Given linearly independent vectors A1 ... Am , How do you show that when small enough d is added to (A1)1, they still maintain linear independence? It seems intuitive, but I can't figure out what to do after comparing two sets of vectors. I'd appreciate if someone can give me some hints. Thanks!
How to show linear independence of slightly perturbed vectors
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linear-algebra
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0Robert's answer is correct but caution is advised if you want to implement it in a computer program. If the vectors are orthogonal then there is no problem. However if two vectors are almost parallel then they might not be able to tolerate a change that you think is small. – 2012-11-08
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0How small? If matrix $M$ is invertible and $\|B - M\| < 1/\|M^{-1}\|$ where $\|\cdot\|$ is a (Banach algebra) matrix norm, then $B$ is invertible. – 2012-11-09
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Suppose these vectors are in ${\mathbb R}^n$ or ${\mathbb C}^n$. If they are linearly independent, there is a certain $m \times m$ matrix that is invertible. The determinant is a continuous function on $m \times m$ matrices.
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0How do we show that the entries of this matrix are continuous on the entries of $A_1$? – 2012-11-08
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0Can you explain how the continuity of determinant works with setting d to be small? – 2012-11-08
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0@lee the determinant is a continuous function of the entries of your matrix. This means that for "small enough" perturbations to the entries of your matrix (in particular, to those corresponding to $A_1$) the determinant will remain close enough to its original value which is different from 0, and thus your vectors remain linearly independent. – 2012-11-08
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0I get it now. Thanks a lot. – 2012-11-08