I have 3 vectors, $(0,3,1,-1), (6,0,5,1), (4,-7,1,3)$, and using Gaussian elimination I found that they are linearly dependent. The next question is to express each vector as a linear combination of the other two. Different resources say just to use Gaussian elimination, but I just end up with a matrix in RREF. How can I find different vectors as a linear combination of others?
[Math] How to express a vector as a linear combination of others
linear algebra
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You wish to determine how to solve $\lambda_1 u+\lambda_2 v=w$ where \begin{align*} u &= \langle 1,1,1\rangle & v &= \langle 2,1,1\rangle & w &= \langle a,b,c\rangle \end{align*} To do so, form the augmented system $$ M = \left[\begin{array}{rr|r} 1 & 2 & a \\ 1 & 1 & b \\ 1 & 1 & c \end{array}\right] $$ The system $M$ can be row-reduced with the following steps
- add $-1$ times row 1 to row 2
- add $-1$ times row 1 to row 3
- scale row 2 by $-1$
- add $-2$ times row 2 to row 1
- add $1$ times row 2 to row 3
These row-reductions gives $$ \left[\begin{array}{rr|r} 1 & 0 & -a + 2 \, b \\ 0 & 1 & a - b \\ 0 & 0 & -b + c \end{array}\right] $$ This shows that $\lambda_1 u+\lambda_2 v=w$ has a solution if and only if $b=c$. If $b=c$ then $\lambda_1 u+\lambda_2 v=w$ is solved by \begin{align*} \lambda_1 &=-a+2\,b & \lambda_2 &= a-b \end{align*}
In the comments you ask how to express $v=(6,7,6)$ in terms of \begin{align*} u_1 &= (1,1,1) & u_2 &= (1,1,2) & u_3 &= (1,2,1) & u_4 &= (2,1,1) \end{align*}
This is equivalent to solving the system $$ \left[\begin{array}{rrrr|r} 1 & 1 & 1 & 2 & 6 \\ 1 & 1 & 2 & 1 & 7 \\ 1 & 2 & 1 & 1 & 6 \end{array}\right] $$ Row-reducing gives $$ \left[\begin{array}{rrrr|r} 1 & 0 & 0 & 4 & 5 \\ 0 & 1 & 0 & -1 & 0 \\ 0 & 0 & 1 & -1 & 1 \end{array}\right] $$ This shows that $$ (5-4\,\lambda) u_1 + \lambda u_2 + (1+\lambda)u_3 + \lambda u_4=v $$ for any $\lambda\in\Bbb R$.
This is the definition of linear independence. It is equivalent to the statement that none of the vectors $v_1,\dots,v_n$ can be written as a linear combination of the remaining vectors.
Here's why $0$ is special: You can always write it as a linear combination of any vectors, by taking $0v_1+\dots+0v_n$. Linear independence is the criterion for that to be the only solution. If you picked $(1,0,\dots,0)$, you wouldn't know, first of all, that it is a linear combination of $v_1,\dots,v_n$ at all, and, even if it were, you wouldn't know what linear combination had to work.
Here's an equivalent formulation of linear independence: Any vector $v$ that is in the span of $v_1,\dots,v_n$ (i.e., that can be written as a linear combination of them) must be a unique linear combination of them. There cannot be two different ways of writing it as a linear combination.
Best Answer
Let's look at Gaussian elimination: \begin{align} \begin{bmatrix} 0 & 6 & 4 \\ 3 & 0 & -7 \\ 1 & 5 & 1 \\ -1 & 1 & 3 \end{bmatrix} \xrightarrow{\text{swap row 1 and 3}}{}& \begin{bmatrix} 1 & 5 & 1 \\ 3 & 0 & -7 \\ 0 & 6 & 4 \\ -1 & 1 & 3 \end{bmatrix}\\ \xrightarrow{R_2-3R_1}{}& \begin{bmatrix} 1 & 5 & 1 \\ 0 & -15 & -10 \\ 0 & 6 & 4 \\ -1 & 1 & 3 \end{bmatrix}\\ \xrightarrow{R_4+R_1}{}& \begin{bmatrix} 1 & 5 & 1 \\ 0 & -15 & -10 \\ 0 & 6 & 4 \\ 0 & 6 & 4 \end{bmatrix}\\ \xrightarrow{-\frac{1}{15}R_2}{}& \begin{bmatrix} 1 & 5 & 1 \\ 0 & 1 & 2/3 \\ 0 & 6 & 4 \\ 0 & 6 & 4 \end{bmatrix}\\ \xrightarrow{R_3-6R_2}{}& \begin{bmatrix} 1 & 5 & 1 \\ 0 & 1 & 2/3 \\ 0 & 0 & 0 \\ 0 & 6 & 4 \end{bmatrix}\\ \xrightarrow{R_4-6R_2}{}& \begin{bmatrix} 1 & 5 & 1 \\ 0 & 1 & 2/3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\\ \xrightarrow{R_1-5R_2}{}& \begin{bmatrix} 1 & 0 & -7/3 \\ 0 & 1 & 2/3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}\\ \end{align} If $v_1$, $v_2$ and $v_3$ are your vectors, this says that $$ v_3=-\frac{7}{3}v_1+\frac{2}{3}v_2 $$ because elementary row operations don't change linear relations between the columns.