Given an $m$ x $k$ orthogonal matrix $Q$ and an $m$ x $m$ matrix A, I know that the matrix $QQ^{T}A$ is the orthogonal projection of $A$ onto the column space of $Q$.
Recently, in a paper, I read that given the same orthogonal matrix $Q$, the equation $Q^{T}AQ$ defines the orthogonal projection of matrix $A$ onto the subspace that is spanned by the columns of matrix $Q$. Isn't that the column space of $Q$?
So, how the above two statements are equivalent? In the first case, the result will be an $m \times m$ while in the second it will be a $k \times k$ matrix.
Best Answer
$\newcommand{\Reals}{\mathbf{R}}$If $m \leq k$, then $QQ^{T} = I_{m}$, so $QQ^{T}A = A$. By contrast, $Q^{T}AQ$ "applies $A$ to the column space of $Q^{T}$" by mapping the column space of $Q^{T}$ to $\Reals^{m} \subseteq \Reals^{k}$ by $Q$, applying $A$ to $\Reals^{m}$, then embedding $\Reals^{m}$ into $\Reals^{k}$ as the column space of $Q^{T}$; if $m < k$, the operator $Q^{T}AQ$ is zero on the orthogonal complement of the column space of $Q^{T}$ in $\Reals^{k}$.
If $k < m$, then $QQ^{T} = I_{k}$ (the $m \times m$ matrix whose upper-left $k \times k$ block is the identity), so $QQ^{T}A$ consists of the first $k$ rows of $A$. By contrast, $Q^{T}AQ$ "embeds $\Reals^{k}$ into $\Reals^{m}$ as the column space of $Q$, applies $A$, then uses dot products with the rows of $Q$ to project back to $\Reals^{k}$".
In either case, $QQ^{T}A = Q(Q^{T}A)$ and $Q^{T}AQ = (Q^{T}A)Q$ are generally not the same. "Orthogonal projection of $A$ to the column space of $Q$" looks to me like loose use of language.