component of a vector $\mathbf a$ onto vector $\mathbf b$ Just a bit unsure about the definition. When I look online and at other questions on this site it says that the component of $\mathbf a$ onto $\mathbf b$ is $\dfrac{\mathbf a\cdot\mathbf b}{|\mathbf b|}$ but when I look in my notes and at my lecture slides it says that the component of $\mathbf a$ onto $\mathbf b$ is $\dfrac{\mathbf a\cdot\mathbf b}{\mathbf b\cdot\mathbf b}$. Also is the projection of $\mathbf a$ onto $\mathbf b$:
$$\frac{\mathbf a\cdot\mathbf b}{\mathbf b\cdot\mathbf b}\mathbf b?$$

Let ${\bf a} \in V^{n+1}$ and define $A=\text{span}({\bf a})$ and $A^{\perp} = \{ {\bf w} \in V : \langle {\bf w},{\bf a}\rangle=0 \}$.
Notice that $V = A \oplus A^{\perp}$. For any ${\bf b} \in V$ we can uniquely decomposed ${\bf b}$ as a component of $A$ and a component of $A^{\perp}$. By asking for the component of ${\bf b}$ onto the vector ${\bf a}$ you are asking for the image of ${\bf b}$ under the canonical projection $\pi : A \oplus A^{\perp} \twoheadrightarrow A$.
Take $\{{\bf a}\}$ as a basis for $A$ and $\{{\bf w}_1,{\bf w}_2,\ldots,{\bf w}_n\}$ as a basis for $A^{\perp}$. We can write ${\bf b}$ as follows:
$${\bf b} = \lambda {\bf a} + \mu_1{\bf w}_1 + \mu_2{\bf w}_2 + \cdots + \mu_n{\bf w}_n$$
Notice that $\pi({\bf b})=\lambda{\bf a}$ and $\lambda$ is the component of ${\bf b}$ on ${\bf a}$. Taking the scalar product:
$$
\begin{eqnarray*}
\langle {\bf a},{\bf b}\rangle &=& \langle {\bf a},  \lambda {\bf a} + \mu_1{\bf w}_1 + \cdots + \mu_n{\bf w}_n\rangle \\ \\
&=& \langle {\bf a},\lambda {\bf a}\rangle + \langle {\bf a},\mu_1 {\bf w}_1\rangle + \cdots + \langle {\bf a},\mu_n {\bf w}_n\rangle \\ \\
&=& \lambda\langle {\bf a},{\bf a}\rangle + \mu_1\langle {\bf a},{\bf w}_1\rangle + \cdots + \mu_n\langle {\bf a},{\bf w}_n\rangle \\ \\
&=& \lambda\langle {\bf a},{\bf a}\rangle + 0 + \cdots + 0 \\ \\
&=& \lambda\langle {\bf a},{\bf a}\rangle \\ \\
\end{eqnarray*}$$
It follows that
$$\lambda = \frac{\langle {\bf a},{\bf b}\rangle}{\langle {\bf a},{\bf a}\rangle}$$
Moreover, we also have
$$\pi({\bf b}) = \frac{\langle {\bf a},{\bf b}\rangle}{\langle {\bf a},{\bf a}\rangle}{\bf a}$$
