Find the projection matrix that projects every vector of $\mathbb{R}^3$ on to the plane of 2 vectors

Question

The two vectors that are mentioned above are $$\begin{pmatrix}1 \\1 \\0\end{pmatrix}$$ $$\begin{pmatrix}0 \\ 1 \\1 \end{pmatrix}$$

I am not really sure what i am doing but i tried finding the projection matrix by putting the 2 vectors into a different matrix A =$$\begin{pmatrix}1 & 0 \\1 &1 \\0&1\end{pmatrix}$$ and then calclulating the projection matrix $$ P = A (A^TA)^{-1} A^T$$ .My result is that the P matrix is $$ 1/2\begin{pmatrix}1/2 & 1 & -1\\1&2&1\\-1 & 1 &2\end{pmatrix}$$ Is my answer correct? If not what am i doing wrong? Thanks in advance

Answer 1

Your answer is not correct, as can be seen by checking $P\begin{pmatrix}1\\1\\0\end{pmatrix}$.

Here's a suggestion how to find such a projection matrix easily. Use the standard basis for $\mathbb R^3$.

The columns of the matrix are the images of the basis vectors.

For example, the following matrix will map all vectors in $\mathbb R^3$

to the plane spanned by $\begin{pmatrix}1\\1\\0\end{pmatrix}$ and $\begin{pmatrix}0\\1\\1\end{pmatrix},$ because it maps the standard basis vectors there:

$\begin{pmatrix}1 & 0 &0 \\ 1 &1 &1 \\ 0 &1 &1 \end{pmatrix}.$ As long as the columns are vectors in the plane spanned by $\begin{pmatrix}1\\1\\0\end{pmatrix}$ and $\begin{pmatrix}0\\1\\1\end{pmatrix},$

the image will be in that plane.

Answer 2

You must have made a mistake, as can be seen by checking $P\begin{pmatrix}1\\1\\0\end{pmatrix}$.

Using your approach, $ A =\begin{pmatrix}1 & 0 \\1 &1 \\0&1\end{pmatrix},$ $A^T=\begin{pmatrix}1 &1 &0 \\0 &1 &1 \end{pmatrix},$ $ A^TA=\begin{pmatrix}2&1\\1&2\end{pmatrix}$, $(A^TA)^{-1}=\dfrac13\begin{pmatrix}2 & -1 \\-1 & 2\end{pmatrix},$ $A(A^TA)^{-1}=\dfrac13\begin{pmatrix}2 &-1 \\ 1& 1 \\ -1 &2\end{pmatrix}$, and $P=A(A^TA)^{-1}A^T=\dfrac13\begin{pmatrix}2 &1 &-1\\1 &2 &1 \\-1 &1& 2\end{pmatrix}$. Can you check that $P\begin{pmatrix}1\\1\\0\end{pmatrix}=\begin{pmatrix}1\\1\\0\end{pmatrix}?$

Answer 3

Your approach was correct, but as another answer explains, it looks like you made an error in your matrix multiplications or inversion. Here’s a related way to solve this problem that might afford fewer opportunities for error. It’s definitely less work.

The orthogonal projection of a vector onto a subspace is equal to the orthogonal rejection from its orthogonal complement. Here, the orthogonal complement is a line, so you just need to find a single nonzero vector $\mathbf v$ orthogonal to your plane and then compute $$P=I-{\mathbf v\mathbf v^T\over\mathbf v^T\mathbf v}.$$ In this problem a suitable $\mathbf v$ can be found by inspection or you can compute the cross product of the two vectors that define the plane.