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For this particular problem you will not be able to prove that all the constants are equal to the zero element, because we have both told you there exists a vector that is a linear combination of the remaining vectors.
Overall, yes - if you prove that zero vector is produced only by a trivial linear combination (it means all constants are equal to the zero element), then you will have proved that the vectors are linearly independent.Let us consider the vectors [itex]\sin x\cos x, \sin 2x,\cos 2x, \sin ^2x,\cos ^2x[/itex]
What is the rank (number of linearly independent vectors) of this system?
We have determined that [itex]\cos 2x = \cos ^2x -\sin ^2x + 0\ldots[/itex]
The rank can therefore be, at most, four. The vector [itex]\cos 2x[/itex] offers us no new information, so we "remove it from play".
We are left to consider [itex]\sin x\cos x, \sin 2x, \sin ^2x,\cos ^2x[/itex]
[itex]\sin x\cos x = \frac{1}{2}\sin 2x + 0\ldots[/itex]. Bam! Another one!Consider now
[itex]\sin 2x, \sin ^2x,\cos ^2x[/itex]
and continue analogously.
Note that if a system consists of only one vector, then it is linearly independent if and only if that vector is a zero vector.
The dimension of the spanned subspace is 3.
In a general case. If we are given [itex]n[/itex] vectors [itex]a_k\in V_F, k\in \{1,2,\ldots ,n\}[/itex] we want to know the dimension of the subspace spanned by these vectors, we must determine the number of linearly independent vectors in that system of vectors. It's a recursive method:
1) If the number of L.I vectors is [itex]n[/itex], then dimension of spanned subspace is also [itex]n[/itex]
2) Assume [itex]r[/itex]-th vector is a linear combination of the remaining vectors, that means [itex]a_r = \sum_{j=1}^{r-1} a_j + \sum_{j=r+1}^n a_j[/itex]. "Remove the vector [itex]a_r[/itex] from play. Left with [itex]n-1[/itex] vectors. If 1), then result, else step2).
Things to note: we can never get rid of all vectors in the system, because ultimately we will come down to there being one vector left, but then it is linearly independent if and only if it's a zero vector. Hang on, that means the Entire system is linearly dependent all the way.
What dimension subspace does a zero vector span?
If you can wrap your head around all that then I have nothing left to teach you. You are now a Jedi!
Overall, yes - if you prove that zero vector is produced only by a trivial linear combination (it means all constants are equal to the zero element), then you will have proved that the vectors are linearly independent.Let us consider the vectors [itex]\sin x\cos x, \sin 2x,\cos 2x, \sin ^2x,\cos ^2x[/itex]
What is the rank (number of linearly independent vectors) of this system?
We have determined that [itex]\cos 2x = \cos ^2x -\sin ^2x + 0\ldots[/itex]
The rank can therefore be, at most, four. The vector [itex]\cos 2x[/itex] offers us no new information, so we "remove it from play".
We are left to consider [itex]\sin x\cos x, \sin 2x, \sin ^2x,\cos ^2x[/itex]
[itex]\sin x\cos x = \frac{1}{2}\sin 2x + 0\ldots[/itex]. Bam! Another one!Consider now
[itex]\sin 2x, \sin ^2x,\cos ^2x[/itex]
and continue analogously.
Note that if a system consists of only one vector, then it is linearly independent if and only if that vector is a zero vector.
The dimension of the spanned subspace is 3.
In a general case. If we are given [itex]n[/itex] vectors [itex]a_k\in V_F, k\in \{1,2,\ldots ,n\}[/itex] we want to know the dimension of the subspace spanned by these vectors, we must determine the number of linearly independent vectors in that system of vectors. It's a recursive method:
1) If the number of L.I vectors is [itex]n[/itex], then dimension of spanned subspace is also [itex]n[/itex]
2) Assume [itex]r[/itex]-th vector is a linear combination of the remaining vectors, that means [itex]a_r = \sum_{j=1}^{r-1} a_j + \sum_{j=r+1}^n a_j[/itex]. "Remove the vector [itex]a_r[/itex] from play. Left with [itex]n-1[/itex] vectors. If 1), then result, else step2).
Things to note: we can never get rid of all vectors in the system, because ultimately we will come down to there being one vector left, but then it is linearly independent if and only if it's a zero vector. Hang on, that means the Entire system is linearly dependent all the way.
What dimension subspace does a zero vector span?
If you can wrap your head around all that then I have nothing left to teach you. You are now a Jedi!
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