Prove Identity: $\forall x,y\in\mathbb{R^d}\setminus\{0\}\,\forall z\in\mathbb{R^d}\colon x^Tz=x^T \operatorname{Proj}_{\,\mathrm{span}\{x,y\}}(z)$

Question

Let $x$ and $y$ be two arbitrary non-zero vectors in $\mathbb{R}^d$

Show that for any $z\in\mathbb{R}^d$, it holds that

$$ z^Tx = \overline{z}^T x $$

where $\overline{z}$ is the projection of $z$ onto the plane spanned by $x$ and $y$.

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Here's my current incomplete version of the proof.

Let $A$ be a matrix with $x$ and $y$ as columns. Then the projection of $z$ onto the $span\{x,y\}$ is just the projection of $z$ onto the column space of $A$. We know that this can be obtained through the least squares solution:

$\operatorname{Proj}_{\operatorname{span}\{x,y\}}(z)=(A^TA)^{-1}A^Tz$

Hence, we have

$$ \operatorname{Proj}_{\operatorname{span}\{x,y\}}(z)^T x = ((A^TA)^{-1}A^Tz)^Tx = z^TA(A^TA)^{-1}x $$

Now, if $x$ is the first column in $A$, we may just rewrite $x=Ae_1$ where $e_1$ is the first unit vector. So

$$ \operatorname{Proj}_{\operatorname{span}\{x,y\}}(z)^T x = z^TA(A^TA)^{-1}Ae_1 = z^TA(A^TA)^{-1}Ae_1 $$ Now I don't know how to continue from here.

Answer 1

It suffices to prove that $(z-\bar{z})^T x = 0$. In this case, $e = z-\bar{z}$ is the error vector, which is perpendicular to $span<x,y>$. Hence, the dot product is zero.

Answer 2

Let $A=\operatorname{Span}(x,y)$, $\exists (z_1,z_2)\in A \times A^\bot \colon z = z_1+z_2$.

Here $z_1=Proj_A(z)$ by definition.

Now

$$x^Tz=x^T(z_1+z_2)=x^Tz_1+x^Tz_2=x^Tz_1$$ since $x\in A$ and $z_2 \in A^\bot$