Proof Check: If $a, b, c$ is a linearly independent system of vectors, is the system $a+b+c, a+b, a$ linearly dependent? Here is my attempt at andwering this question. I would appreciate any feedback.
This system of vector is linearly independent. Assume, to the contrary, that the system is linearly dependent, i.e., $\lambda_1 (a+b+c) + \lambda_2 (a+b) + \lambda_3 a = 0$ implies that at least one of the coefficients is non-zero, say $\lambda_3 \neq 0.$ Note that $\lambda_3 a = (-\lambda_1)(a+b+c) + (-\lambda_2)(a+b)$. Since the system of vectors $a, b, c$ is linearly independent, the sums $a+b+c$ and $a+b$ are both non-zero. Furthermore, since $\lambda_3 \neq 0,$ it follows that at least $\lambda_1$ or $\lambda_2$ has to be non-zero, for if both of them were equal to zero, then we would have $\lambda_3 a = 0$ and so $\lambda_3 = 0,$ a contradiction; hence, assume $\lambda_1 \neq 0.$ Thus, since $(-\lambda_2)(a+b) = \lambda_3 a + \lambda_1(a+b+c) \neq 0,$ it is clearly that $\lambda_2 \neq 0.$ But this leads to absurdity since $\lambda_1,\lambda_2, \lambda_3 \neq 0$ implies that $\lambda_1 (a+b+c) + \lambda_2 (a+b) + \lambda_3 a \neq 0$. Therefore, we retrieve our assumption that the given system of vectors is linearly dependent, and conclude that the system must be linearly independent.
 A: It's a little easier to see what's happening if you look at things from a slightly different perspective.
$$\lambda_1 (a+b+c) + \lambda_2 (a+b) + \lambda_3 a = 0 \\
\Rightarrow (\lambda_1+\lambda_2+\lambda_3)a + (\lambda_1+\lambda_2)b +\lambda_1 c=0
\\ \Rightarrow \lambda_1 =0 \Rightarrow \lambda_2=0 \Rightarrow \lambda_3 =0. $$
The second implication is true because $\{a, b, c \}$ linearly independent forces the coefficient of $c$ to be $0$.  Once you know that, $\lambda_2=0$ because the coefficient of $b$ must be $0$, and similarly $\lambda_3=0$.
A good general approach is to collect coefficients of your linearly independent set and then note that they each must be zero.
A: $\lambda_1 a + \lambda_2 (a+b) + \lambda_(a+b+c) = (\lambda_1+\lambda_2+\lambda_3) a + (\lambda_2+\lambda_3) b + \lambda_3 c$
As $a,b,c$ are lineraly indepndent:
$\lambda_1+\lambda_2+\lambda_3 = 0\\
\lambda_2+\lambda_3 = 0\\
\lambda_3 = 0$
A completely different approach:
If a,b,c are linearly independent, then the matrix [a,b,c] with a,b,c as column vectors has a non-zero determinant.
$[a,b,c]\begin{bmatrix} 1&1&1\\&1&1\\&&1\end{bmatrix} = [a,a+b,a+b+c]$
$\det \begin{bmatrix} 1&1&1\\&1&1\\&&1\end{bmatrix} = 1$
$\det [a,a+b,a+b+c] = \det [a,b,c]$
$[a,a+b,a+b+c]$ is non-sigular and $a,a+b,a+b+c$ are linearly independent.
