I have a matrix A,

This matrix is equal to the following, where ^T = transpose

A= [u1 u2 u3]

u1 = (1, 1, 1, 3)^T

u2 = (1, 0, 2, 2)^T

u3 = (1, −4, 1, 4)^T

I'm looking to determine whether or not the columns of A are linearly independent in R4

Here is where I'm uncertain, with the transpose of the vectors, this is the resulting matrix

 1   1   1 
 1   0  -4 
 1   2   1 
 3   2   4 

If i'm looking for the values of each, c1,c2,c3,c4, are these values now based on the rows because of the transposition?

 1c1     1c1     1c1 
 1c2     0c2    -4c2 
 1c3     2c3     1c3 
 3c4     2c4     4c4 

Or does it stay normalized?

 1c1     1c2     1c3 
 1c1     0c2    -4c3 
 1c1     2c2     1c3 
 3c1     2c2     4c3 

I realize that this may be an odd question, but the math department at my school hires grad students and adjuncts to instruct class.

  • $\begingroup$ Second one is correct because your are checking for $\Bbb R^4$ vectors. $\endgroup$
    – user312097
    Oct 7 '17 at 22:05

Your vectors are $u_1, u_2, u_3$. To determine whether these are linearly dependent or not, we need to see if $$c_1 u_1 + c_2u_2 + c_3 u_3 = 0$$

has a non-trival solution or not for $c_k \in \Bbb R$ (since you did not specify the field I assumed it to be real numbers).

Therefore we have,

$$c_1 (1,1,1,3)^T + c_2(1,0,2,2)^T + c_3 (1, -4, 1, 4)^T = 0$$

$$ \begin{bmatrix}c_1 + c_2 + c_3 \\ c_1-4c_3 \\ c_1 + 2c_2 + c_3\\3c_1 + 2c_2 + 4c_3\end{bmatrix} = 0$$

Solving the system you can see $c_1 = c_2 = c_3 = 0$.

Hence this is a linearly independent system.


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