Question

Determine all values of the constant k for which the vectors are linearly independent in R^4
(1,0,1,k ), (-1,0,k,1) and ( 2,0,1,3)
I don't get the concept. Please, explain me

Answer 1

the matrix is \[\left[\begin{matrix}1&-1&2\\0&0&0\\1&k&1\\k&1&3\end {matrix}\right]\] we have a zero row there, it mean they are linearly dependent. How value of k can turn it independent?

Answer 2

Yes, these are linearly dependent but that does not mean that the matrix does not have a dimension. The Rank of the row space is 3 and equals the dimension. But the columns are in \(\frak{R}^4\) space but can never full that space because there are not enough columns.

So, we have these vectors in \(\frak{R}^4\), that can only span \(\frak{R}^3\) , to make these independent, we need to ensure that the columns are independent.

What values of \(k\) will do that?

Answer 3

Is it k \(\neq 0\) and k\(\neq2\)?

Answer 4

How about \(k\ne-1\) that takes care of independence from 1st column, but how can we ensure that it is independent of column 3?

Answer 5

I take \(C_1 V_1 + C_2 V_2 = V_3\) then convert the result

Answer 6

You can take the dot product with each vector and make it equal to zero if possible. Then you can guarantee that they will be independent. That is another approach.

Answer 7

yea, if k = -1 then \(V_2 = -V_1\)

Answer 8

so? what should I do to this problem? consider pair of them?

Answer 9

I row-reduced this matrix and got the following

Answer 10

$$
\left[\begin{matrix}1&-1&2\\0&0&0\\1&k&1\\k&1&3\end {matrix}\right]\\
=\left[\begin{matrix}1&-1&2\\1&k&1\\k&1&3\\0&0&0\end {matrix}\right]\\
=\left[\begin{matrix}1&-1&2\\0&-1-k&1\\0&-1-k&2k-3\\0&0&0\end {matrix}\right]\\
=\left[\begin{matrix}1&-1&2\\0&-1-k&1\\0&0&2k-4\\0&0&0\end {matrix}\right]\\
$$
This imposes some requirements to make these vectors independent. Do you see them?

Answer 11

I see that \(k\ne -1\) and \(k\ne 2\). That's if I did the reduction correctly. With these requirements, the pivots will NOT be zero and the vectors will be independent.

Answer 12

yes, from row 2 , we get -1-k=-1 so, k = 2
from row 3 , we get 2k-4 = 0 so, k =2 , that mean k =2 is solution for it be dependent, just convert?

Answer 13

oops, k \(\neq 0\)

Answer 14

I don't want \(-1-k\) to be zero and I don't want \(2k-4\) to be zero. If they ARE zero, then there is NO chance that these three vectors are independent. We need nonzero values at the Pivots.

Answer 15

so, k \(\neq 0 \) and k\(\neq 2\) how about your -1, it works , too.

Answer 16

because it makes \(V_1 = -V_2\) and the system becomes dependent, right?
therefore, the solution must be 3 :k\(\neq\) -1,0,2

Answer 17

I am not sure about \(k\ne 0\). It seems like it CAN be part of the solution. Did you check my work?

Answer 18

yea, that's your work, row 2 said that -1-k = -1 \(\rightarrow k =0\)

Answer 19

No. We don't need -1-k = -1 , we need -1-k to NOT equal zero.

Answer 20

hey, you confused yourself. hehehe. the whole row cannot =0, right? so let it =0 then negate it, done. right?

Answer 21

you said "We don't need -1-k =-1" yes, we don't want it, if k =0 , we have it. so, k must be NOT =0 , right?

Answer 22

Good point. We need 2 conditions then. The row can not equal zero AND there needs to be a nonzero pivot.

Answer 23

But please check my row reduction, there might be an error.

Answer 24

nope, yours is good, I checked.

Answer 25

Thanks a lot, friend.

Answer 26

No problem :)