under what conditions do we have the range of two matrices identical? Let $O_1,\ldots,O_k$ be a set of $n \times n$ matrices, where the dimension of the range of each $O_i$ is of size $m$. Let $U$ be an $n \times m$ matrix.
Let $\alpha_1,\ldots,\alpha_k$ be positive values that sum to 1.
Define $O = \sum_{i=1}^k \alpha_i O_i$. We know that the range of $U$ is identical to the range of $O$.
I have two questions:


*

*If the $O_i$ are independent matrices, does it mean that for each $i$, we have: $O_i^T U (U^T U)^{-1} U^T = O_i^T$?

*If this is not true, or even if it is true, what are some other conditions under which we have the equality for all $i$ that $O_i^T U (U^T U)^{-1} U^T = O_i^T$?

*I know this equality will be true for all $i$ if the range of $O_i$ is contained in the range of $O$. Under what conditions on $\alpha_i$ or some other condition will we get this?
($A^T$ is the transpose of $A$)
Thanks.
 A: EDITED:
By "range", do you mean the "image" of the matrix, as interpreted as a linear transformation?
The answer to your question 1 is sometimes.  For an $n \times m$ matrix $U$, the $m \times n$ matrix $U^+ = (U^TU)^{-1}U^T$ is the Penrose inverse, or pseudo-inverse of the matrix $U$.  Among other properties, $UU^+ = U(U^TU)^{-1}U^T$ is the matrix of the orthogonal projection from $\mathbb{R}^n$ to the column space of $U$.  So we cannot expect $UU^+ = I$, and by extension, $O_i^TUU^+ \neq O_i^T$ in general.

Now in your problem, you state that $im(O) = im(U)$.  This implies that $coker(O) = coker(U)$.  Now $coker(O) = ker(O^T)$.  A similar statement is true for $coker(U)$, however I'd like to use an alternate interpretation:  $coker(U) = col(U)^{\perp}$, the orthogonal complement of $col(U)$.  So we have: $col(U)^{\perp} = ker(O^T)$.  Consider any $\mathbf{v} \in \mathbb{R}^n$.  Decompose $\mathbf{v} = \mathbf{w} + \mathbf{y}$ where $\mathbf{w} \in col(U)$ and $\mathbf{y} \in col(U)^{\perp}$.  So $\mathbf{w} = UU^+\mathbf{v}$,
and $O^T\mathbf{y} = 0$.
$$ O^T\mathbf{v} = O^T(\mathbf{w} + \mathbf{y}) = O^T\mathbf{w} + \mathbf{0}
 = O^TUU^+\mathbf{v}.$$
Therefore, $O^T = O^TUU^+$.
(Note, I only used the fact that $im(O) = im(U)$ in this proof).
Hope this helps!
