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I propose the following lemma and its proof. It is related to row-null and column-null matrices - i.e. matrices whose rows and columns both sum to zero. Could you please give your opinion on the plausibility of the lemma, and the validity of the proof?

Lemma: Let $Z\in\text{GL}(n,\mathbb{R})$ be a general $n\times n$ real matrix, and let $Y\in\mathcal{S}(n,\mathbb{R})$, where $\mathcal{S}(n,\mathbb{R})$ is the space of row-null column-null $n\times n$ real matrices. Then $\text{Tr}(ZY)=0$ for all $Y$ in $\mathcal{S}(n,\mathbb{R})$ if and only if $Z$ has the form $$Z_{ij}=\left(p_{j}-p_{i}\right)+\left(q_{j}+q_{i}\right)$$.

Proof: Consider the space of row-null and column-null matrices

$$\mathcal{S}(n,\mathbb{R})= \left\{ Y_{ij}\in GL(n,\mathbb{R}):\sum_{i}Y_{ij}=0,\sum_{j}Y_{ij}=0 \right\} $$

Its dimension is $$\text{dim}(S(n,\mathbb{R}))=N^{2}-2N+1$$ since the row-nullness and column-nullness are defined by $2N$ equations, only $2N-1$ of which are linearly independent. Consider the following space

$$\mathcal{G}(n,\mathbb{R})=\left\{ Z_{ij}\in GL(n,\mathbb{R}):Z_{ij}=\left(p_{j}-p_{i}\right)+\left(q_{j}+q_{i}\right)\right\}$$

Its dimension is

$$\text{dim}(\mathcal{G}(n,\mathbb{R}))=2N-1$$ where $N-1$ is the contribution from the antisymmetric part and $N$ is from the symmetric part.

Assume $Y\in\mathcal{S}$ and $Z\in\mathcal{G}$, then the Frobenius inner product of two such elements is $$ \text{Tr}(ZY) =\sum_{ij}\left[\left(p_{j}-p_{i}\right)Y_{ji}+\left(q_{j}+q_{i}\right)Y_{ji}\right] $$ $$ =\sum_{j}(q_{j}+p_{j})\sum_{i}Y_{ji}+\sum_{i}(q_{i}-p_{i})\sum_{j}Y_{ji}=0 $$ Since $\text{dim}(\mathcal{G})+\text{dim}(\mathcal{S})=\text{dim}(GL)$ and $\mathcal{G}\perp\mathcal{S}$, then $\mathcal{G}$ and $\mathcal{S}$ must be complementary in $GL$. Therfore, if $Y$ is orthogonal to all the matrices in $\mathcal{S}$, it must lie in $\mathcal{G}$.

PS: How can I get the curly brackets {} to render in latex mode?

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@Celil: Please use \\{ \\} along with the $ sign for curly bracket to work. –  anonymous Sep 3 '10 at 5:41
    
Chandru: Only a single backslash is needed, e.g. my question for the OP: "Where did the $\{p_i, q_i\}$ come from?" –  J. M. Sep 3 '10 at 5:48
    
-1: This is essentially asking for someone to proofread your work. –  Douglas S. Stones Sep 3 '10 at 9:03
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You seemed to mean tr(ZY)=0 FOR ALL Y in S, but the "FOR ALL" qualifier is not in your statement. –  user1551 Sep 3 '10 at 9:36
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+1: This is essentially asking for someone to proofread your work. (This is a.k.a. collaborative research.) –  Tom Stephens Sep 3 '10 at 18:30

2 Answers 2

up vote 4 down vote accepted

Here is an alternate way of proving your Lemma. I'm not sure if its any simpler than your proof -- but it's different, and hopefully interesting to some.

Let $S$ be the set of $n\times n$ matrices which are row-null and column-null. We can write this set as: $$ S = \left\{ Y\in \mathbb{R}^{n\times n} \,\mid\, Y1 = 0 \text{ and }1^TY=0\right\} $$ where $1$ is the $n\times 1$ vector of all-ones. The objective is the characterize the set $S^\perp$ of matrices orthogonal to every matrix in $S$, using the Frobenius inner product.

One approach is to vectorize. If $Y$ is any matrix in $S$, we can turn it into a vector by taking all of its columns and stacking them into one long vector, which is now in $\mathbb{R}^{n^2\times 1}$. Then $\mathop{\mathrm{vec}}(S)$ is also a subspace, satisfying: $$ \mathop{\mathrm{vec}}(S) = \left\{ y \in \mathbb{R}^{n^2\times 1} \,\mid\, (\mathbf{1}^T\otimes I)y = 0 \text{ and } (I \otimes \mathbf{1}^T)y = 0 \right\} $$ where $\otimes$ denotes the Kronecker product. In other words, $$ \mathop{\mathrm{vec}}(S) = \mathop{\mathrm{Null}}(A),\qquad\text{where: } A = \left[ \begin{array}{c} \mathbf{1}^T\otimes I \\ I \otimes \mathbf{1}^T \end{array}\right] $$ Note that vectorization turns the Frobenius inner product into the standard Euclidean inner product. Namely: $\mathop{\mathrm{Trace}}(A^T B) = \mathop{\mathrm{vec}}(A)^T \mathop{\mathrm{vec}}(B)$. Therefore, we can apply the range-nullspace duality and obtain: $$ \mathop{\mathrm{vec}}(S^\perp) = \mathop{\mathrm{vec}}(S)^\perp = \mathop{\mathrm{Null}}(A)^\perp = \mathop{\mathrm{Range}}(A^T) $$ So every vector in $vec(S^\perp)$ is of the form $(\mathbf{1}\otimes I)a + (I\otimes \mathbf{1})b$ for some vectors $a$ and $b$ in $R^{n\times 1}$. It follows that every matrix in $S^\perp$ is of the form $a1^T + 1b^T$. This parametrization is equivalent to the one you presented if you set $a_i = q_i-p_i$ and $b_j = q_j + p_j$.

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I've checked your proof. It is correct.

I want to say obvious thing: I know two methods for checking the proof is correct.
1. Check the answer, if you can get it from something else.
1'. If you can't get the answer independently, check some of it's properties.
2. Check the proof line by line.
2'. Divide your proof on parts and check each of them, using one of the above.

Applying second method is obvious. If you want, to apply first, you can try to prove the following.
1. Every symmetric matrix, that is orthogonal to S(n,R), is of the form $q_i+q_j$.
2. Every symmetric matrix, that is orthogonal to S(n,R), is of the form $q_i+q_j$.

This can be done by using equations of the form Tr(ZY)=0 with matrices $Y\in S(n,\mathbb{R})$ of the form $Y_{ij}=a_i b_j$, where a and b are vectors of the form vectors with one coordinate equal to 1, another --- equal to -1, all other coordinates equal to 0 (like $(0,\dots,0,1,0,\dots,0,-1,0,\dots,0)$).

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