How to find a linearly independent vector?

Question

Given two vectors $(1,2,8),(0,1,9)$ find a 3rd vector that is linearly independent from these two vectors.

I sort of have an idea how to go about solving the problem but I'm not 100% sure. I'm know we just want to find a vector that can't be written as the sum of the two given vectors but how exactly do I go about finding one such vector?

Got it now many thanks to all the helpers some brilliant explanations.

Answer 1

I like your idea about finding a vector that can't be written as a sum of the two vectors above. Let's take a look at what that would look like.

Every possible sum of these two vectors can be expressed as $c_{1}(1,2,8) + c_{2}(0,1,9)$ for some $c_{1}, c_{2}$ in $\Bbb R$. So, all possible sums can be expressed in the form $(c_{1}, 2c_{1} + c_{2}, 8c_{1} + 9c_{2})$.

We want to come up with a third vector $(v_{1}, v_{2}, v_{3})$ that can't be expressed in the above form. Whatever $c_{1}$ and $c_{2}$ you pick for the linear combination above, we need that the third component $v_{3}$ is exactly $8c_{1} + 9c_{2}$. Let's pick a vector whose third component is different from this (i.e., pick $c_{1}$ and $c_{2}$, and fill in the first two components of $(c_{1}, 2c_{1} + c_{2}, 8c_{1} + 9c_{2})$, but make the third component different from this).

So, even though you can pick $c_{1}$ and $c_{2}$ to be anything, I will pick $c_{1} = c_{2} = 1$. Then the vector I will construct will be:

$(1c_{1}, 2c_{1} + 1c_{2}, 11) = (1, 2 + 1, 11) = (1, 3, 11)$

Notice that I made the third component different from $8c_{1} + 9c_{2} = 8 + 9 = 17$. Then this new vector can't be written as a linear combination of the previous two vectors, because that's how we constructed it.

Answer 2

Here's a simple method: you want to find a vector $(a,b,c)$ such that the system $$ x(1,2,8)+y(0,1,9)=(a,b,c) $$ has no solution. It's a linear system because it can be written as $$ \begin{cases} x=a\\ 2x+y=b\\ 8x+9y=c \end{cases} $$ The matrix of this system is $$ \left[\!\begin{array}{cc|c} 1 & 0 & a \\ 2 & 1 & b \\ 8 & 9 & c \end{array}\!\right] $$ If we proceed with Gaussian elimination we get \begin{align} \left[\!\begin{array}{cc|c} 1 & 0 & a \\ 2 & 1 & b \\ 8 & 9 & c \end{array}\!\right] &\to \left[\!\begin{array}{cc|c} 1 & 0 & a \\ 0 & 1 & b-a \\ 0 & 9 & c-8a \end{array}\!\right] \\&\to \left[\!\begin{array}{cc|c} 1 & 0 & a \\ 0 & 1 & b-a \\ 0 & 0 & c-8a-9b+9a \end{array}\!\right] \end{align} so we just need to have $$ a-9b+c\ne0 $$ and we can choose whatever values of the parameters, so for example $a=1$, $b=0$ and $c=0$.

Of course infinitely many other choices are possible.

A different method is finding a non zero vector which is orthogonal to the two given vectors; if the vector is $(a,b,c)$ we get $$ \begin{cases} a+2b+8c=0\\ b+9c=0 \end{cases} $$ This gives $a=-2b-8c$ and $b=-9c$. So we can set $c=1$ and get $b=-9$, $a=10$.

With the first method we find all vectors that solve our problem, but your task is just finding one, so take your pick.

Answer 3

Sorry to add to an old question, but it came up for me and I think this method may be useful for others. The method is equivalent to @egreg's answer, but maybe a bit easier to implement programmatically when you need the last vector to make a complete basis - a square matrix of linearly independent rows and columns - as in this question.

A matrix with linearly independent rows has a non-zero determinant, so if you want to find the vector $(a,b,c)$ to make a complete basis, we need

$$ \begin{vmatrix} a & b & c \\ 1 & 2 & 8 \\ 0 & 1 & 9 \\ \end{vmatrix} \neq 0 $$

The determinant can be computed with the elements in the top row and the minors (the Laplace expansion), so we get

$$ a \begin{vmatrix} 2 & 8 \\ 1 & 9 \\ \end{vmatrix} - b \begin{vmatrix} 1 & 8 \\ 0 & 9 \\ \end{vmatrix} + c \begin{vmatrix} 1 & 2 \\ 0 & 1 \\ \end{vmatrix} \neq 0 $$

which again is

$$ 10a - 9b + c \neq 0 $$

Any $(a,b,c)$ that satisfies that will be linearly independent. A really simple approach would be just to pick one of the elements with non-zero coefficients and set it to $1$, and set the other elements to zero. In this case none of the coefficients are zero, so $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$ are all linearly independent with the first two vectors you gave.

Answer 4

You can calculate the orthogonal complement for any n dimensional space. In MATLAB:

A = [1 2 8 ; 0 1 9];
p = null(sym(A))
10
-9
 1
dot(p,A(1,:))
0
dot(p,A(2,:))
0