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 &lt; 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 &lt; 0 and y &gt; 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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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  &amp;lt;  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&#039;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  &amp;lt;  0 and   y  &amp;gt;  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&#039;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.

Gia Schaefer · Accepted Answer

The business about passing through the origin is not relevant. Your definition is perfectly fine. The issue is whether there exist constants   a and   b, not both 0, that (simultaneously) work for all   t. For the purpose of finding such constants, or showing that they do not exist,  t  =  0 is completely useless. But you could for example take the two values   t  =  1 and   t  =  −  1 (or −17). You obtain the two linear equations   a  +  b  =  0 and   −  a  +  b  =  0, which have only the solution   a  =  b  =  0.

I am trying to show that the functions t 3 </msup> and b are independent on the w

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