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I'm learning Linear Algebra on my own time. Currently studying projections. I understand all of the derivation, but there is one piece of it that bothers me.

Say I have a matrix $A$ that represents a plane in $3D$, (matrix has 2 columns and 3 rows), $b$ is a vector I'm trying to project onto a plane, $p$ is its projection and $e$ (error) is the vector normal to the plane, $e = b - p$

Given that $e$ is normal to the plane then it is normal to every vector in the plane, meaning the inner product with any vector in the plane is 0, i.e. $A^Te = 0$, substituting $e = b - p$ we get $$ A^T(b - p)= 0 $$ $$ A^Tb - A^Tp = 0 $$ $$ A^Tb = A^Tp $$ $$ b = p $$

My derivation clearly shows that projection equals to the vector being projected, but geometrically it is not so. What am I missing?

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You claim that $(A^Tb = A^Tp) \Rightarrow (b=p)$. This isn't true.

E.g. let $A = \begin{pmatrix}1 & 0\\ 0 & 1\\ 0 & 0\end{pmatrix}$. This has two columns and three rows as required.

Let $b = \begin{pmatrix}1 \\ 1\\ 1\end{pmatrix}, c = \begin{pmatrix}1\\ 1\\ 2\end{pmatrix}$. Clearly $b \neq c$.

However, $A^Tb = \begin{pmatrix}1 & 0 & 0\\ 0 & 1 & 0\end{pmatrix}\begin{pmatrix}1 \\ 1\\ 1\end{pmatrix} = \begin{pmatrix}1 \\ 1\end{pmatrix} = \begin{pmatrix}1 & 0 & 0\\ 0 & 1 & 0\end{pmatrix}\begin{pmatrix}1 \\ 1\\ 2\end{pmatrix} = A^Tc$.

Intuitively this happens because $b, c$ are vectors in $\mathbb{R}^3$, but $A^Tb, A^Tc$ are vectors in $\mathbb{R}^2$, meaning you "lose information" in the projection.

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  • $\begingroup$ I understand your explanation. But how come the following is not true. $A^Tb - A^Tp = 0$, add $A^Tp$ to both sides. $A^Tb - A^Tp + A^Tp = 0 + A^Tp$ and this leads to $A^Tb=A^Tp$. This type of reasoning leaves "uneasy" feeling. What's wrong with this logic? $\endgroup$
    – flashburn
    Nov 5 '17 at 1:07
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    $\begingroup$ Nothing is wrong up to there. What you've shown is that two different vectors in $\mathbb{R}^3$ can project to the same vector in $\mathbb{R}^2$. This is completely okay! It's like saying that a cylinder and a sphere both cast the same shadow. $\endgroup$
    – Harambe
    Nov 5 '17 at 1:17

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