Linear algebra questions and answers

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lurtzslikgtgjd 2022-05-17 Answered

I am trying to show that the functions t 3 and b are independent on the whole real line. To do this, I try and prove it by contradiction. So assume that they are dependent. So then there must exists constants a and b such that a t 3 + b | t | 3 = 0 for all t ( , ). Now pick two points x and y in this interval and assume without loss of generality that x < 0, y 0. Now form the simultaneous linear equations
a y 3 + b | y | 3 = 0, viz.
[ x 3 | x | 3 y 3 | y | 3 ] [ a b ] = [ 0 0 ]
Now here's my problem. If I look at the determinant of the coefficient matrix of this system of linear equations, namely x 3 | y | 3 y 3 | x | 3 and noting that x < 0 and y > 0, I have that the determinant is non-zero which implies that the only solution is a = b = 0, i.e. the functions t 3 and | | t | 3 are linearly independent. However what happens if indeed y = 0? Then the determinant of the matrix is 0 and I have got a problem.
Is there something that I am not getting from the definition of linear independence?
The definition (I hope I state this correctly) is: If f and g are two functions such that the only solution to a f + b g = 0     t in an interval I is a = b = 0, then the two functions are linearly independent.
But what happens if my functions pass through the origin, like the above? Then I've just shown that there exists a t in an interval containing zero such that the two functions are zero, viz. I can plug in any b and a such that a f + b g = 0.





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