[Math] Express vector as linear combination of vectors

linear algebravectors

let's say I have two vectors:

$u = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$ and $v = \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}$

I can easily express $\; w = \begin{pmatrix} 3 \\ 2 \\ 2 \end{pmatrix}$ as a linear combination of u and v.

My problem is: how can I express a vector v as a linear combination of n vectors?

I searched all around and I just found about expressing a $R^n$ vector as linear combination of exactly $n$ vectors from $R^n$.

  • $[u_1 \; u_2 \; u_3 \; | \; v]$
  • apply gaussian elimination;

But I didn't discover how to express it with k vectors ($ k \ne n$), for example, if I apply this method with u and v to find w, this is what happens:

$ \left[
\begin{array}{cc|c}
1&2&3\\
1&1&2\\
1&1&2
\end{array}
\right] $ applying elimination: $ \left[
\begin{array}{cc|c}
1&2&3\\
0&-1&-1\\
0&0&0
\end{array}
\right] $

And I know the answer for $\;au + bv = w\;$ is $\;(a = 1, \;b = 1)$ but I couldn't extract this information from the matrix.

Formalizing: Given a $R^n$ vector, how can I express it as a linear combination of $k$ vectors, where $k \ne n$ ?

Best Answer

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 $\lambda\in\Bbb R$.

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