Thread: Linear Independence

1. Linear Independence

Hi everyone,

I have the following question that needs confirmation and clarification. This is a review for my elementary differential equations course.

Question: Which of the following sets of vectors are linearly independent?

A. [ 18 ] , [ -9 ]
[ -8 ] , [ 4 ]

B. [ 5 ] , [ -6 ] , [ -1 ]
[ -9 ] , [ -5 ] , [ -14 ]
[ -3 ] , [ 7 ] , [ 4 ]

C. [ 6 ] , [ 3 ] , [ 2 ]
[ -7 ] , [ 8 ] , [ 4 ]
[ 0 ] , [ 0 ] , [ 0 ]

D. [ 1 ] , [ 6 ]
[ 3 ] , [ 1 ]

E. [ 8 ]
[ 2 ]

F. [ 0 ] , [ 5 ]
[ 0 ] , [ -7 ]

Attempt At solving:

A. Since the second matrix is a scalar multiple of the first matrix, the must be linearly dependent so that one is out.

B. If you add together matrix one and matrix 2 together, the result is the 3rd vector which is a linear combination of other vectors in the set so therefore must be linearly dependent as well.

C. None of the vectors in this set are deemed a scalar multiple or a linear combination so must be said as linearly independent.

D. Same as the C.

E. I'm not entirely sure on this one, but would it require at least 2 vectors to be considered as linearly independent?

F. F should not be because there is not a leading 1 so it cannot be linearly independent.

I would like to know which ones I have chosen are correct and which reasoning I have stated above are wrong as this is a review for my differential equations course. I did not previously learnt his material in linear algebra so i am not entirely sure!

2. Your answer for C must be wrong, because the dimension of the space spanned by the vectors is only 2 (they all have zeros in the last component), but there are three vectors. Try to solve the linear system

$x\begin{bmatrix}6\\-7\\0\end{bmatrix}+y\begin{bmatrix}3\\8\\0\end{bmat rix}=\begin{bmatrix}2\\4\\0\end{bmatrix}$

for $x$ and $y$. If you can find a nonzero solution, then your vectors are linearly dependent.

For E, what is the dimension of the space?

For F, I think your conclusion is correct, but I'm not so sure about your line of reasoning. If you include the zero vector in any group of vectors, you're going to have linear dependence, because you can set a linear combination of vectors in the group equal to zero, with all the coefficients of the nonzero vectors being zero, and the zero vector having any nonzero coefficient. Does that make sense?

3. Originally Posted by Belowzero78
Hi everyone,

I have the following question that needs confirmation and clarification. This is a review for my elementary differential equations course.

Question: Which of the following sets of vectors are linearly independent?

A. [ 18 ] , [ -9 ]
[ -8 ] , [ 4 ]

B. [ 5 ] , [ -6 ] , [ -1 ]
[ -9 ] , [ -5 ] , [ -14 ]
[ -3 ] , [ 7 ] , [ 4 ]

C. [ 6 ] , [ 3 ] , [ 2 ]
[ -7 ] , [ 8 ] , [ 4 ]
[ 0 ] , [ 0 ] , [ 0 ]
a<6, -7, 0>+ b<3, 8, 0>+ c<2, 4, 0>= <6a+ 3b+ 2c, -7a+ 8b+ 4c, 0>= <0, 0, 0>
gives the two equations 6a+ 3b+ 2c= 0, -7a+ 8b+ 4c= 0. Multiply the first equation by 2 and subtract the second equation from that to get (12-(-7))a+ (6- 8)b+ (4- 4)c= 19a- 2b= 0. Then b= (19/2)a satisifies that for all a. In particular, taking a= 2 gives b= 19. Putting a= 2, b= 19 into 6a+ 3b+ 2c= 0 gives 12+ 57+ 2c= 0 or 2c= -69 or c= -69/2.

2<6, -7, 0>+ 19<3, 8, 0>- (69/2)<2, 4, 0>= <0, 0, 0>.

D. [ 1 ] , [ 6 ]
[ 3 ] , [ 1 ]

E. [ 8 ]
[ 2 ]

F. [ 0 ] , [ 5 ]
[ 0 ] , [ -7 ]

Attempt At solving:

A. Since the second matrix is a scalar multiple of the first matrix, the must be linearly dependent so that one is out.

B. If you add together matrix one and matrix 2 together, the result is the 3rd vector which is a linear combination of other vectors in the set so therefore must be linearly dependent as well.

C. None of the vectors in this set are deemed a scalar multiple or a linear combination so must be said as linearly independent.

D. Same as the C.

E. I'm not entirely sure on this one, but would it require at least 2 vectors to be considered as linearly independent?
Not exactly- any set consisting of a single non-zero vector is "independent".

F. F should not be because there is not a leading 1 so it cannot be linearly independent.
On the other hand, any set of vectors containing the 0 vector cannot be independent.:
1<0,0>+ 0<5, -7>= <0, 0>.

I would like to know which ones I have chosen are correct and which reasoning I have stated above are wrong as this is a review for my differential equations course. I did not previously learnt his material in linear algebra so i am not entirely sure!

4. Thanks alot guys, So based on reading your conclusions, the correct answer would just be D?

5. Correct. I believe that D is the only linearly independent set of vectors.

6. No, E, consisting of a single non-zero vector, is also and independent set.

I know of no definition of "independent set of vectors" that requires there be more than one vector in the set.

If there were, how would we deal with one-dimensional spaces?

7. I was thinking of that entry as two 1-dimensional vectors. Silly me.