If {C,D} was not a set of independent matrices then what sort of relation would C and D satisfy? What does taking the transpose of both sides of said relation tell you?
#3
jumbogala
414
4
If {C,D} were not a set of independent matrices, then I would be able to find some numbers a and b not both zero, such that aC + bD = 0.
Transpose of this... aCT + bDT = 0?
#4
Mandark
42
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Right, now use the relations given in the question to rewrite that last equation.
#5
jumbogala
414
4
Okay, so that means
aC + b(-D) = 0.
Now that means that aC + b(-D) = aC + bD
and can be rewritten aC -b(D) = aC + bD... which would mean that a = a, and -b = b. But the only way -b = b can be true is if b = 0. So therefore, b is zero and my initial condition is false, and they are independent! right?
#6
jumbogala
414
4
oh wait, but a can still be something other than zero... my condition is that they are not BOTH zero. so my initial condition is not necessarily false =/
#7
Mandark
42
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Sort of. You can't exactly just equate coefficients if that's what you did. The last equation you wrote implies that bD = 0 by just rearranging. You can't cancel the D here either, but you can use one property given in the question which you haven't used yet. :)
#8
jumbogala
414
4
Wait, why can't I cancel the D? If bD = 0 and D is not zero, then isn't b zero?
Am I supposed to use the fact that they are n x n?
#9
jumbogala
414
4
I just realized the question is not supposed to say that C and D are independent matrices, just that they are n x n and nonzero, if that makes a difference.
Sure it makes a difference. You've got aC+bD=0 and aC-bD=0. To show C and D are independent, you want to show that a and b are both zero. Suppose a is nonzero?
#11
jumbogala
414
4
If it's still true that bD is zero, then if a is nonzero, you'd have aC is not zero.
Then aC + bD = 0 --> aC + 0 = 0 but if aC is not zero this is obviously not true.