Linear independence

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  • #1
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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.

Homework Equations




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.
 
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  • #2
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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.

Homework Equations




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.
 
  • #3
Ray Vickson
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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?
 
  • #4
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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##
 
  • #5
Ray Vickson
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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##

No, I was not referring to Cramer's rule (which is rarely actually used when solving equations). I was referring to the theorem that if det(A) ≠ 0 then the n ×n system Ak = 0 has k = 0 as its only solution.
 

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