Linear Independence: Answer to Homework

[nuuskur]

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.

 

[Mark44]

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.

 

[Ray Vickson]

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?

 

[nuuskur]

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]

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.