros. If a set of vectors contains the zero vector, it is linearly dependent..but would this affect the linear independence of the columns of the matrix? If you augment the matrix with the zero vector, then the third row says that the only solution is the trivial one, which means that the columns of the matrix are linearly independent.
 
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If the columns are linearly dependent, then the third column can be made by adding multiples of the first and second columns together. Can this be done?
 
No it can't, so the fact that there is a row of zeros doesn't matter for the columns of the matrix..
 
Well no, because you're not looking at the rows here. To put the question another way, is the set of vectors {(1 0 0 0), (3 -8 0 0), (-2 11 1 0)} linearly independent?
 
A row of zeros is significant because it immediately tells us that the dimension of the space spanned by the column vectors must be less than the number of columns (i.e. the number of components of the vectors). In this case however, that only means that the space spanned by the vectors is at most three-dimensional, but you knew that already since there are only three vectors in this problem.

In other problems, a row of zeros may immediately give away the answer. For example: Are (a b 0), (c d 0) and (e f 0) linearly independent? No, they must be linearly dependent because there are three of them and they're all in the x-y plane, which is two-dimensional.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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