How to prove formula for contraction of a vector with a Multivector? I am currently reading "Space-Time Algebra" by David Hestenes and the following proof is given for the formula of contraction of a vector $a$ and multivector $b_1 \wedge b_2 \wedge b_3 \wedge \ldots \wedge b_k$. To prove is
$$
a \cdot (b_1 \wedge b_2 \wedge b_3\wedge \ldots \wedge b_k) = \sum_r (-1)^{k+1} (a\cdot b_r)(b_1 \wedge \ldots \wedge  b_{r-1} \wedge b_{r+1} \wedge  \ldots \wedge b_k).
$$
We will prove that
$$
a\cdot(b_1b_2b_3 \ldots b_k) = \sum_r (-1)^{k+1} (a\cdot b_r) (b_1\ldots b_{r-1}b_{r+1} \ldots b_k).
$$
Expand both sides and equate multivectors of same grade:
$$
\frac{1}2 (ab_1b_2\ldots b_k + b_1ab_2\ldots b_k) = (a\cdot b_1)(b_2b_3\ldots b_k) \\
-\frac{1}2 (b_1ab_2\ldots b_k + b_1b_2a\ldots b_k) = (a\cdot b_2)(b_2b_3\ldots b_k) \\
\vdots \\
(-1)^{k+1} \frac{1}2 (b_1\ldots b_{k-1}ab_k + b_1\ldots b_{k-1}b_ka) = (-1)^{k+1} (a\cdot b_k)(b_1\ldots b_{k-1}).
$$
Then, the textbook states "The largest multivector [in the above series of equalities] gives the formula [to prove]". A few things I don't understand -

*

*The right hand side of the equation to prove has a sum over $(k-1)$-grade vectors, so the left hand side must also be a grade $(k-1)$-vector. How are we expanding the contraction $a\cdot (b_1b_2\ldots b_k)$? It should be the grade $(k-1)$ part of the geometric product of the vector and multivector $\langle ab_1b_2\ldots b_k\rangle_{k-1}$. How do we write that explicitly?

*What does the book mean by "largest multivector"? I thought both sides of the equations always have $(k-1)$-grade multivectors?

 A: The precise wording in the textbook is a bit different to your paraphrase - this may be where some of the confusion has arisen.
You have a typo on the right hand side in your second equation. It should be:
$$\frac{1}2 (ab_1b_2\ldots b_k + b_1ab_2\ldots b_k) = (a\cdot b_1)(b_2b_3\ldots b_k) \\
-\frac{1}2 (b_1ab_2\ldots b_k + b_1b_2a\ldots b_k) = -(a\cdot b_2)(b_1b_3\ldots b_k) \\
\vdots \\
(-1)^{k+1} \frac{1}2 (b_1\ldots b_{k-1}ab_k + b_1\ldots b_{k-1}b_ka) = (-1)^{k+1} (a\cdot b_k)(b_1\ldots b_{k-1})$$
In the list of equations above, the left hand sides are of the form $\frac{1}{2}(m_1+m_2)$, $-\frac{1}{2}(m_2+m_3)$, $\frac{1}{2}(m_3+m_4)$, $-\frac{1}{2}(m_4+m_5)$, ... $(-1)^{k+1}\frac{1}{2}(m_{k-1}+m_k)$. The second term in each equation is the same as the first term in the next equation. So if you add up all the left hand sides, all but the first and last terms cancel. Thus, the sum of the left hand sides is:
$$\frac{1}{2} (a(b_1b_2\ldots b_k) + (-1)^{k+1} (b_1b_2\ldots b_k)a)$$
The right hand side of each equation is found from the left hand side by picking out the two terms that get swapped over, using the distributive law to write as $b_1b_2\ldots b_{r-1}(ab_r+b_ra)b_{r+1}\ldots b_k$ and moving the (scalar) dot product of two vectors to the front. The sum of the right hand sides is thus the right hand side of:
$$\frac{1}{2} (a(b_1b_2\ldots b_k) + (-1)^{k+1} (b_1b_2\ldots b_k)a) = \sum_r (-1)^{k+1} (a\cdot b_r) (b_1\ldots b_{r-1}b_{r+1} \ldots b_k)$$
We have from equation 3.8 of Hestenes that the inner product of a vector and r-vector is:
$$a\cdot A_r=\frac{1}{2}(aA_r+(-1)^{r+1}A_ra)$$
That looks a lot like the left hand side of the sum of equations. Unfortunately, the product of the $b_1b_2\ldots b_k$ is not in general a $k$-vector, because the $b$ vectors are not necessarily orthogonal, but can be a sum of vectors of different grades. If the multivector $A$ has a mixture of odd and even grades, then $a\cdot A=a\cdot (A_o+A_e)=\frac{1}{2}(aA_o+A_oa)+\frac{1}{2}(aA_e-A_ea)$. Fortunately, being a product of vectors, the components of $b_1b_2\ldots b_k$ are all of the same parity (either all odd-grade or all even-grade), and so we can say we have proved:
$$a\cdot(b_1b_2b_3 \ldots b_k) = \sum_r (-1)^{k+1} (a\cdot b_r) (b_1\ldots b_{r-1}b_{r+1} \ldots b_k)$$
This expression can have components of several different grades, because the $b$ vectors are not orthogonal in general. The $(k-1)$-grade component (the highest grade) corresponds to the outer product of the $b$s on either side, and that's the identity we are trying to prove.
