Why doesn't the dot product give you the coefficients of the linear combination?

Solution 1:

Your $v_1$ and $v_2$ need to be orthonormal. To expand on learnmore's answer, essentially, the reason you need orthogonality for this to work is that if your $v_1$ and $v_2$ are not orthogonal, then they will have a non-zero dot-product $v_1\cdot v_2$. This means that $v_2$ carries some weight "in the direction" $v_1$. Your intuition that $c = a\cdot v_1$ is the "amount of $a$ in the direction $v_1$" is correct - keep that intuition! Similarly, $d=a\cdot v_2$ is the amount of $a$ in the direction of $v_2$.

However - since $v_1$ and $v_2$ are not perpendicular, the number $c$ has "piece" of $v_2$ in it, and the number $d$ has a "piece" of $v_1$ in it. So, when you try to expand $a$ in the basis $\{v_1,v_2\}$, you would need an extra term to compensate for the "non-orthogonal mixing" between $v_1$ and $v_2$.

The technical details are as follows. Since $v_1$ and $v_2$ are linearly independent, we can write

$$ a = \alpha v_1+\beta v_2 $$ for some scalars $\alpha, \beta$. Now, take the dot product of $a$ with $v_1$ and expand it out:

$$ a\cdot v_1 = (\alpha v_1+\beta v_2)\cdot v_1 = \alpha v_1\cdot v_1 + \beta v_1\cdot v_2 = \alpha + \beta v_1\cdot v_2 $$ similarly, expand out $a\cdot v_2$:

$$ a\cdot v_2 = \alpha v_1\cdot v_2 + \beta $$

Those extra terms ($\beta v_1\cdot v_2$ and $\alpha v_1\cdot v_2$) express the non-orthogonality. Written another way, we have

$$ \alpha = a\cdot v_1 - \beta v_1\cdot v_2 $$ and

$$ \beta = a\cdot v_2 - \alpha v_1\cdot v_2 $$ which shows clearly that the correct expansion coefficients have $a\cdot v_j$, but also another piece compensating for the non-orthogonality. I could go on - you can use matrices and such, but hopefully this is enough to convince you.

Solution 2:

Your intuition is mostly correct, and you would probably have seen the flaws in your reasoning if you had drawn a picture like this: enter image description here

We have two linearly-independent unit vectors $\mathbf{U}$ and $\mathbf{V}$, and a third vector $\mathbf{W}$ (the green one). We want to write $\mathbf{W}$ as a linear combination of $\mathbf{U}$ and $\mathbf{V}$. The picture shows the projections $(\mathbf{W} \cdot \mathbf{U})\mathbf{U}$ (in red) and $(\mathbf{W} \cdot \mathbf{V})\mathbf{V}$ (in blue). These are the things you call "shadows", and that's a good name. As you can see, when you add them together using the parallelogram rule, you get the black vector, which is obviously not equal to the original vector $\mathbf{W}$. In other words $$ \mathbf{W} \ne (\mathbf{W} \cdot \mathbf{U})\mathbf{U} + (\mathbf{W} \cdot \mathbf{V})\mathbf{V} $$ You certainly can write $\mathbf{W}$ in the form $\mathbf{W} = \alpha\mathbf{U} + \beta\mathbf{V}$, but $\alpha = \mathbf{W} \cdot \mathbf{U}$ and $\beta = \mathbf{W} \cdot \mathbf{V}$ are not the correct coefficients unless $\mathbf{U}$ and $\mathbf{V}$ are orthogonal. And you can even calculate the coefficients $\alpha$ and $\beta$ using dot products, as you expected. It turns out that $$ \mathbf{W} = (\mathbf{W} \cdot \bar{\mathbf{U}})\mathbf{U} + (\mathbf{W} \cdot \bar{\mathbf{V}})\mathbf{V} $$ where $(\bar{\mathbf{U}}, \bar{\mathbf{V}})$ is the so-called dual basis of $(\mathbf{U}, \mathbf{V})$. You can learn more here.