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The question I am currently working on is, "...find $a$ and $b$ such that $\vec{v} = a \vec{u} + b \vec{w}$, where $ \vec{u} = \langle 1,2 \rangle$, $\vec{w} = \langle 1,-1 \rangle$, and $\vec{v} = \langle 2,1 \rangle$ Before I asked a question on this forum, I did a google search to see if it had been previously asked. Indeed, I found one on yahoo answers: However, it wasn't exactly helpful. For the best answer, I understand the solution up until the part, "comparing LHS and RHS." What did the person do, why can someone just split up the equation like the person did?

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Read more about vectors? – hjpotter92 Jan 27 '13 at 0:48
To him/her, $i$ and $j$ are the vectors from the basis. So you can do that. – Sigur Jan 27 '13 at 0:48
up vote 2 down vote accepted

$\alpha \vec{u} + \beta \vec{w}=\langle \alpha+\beta,2 \alpha-\beta\rangle=\langle2,1\rangle.$ and now $\alpha+\beta =2$ and $2\alpha-\beta=1$. Now it's simple.

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Putting in the values for $\vec{v}, \vec{u}$ and $\vec{w}$ in the equation, we get:

$$ \langle 2, 1 \rangle = a * \langle 1, 2 \rangle + b * \langle 1, -1 \rangle $$

Here, $a$ and $b$ are just scalar multipliers and nothing else. Using the property of vectors and their scalaar multiplication:

$$ f * \vec{v} = f * \left( v_1, v_2, \dots v_n \right) = \left( f*v_1, f*v_2, \dots f*v_n \right) $$

You'll arrive at two linear equations in $a$ and $b$ as follows:

  • $$ a + b = 2 $$
  • $$ 2a - b = 1 $$

Solving which, you get $a = 1$ and $b = 1$.

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Your question is basically one of solving a system $y=Ax$ for a fixed $A$ and $y$. That's exactly what the multiplication of an $n\times n$ matrix with a vector of length $n$ is, a linear combination of the vectors in the matrix.

We'll work explicitly in the 2 dimensional case, as that is the context of your problem. We're given two vectors $\vec v = <a,b>,$ and $\vec w = <c,d>$. We're trying to find $\alpha$ and $\beta$ such that $$\alpha \vec v + \beta \vec w = <e,f>$$

How do we do this? Well we use a wonderful piece of notation called a matrix. This can be represented by the following: $$\left( \begin{array}{ccc} a & c \\ b & d \end{array} \right) \left( \begin{array}{ccc} \alpha \\ \beta \end{array} \right)= \left( \begin{array}{ccc} e \\ f \end{array} \right)$$

or in your case: $$\left( \begin{array}{ccc} 1 & 1 \\ 2 & -1 \end{array} \right) \left( \begin{array}{ccc} \alpha \\ \beta \end{array} \right)= \left( \begin{array}{ccc} 2 \\ 1 \end{array} \right)$$

Then you can invert the matrix using Cramer's rule, which solves the system $y=Ax$ by finding a matrix $A^{-1}$ such that $A^{-1}y=x$. This only works when the determinant of the matrix is non-zero, because the formula for the inverse is: $$A^{-1} = \frac{1}{\det A}\operatorname{adj}A$$

Where $\det$ gives the determinant of $A$, and $\operatorname{adj}$ gives the adjugate matrix. Calculating these values allows us to find the inverse matrix: $$\left( \begin{array}{ccc} \frac{1}{3} & \frac{1}{3} \\ \frac{2}{3} & \frac{-1}{3} \end{array} \right)$$ And to find $\alpha, \beta$ you simply multiply this by the vector $\vec y$, and get $$\left( \begin{array}{ccc} \alpha \\ \beta \end{array} \right) = \left( \begin{array}{ccc} \frac{1}{3} & \frac{1}{3} \\ \frac{2}{3} & \frac{-1}{3} \end{array} \right)\left( \begin{array}{ccc} 2 \\ 1 \end{array} \right)\\ \left( \begin{array}{ccc} \alpha \\ \beta \end{array} \right)= \left( \begin{array}{ccc} 1 \\ 1 \end{array} \right)$$

Which gives your solution. There are easier solutions for such a simple case, but this generalizes far better.

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