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It appears to me to be self evident that in a three dimensional Euclidean space, the component of two or more vectors in the direction of the vector ${\bf w}$, is just the summed component of each of the vectors in the direction of ${\bf w}$: $$ (\bf u + \bf v + \bf \dots)_{w} = \bf u_{w} + \bf v_{w} + \dots $$ Is further work needed here to prove this?

category
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    Projection on a vector (here $\mathbb{w}$) is a linear operator, so mentioning that is enough. – Tanny Sieben Jun 10 '21 at 15:46
  • Hi category. You have not given any context here, so that makes it hard to answer. I guess you are interested in vector spaces such as the plane of high-school geometry, or real 3-space as seen in mechanics. You ask about proving. It depends what is given. Is an inner product space given? Or, are you trying to prove that real 3-space is an inner product space? – 311411 Jun 10 '21 at 16:11
  • https://math.stackexchange.com/a/185690 and also https://math.stackexchange.com/a/698483 – 311411 Jun 10 '21 at 16:13
  • Hi @311411, the vector space here is a three dimensional Euclidean space. I will update the question. – category Jun 10 '21 at 16:22
  • @311411 I'm not sure if I'm trying to prove the latter, that this space is an inner product space - I think this may be different, since I'm not mapping vectors to a scalar here. – category Jun 10 '21 at 16:31
  • I suppose that the projection matrix would be ${\mathbf{\hat{w}} \mathbf{\hat{w}}^{\mathsf{T}}}$ to get the projection onto ${\mathbf{w}}$. – category Jun 10 '21 at 16:44
  • Then ${\mathbf{\hat{w}} \mathbf{\hat{w}}^{\mathsf{T}}(\mathbf{u} + \mathbf{v}) = \mathbf{\hat{w}} \mathbf{\hat{w}}^{\mathsf{T}}\mathbf{u} + \mathbf{\hat{w}} \mathbf{\hat{w}}^{\mathsf{T}}\mathbf{v}}$ as ${\mathbf{\hat{w}} \mathbf{\hat{w}}^{\mathsf{T}}}$ is a Linear Operator – category Jun 10 '21 at 17:00

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Building on the comment that the 'Projection on a vector' is a linear operator, the component of the vector ${(\mathbf u + \mathbf v + \dots)}$ in the direction of ${\mathbf w}$ can be given by multiplying the said vector by the projection matrix ${\mathbf {\mathbf {\hat{w}}} {\mathbf {\hat{w}^{\mathsf T}}}}$ : $$ \mathbf {\mathbf {\hat{w}}} {\mathbf {\hat{w}^{\mathsf T}}} (\mathbf u + \mathbf v \dots) $$ where the unit vector ${\mathbf {\hat{w}} = \frac {\mathbf{w}} {\|\mathbf{w}\|}}$.

Since the projection matrix represents a Linear Transformation, this can be written as: $$ \mathbf {\mathbf {\hat{w}}} {\mathbf {\hat{w}^{\mathsf T}}} \mathbf u + \mathbf {\mathbf {\hat{w}}} {\mathbf {\hat{w}^{\mathsf T}}} \mathbf v + \dots $$

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    I guess $\mathbf {\hat w}$ is the unit vector in the direction of $\mathbf w.$ That's the usual convention, but if you're actually writing up a proof I would define the vector explicitly and not rely on convention. – David K Jun 12 '21 at 01:57
  • Updated to include the definition of the unit vector ${\mathbf {\hat{w}}}$. – category Jun 13 '21 at 07:45