How to find the projection matrix onto a subspace

Projecting onto a subspace finds the point in that subspace closest to the vector we start with. We can build one matrix that performs the projection for every input. Let’s form it from a basis for a plane in three dimensions.

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Put the basis vectors into a matrix

Use the plane spanned by $u=(1,-2,0)$ and $v=(2,1,-2)$. Put these independent vectors into the columns of $A$.

$A=\left[\begin{array}{rr}1& 2\\ -2& 1\\ 0& -2\end{array}\right]$

The projection matrix is $P=A(A^{T}A{)}^{-1}{A}^T$. The independent columns ensure that $A^{T}A$ is invertible.

Find the small matrix inside the formula

The basis vectors are perpendicular, so their mixed dot products are zero. The diagonal entries are their squared lengths.

$A^{T}A=\left[\begin{array}{rr}5& 0\\ 0& 9\end{array}\right]$

$(A^{T}A{)}^{-1}=\left[\begin{array}{rr}\frac{1}{5}& 0\\ 0& \frac{1}{9}\end{array}\right]$

Multiply to get the projection matrix

Multiplying the three factors gives the following matrix after multiplying every entry by 45.

$45P=\left[\begin{array}{rrr}29& -8& -20\\ -8& 41& -10\\ -20& -10& 20\end{array}\right]$

Check an input and its perpendicular residual

For $x=(1,0,0)$, the projection is $Px=\frac{1}{45}(29,-8,-20)$. The remaining part is $x-Px=\frac{1}{45}(16,8,20)$. Its dot products with the two basis vectors are zero, so the residual is perpendicular to the whole plane.

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