Linear Algebra 8h: A Linear System with Zero Columns

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[@MathTheBeautiful]

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[@rovshanabdurrahimov5892]

amazing!!!!!!!!

[@anantgairola3394]

Great lecture series as always, Professor Grinfeld! Is it correct to think about this the following way? C1=[1 2 3]', C2= [0 0 0]', C3=[2 4 3]' and C4=[0 0 0] are actually a proper sub-space of R3 since the second entry is twice the first entry so C1, C2, C3 and C4 can at most be R2. The right hand side is "most likely" going to belong to R3 and will therefore lie outside the span of LHS. I am trying to think of this as having two geometric vectors in a plane and then hoping to resolve any arbitrary vector in 3-D space (could be possible but most likely not).
PS I am watching this entire series not for "conventional academic purpose" but as a lot more entertaining substitute for watching a show on Netflix at night haha! Watching your lectures is pure joy! So insightful :D Thank you for making these!

[@vmkankku]

How could I use this fact in real life applications? What kind of examples this would have? (System with zero columns and their null space construction)

[@MathTheBeautiful]

How about the equation 2x + 3y + 0z = 7?

[@loremazza5191]

Another reason why the (3,16,6) vector doesn't belong to the column space is:
Since we know (3,6,6) does belong to the column space, the vector (3,16,6) is part of the column space only if the difference between those two vectors, (0,10,0) is itself part of the column space, but this is not true.

[@MathTheBeautiful]

Good point!

[@ulagy1]

Isn't [2, 0,-1, 0] also a part of null space?

[@SeeTv.]

No because in the third row this would be 6+0-3+0 which is 3 and not 0. But if the third column was [2,4,6] i.e. twice the first one then it would be correct. Then the Null Space would be 3 dimensional by the rank nullity theorem. The rank would be 1 because it's just the span of the first column, i.e. a line. And 4-1=3 = dimension of null space