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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!

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Robert'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. –  tst Nov 8 '12 at 22:49
    
How small? If matrix $M$ is invertible and $\|B - M\| < 1/\|M^{-1}\|$ where $\|\cdot\|$ is a (Banach algebra) matrix norm, then $B$ is invertible. –  Robert Israel Nov 9 '12 at 2:48

1 Answer 1

up vote 4 down vote accepted

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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How do we show that the entries of this matrix are continuous on the entries of $A_1$? –  pedrosorio Nov 8 '12 at 22:52
    
Can you explain how the continuity of determinant works with setting d to be small? –  lee Nov 8 '12 at 23:00
    
@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. –  pedrosorio Nov 8 '12 at 23:04
    
I get it now. Thanks a lot. –  lee Nov 8 '12 at 23:09

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