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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:

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

  2. 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 $O_i^T U (U^T U)^{-1} U^T = O_i^T$?

  3. 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.

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1 Answer 1

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!

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yes, I know it is not true in general, but I have very specific conditions here. And yes, range is the image of a linear transformation. Thanks! –  Alg Oct 20 '11 at 19:13
    
But I am asking about $O_i$, not $O$. I know it is true for $O$. –  Alg Oct 20 '11 at 20:11
    
If each $im(O_i) = im(U)$, then the result holds for $O_i$ –  Shaun Ault Oct 20 '11 at 20:57
    
In fact, if $im(O_i) \subseteq im(U)$, then $coker(O_i) \supseteq coker(U)$, which is all that is really needed to prove $O_i^T\mathbf{v} = O^TUU^+\mathbf{v}$. –  Shaun Ault Oct 20 '11 at 21:02

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