# Linear independence

## Homework Statement

Assume vectors ##a,b,c\in V_{\mathbb{R}}## to be linearly independent. Determine whether vectors ##a+b , b+c , a+c## are linearly independent.

## The Attempt at a Solution

We say the vectors are linearly independent when ##k_1a + k_2b +k_3c = 0## only when every ##k_n = 0## - the only solution is a trivial combination.
Does there exist a non-trivial combination such that
##k_1(a+b) + k_2(b+c) + k_3(a+c) = 0##?. Distributing:
##k_1a + k_1b + k_2b + k_2c + k_3a + k_3c = (k_1 + k_3)a + (k_1+k_2)b + (k_2+k_3)c = 0## Since ##a,b,c## are linearly independent, the only way this result can occur is when:
##k_1+k_3 =0\Rightarrow k_1 = -k_3##
##k_1+k_2 =0##
##k_2+k_3 =0##

Substituting eq 1 into eq 2 we arrive at ##k_2 - k_3 = 0## and according to eq 3 ##k_2 + k_3=0##, which means ##k_2 - k_3 = k_2 + k_3##, therefore ##k_3 = 0##, because ##k=-k## only if ##k=0##. The only solution is a trivial combination, therefore the vectors ##a+b, b+c, a+c## are linearly independent.

Last edited:

Mark44
Mentor

## Homework Statement

Assume vectors ##a,b,c\in V_{\mathbb{R}}## to be linearly independent. Determine whether vectors ##a+b , b+c , a+c## are linearly independent.

## The Attempt at a Solution

We say the vectors are linearly independent when ##k_1a + k_2b +k_3c = 0## only when every ##k_n = 0## - the only solution is a trivial combination.
Does there exist a non-trivial combination such that
##k_1(a+b) + k_2(b+c) + k_3(a+c) = 0##?. Distributing:
##k_1a + k_1b + k_2b + k_2c + k_3a + k_3c = (k_1 + k_3)a + (k_1+k_2)b + (k_2+k_3)c = 0## Since ##a,b,c## are linearly independent, the only way this result can occur is when:
##k_1+k_3 =0\Rightarrow k_1 = -k_3##
##k_1+k_2 =0##
##k_2+k_3 =0##

Substituting eq 1 into eq 2 we arrive at ##k_2 - k_3 = 0## and according to eq 3 ##k_2 + k_3=0##, which means ##k_2 - k_3 = k_2 + k_3##, therefore ##k_3 = 0##, because ##k=-k## only if ##k=0##. The only solution is a trivial combination, therefore the vectors ##a+b, b+c, a+c## are linearly independent.
That works for me. (IOW, I agree that the three new vectors are linearly independent.)

Instead of working with the system of equations, you can set up a matrix and row reduce it. If you end up with the identity matrix, what that says is that ##k_1 = k_2 = k_3 = 0##, and that there are no other solutions.

The matrix looks like this, from your system:
$$\begin{bmatrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1\end{bmatrix}$$
After a few row operations, the final matrix is I3.

Ray Vickson
Homework Helper
Dearly Missed
That works for me. (IOW, I agree that the three new vectors are linearly independent.)

Instead of working with the system of equations, you can set up a matrix and row reduce it. If you end up with the identity matrix, what that says is that ##k_1 = k_2 = k_3 = 0##, and that there are no other solutions.

The matrix looks like this, from your system:
$$\begin{bmatrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1\end{bmatrix}$$
After a few row operations, the final matrix is I3.

Alternatively, you can compute the determinant of the matrix to find that it is nonzero. What would that tell you?

Alternatively, you can compute the determinant of the matrix to find that it is nonzero. What would that tell you?
Oh. Cramer's rule.
##k_n = \frac{D_{k_n}}{D}## and since the determinant of the system is non zero, the corresponding determinants for every ##k_n## would be 0 (a full column of 0-s means det = 0) and therefore ##k_1 = k_2 = k_3 = 0##

Ray Vickson