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[Ciarlet, Problem $1.1-10$] Let $A_k$, $1 \leq k\leq m$, be matrices of order $n$ satisfaying $$\sum_{k=1}^mA_k\ =\ I.$$ Show that the following conditions are equivalent.

  1. $A_k = (A_k)^2$, $1 \leq k \leq m$,
  2. $A_kA_l=0$, for $k\neq l$, $1\leq k,l\leq m$
  3. $\sum\limits_{k=1}^m\mbox{rank}(A_k) = n$.

Remark: We already proved $(2)\to(1)$ and $(1)\to(3)$, but we couldn't prove $(3)\to(2)$ or $\{(3)\to(1)\;\; and\;\; (1)\to(2)\}$. Please help us, thanks so much.

Lord_Farin
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FASCH
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    http://math.stackexchange.com/questions/341357/let-a-and-b-be-n-times-n-real-matrices-such-that-ab-ba-0-and-ab-is-i/341370#341370 - See this for the gist of the proof that (2) implies (3). Not quite what you asked but maybe it will give details as to how to show the converse – muzzlator Mar 27 '13 at 19:08
  • It's difficult, because that gist it doesn't work in backwards :S – FASCH Mar 27 '13 at 19:45

1 Answers1

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Condition 3 implies that your vector space is a direct sum of the images of the operators, i.e. $$V = \bigoplus_{k=1}^m\mathrm{Im}(A_k)$$ Without loss of generality, let us focus on $A_1$. If $\mathbf{x}\in \ker(A_1)$ then $$\mathbf{x} = A_1\mathbf{x} + A_2\mathbf{x} + \cdots + A_m\mathbf{x} = A_2\mathbf{x} + \cdots + A_m\mathbf{x}$$ Therefore it follows that $\ker(A)\subseteq \mathrm{Im}(A_2)\oplus \cdots \oplus \mathrm{Im}(A_m)$. On the other hand, if $\mathbf{x} \in \mathrm{Im}(A_2)\oplus \cdots \oplus \mathrm{Im}(A_m)$ then $$\mathbf{x} = A_2\mathbf{x_2} + \cdots + A_m\mathbf{x_m}$$ for some $\mathbf{x_i}\in V$. But we also have $$\mathbf{x} = A_1\mathbf{x} + A_2\mathbf{x} + \cdots + A_m\mathbf{x}$$ By uniqueness of representation, we must have $A_1\mathbf{x} = \mathbf{0}$ so $\mathbf{x}\in\ker(A_1)$. This shows $$\ker(A_1) = \bigoplus_{k=2}^m\mathrm{Im}(A_k)$$ And in particular, we have $$V = \mathrm{Im}(A_1) \oplus \ker(A_1)$$ For $\mathbf{x}\in \mathrm{Im}(A_1)$ we also have $$\mathbf{x} = A_1\mathbf{x} + A_2\mathbf{x} + \cdots + A_m\mathbf{x}= A\mathbf{x}$$ since by uniqueness of representation, we get that $A_k\mathbf{x} = \mathbf{0}$ for $k>1$. The two properties above characterizes $A_1$ as a projection. That's 3 implies 1.

The above argument also shows 2, since if we have $$\ker(A_1) = \bigoplus_{k=2}^m\mathrm{Im}(A_k)$$ then this trivially implies that for $k\neq 1$ $$A_k\mathbf{x} \in \mathrm{Im}(A_k) \subseteq \ker(A_1)$$ so that we get $A_1A_k\mathbf{x} = \mathbf{0}$ for all $\mathbf{x}\in V$ and all $k\neq 1$.

Finally as a note, 1 to 2 is probably most easily done through 1 to 3 to 2. Condition 1 very easily implies condition 3 since for projections, the trace is equal to the rank. This means $$n = \mathrm{tr}(I) = \sum_{k=1}^m\mathrm{tr}(A_k) = \sum_{k=1}^m\mathrm{rank}(A_k)$$

Edit: I apologize for leaving out a key condition above. Hopefully this will make more sense. Let $A$ be a linear operator on an $n$-dimensional vector space $V$. The two conditions

  1. $\ker(A)\oplus\mathrm{Im}(A) = V$
  2. $A^2\mathbf{x} = A\mathbf{x}$ for $\mathbf{x}\in\mathrm{Im}(A)$

together characterize $A$ as an idempotent operator. I use the term "projection" synonymously with "idempotent linear transformation". This is the standard definition to my knowledge. For example, wikipedia uses it.

To see this, note that for any $\mathbf{v}\in V$ we can write $\mathbf{v}$ uniquely as $$\mathbf{v} = \mathbf{x} + \mathbf{y},\ \ \ \ \mathbf{x}\in\mathrm{Im}(A),\ \ \mathbf{y}\in\ker(A)$$ Then this of course means that $$A^2\mathbf{v} = A^2(\mathbf{x}+\mathbf{y}) = A^2\mathbf{x} = A\mathbf{x} = A(\mathbf{x}+\mathbf{y}) = A\mathbf{v}$$ so that $A$ is indeed idempotent on $V$.

EuYu
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  • How does (3) imply a direct sum? –  Mar 27 '13 at 22:36
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    @SteveD If a sum of subspaces equals an $n$ dimensional vector space and the dimensions of those subspaces total $n$ then the vector space is a direct sum of the subspaces. To see this, fix a basis for each subspace. Then the union of the basis are $n$ vectors which span an $n$-dimensional vector space, meaning the union is itself a basis. In particular, this shows that the subspaces themselves are independent in the first place. – EuYu Mar 27 '13 at 22:51
  • Where do you use the fact of $\sum A_k = I$? – FASCH Mar 27 '13 at 23:05
  • @FASCH The second line. "If $\mathbf{x}\in \ker(A_1)$ then ...". Actually, the fact that $V$ is a direct sum of the images relies directly on the fact that the operators sum to the identity. – EuYu Mar 27 '13 at 23:06
  • Could you please, explain to me, why "That's 3 implies 1."? Thanks – FASCH Mar 28 '13 at 00:02
  • @FASCH I had already mentioned that $\ker(A_1) \oplus \mathrm{Im}(A_1)$ characterizes $A_1$ as a projection. Are you unsure why that condition implies that $A_1$ is a projection? – EuYu Mar 28 '13 at 00:39
  • Yes, and also what is the projection definition that you used – FASCH Mar 28 '13 at 00:42
  • @FASCH I see your confusion. I apologize for leaving out a key condition previously. Hopefully it's more clear now. – EuYu Mar 28 '13 at 01:43