Consider a matrix $A = \begin{bmatrix} A_1 & A_2 \end{bmatrix}\in \mathbb{R}^{n\times(d_1+d_2)}$ with $rank(A) = d_1+d_2=d\leq n$. That is, the columns of $A$ are linearly independent. I want to represent $P_{R(A)}^\perp$, the projection onto the orthogonal complement of $R(A)$ (the range or image of $A$) as the following
$$
P_{R(A)}^\perp = P_{R(A_1)}^\perp P_{R(A_2)}^\perp.
$$
We have that $R(A)= R(A_1)+R(A_2)$ so $R(A)^\perp = R(A_1)^\perp\cap R(A_2)^\perp$. However, in order to have the representation to hold I have read that we need to establish that
$$
P_{R(A_1)}^\perp P_{R(A_2)}^\perp = P_{R(A_2)}^\perp P_{R(A_1)}^\perp,
$$
which I can't seem to prove. This is equivalent with showing that
$$
P_{R(A)} = P_{R(A_1)} + P_{R(A_2)}- P_{R(A_2)}P_{R(A_1)}.
$$
Do I need further assumptions or can the representation be shown to hold under the assumption of linearly independent columns of $A$?
Best Answer
Consider $$A = \left[\begin{array}{c|c}1 & 1\\0 & 1\\0&0\end{array}\right]$$
Then $R(A) = \{\begin{bmatrix}x&y&0\end{bmatrix}^T\mid x,y \in \Bbb R\}$ and $R(A)^\perp = \{\begin{bmatrix}0&0&z\end{bmatrix}^T\mid z \in \Bbb R\}$ so $$P_{R(A)}^\perp = \begin{bmatrix}0&0&0\\0&0&0\\0&0&1\end{bmatrix}$$ $R(A_1) = \{\begin{bmatrix}x&0&0\end{bmatrix}^T\mid x \in \Bbb R\}$ and $R(A_1)^\perp = \{\begin{bmatrix}0&y&z\end{bmatrix}^T\mid y,z \in \Bbb R\}$ so $$P_{R(A_1)}^\perp = \begin{bmatrix}0&0&0\\0&1&0\\0&0&1\end{bmatrix}$$ And $R(A_2) = \{\begin{bmatrix}t&t&0\end{bmatrix}^T\mid t \in \Bbb R\}$ and $R(A_2)^\perp = \{\begin{bmatrix}t&-t&z\end{bmatrix}^T\mid y,z \in \Bbb R\}$ so $$P_{R(A_2)}^\perp = \begin{bmatrix}\frac12&-\frac12&0\\-\frac12&\frac12&0\\0&0&1\end{bmatrix}$$ Obviously $$P_{R(A_2)}^\perp P_{R(A_1)}^\perp \ne P_{R(A)}^\perp \ne P_{R(A_1)}^\perp P_{R(A_2)}^\perp$$ In order for this to work, $R(A_1)$ must be orthogonal to $R(A_2)$.