Proof of $(\vec{A} \cdot \vec{B})\vec{B}=\vec{B}(\vec B{}^T)(\vec{A})$?

Question

Given vectors $\vec{A},\vec{B}$ in $ℝ^3$, does anyone have a nice proof or explanation of why $(\vec{A} \cdot \vec{B})\vec{B}=\vec{B}(\vec B{}^T)(\vec{A})$?

Answer 1

I'm going to answer this question, my own question, thanks to the comment of WalterJ. First, recall that $\vec A \cdot \vec B=\vec A{}^T \vec B$. Using the commutative property of the inner product, $\vec A \cdot \vec B=\vec B \cdot \vec A=\vec B{}^T \vec A$ so that $(\vec A \cdot \vec B)\vec B=(\vec B \cdot \vec A)\vec B = \vec B{}^T \vec A\vec B$. Note that a scalar times a vector is equivalent to the same vector times the same scalar so that $(\vec B \cdot \vec A)\vec B =\vec B(\vec B \cdot \vec A)=\vec B(\vec B{}^T \vec A)$. By the associative property of matrix multiplication, $\vec B(\vec B{}^T \vec A)=\vec B\vec B{}^T \vec A$.

Answer 2

I have been studying Kronecker product. So, if you need a rather complicated proof, here I have one.

I am going to drop the vector sign, for simplicity. For the LHS, we can rewrite

$$(A.B)B=(A.B) \otimes B=B\otimes(A.B)=B\otimes(B^TA)$$

Also, we have the following for the RHS.

$$BB^TA=(B\otimes B^T)A$$

There is a property of Kronecker product that says

$$(A\otimes B)(C\otimes D)=(AC)\otimes (BD)$$

if $AC$ and $BD$ exist.

Using the property, we start from the LHS to get the RHS.

$$B\otimes(B^TA)=(BI)\otimes (B^TA)=(B\otimes B^T)(I\otimes A)=(B\otimes B^T) A$$

Note that $I$ is is the scalar $1$.

Answer 3

$$\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix} \cdot \begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix} =a_1b_1+a_2b_2+a_3b_3$$ $$(a_1b_1+a_2b_2+a_3b_3) \begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix} =\begin{pmatrix}(a_1b_1+a_2b_2+a_3b_3)b_1\\(a_1b_1+a_2b_2+a_3b_3)b_2\\(a_1b_1+a_2b_2+a_3b_3)b_3\end{pmatrix}$$

Going the other way:

$$\begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix} (b_1,b_2,b_3) =\begin{pmatrix}b_1^2+b_1b_2+b_1b_3\\ b_2b_1+b_2^2+b_2b_3\\ b_3b_1+b_3b_2+b_3^2\end{pmatrix}$$ $$\begin{pmatrix}b_1^2+b_1b_2+b_1b_3\\ b_2b_1+b_2^2+b_2b_3\\ b_3b_1+b_3b_2+b_3^2\end{pmatrix} \begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix} =\begin{pmatrix}(a_1b_1+a_2b_2+a_3b_3)b_1\\(a_1b_1+a_2b_2+a_3b_3)b_2\\(a_1b_1+a_2b_2+a_3b_3)b_3\end{pmatrix}$$