$ \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\abs}[2][]{\left\lvert#2\right\rvert_{\text{#1}}} \newcommand{\ket}[1]{\left\lvert#1 \right.\rangle} \newcommand{\bra}[1]{\langle\left. #1\right\rvert} \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\dd}{\text{d}} \newcommand{\dv}[2]{\frac{\dd #1}{\dd #2}} \newcommand{\pdv}[2]{\frac{\partial}{\partial #1}} $
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If $V$ is a vector space and $P_1, \dots, P_k\in\mathcal{L}(V)$ are projections satisfying
  1. $P_iP_j=0$ if $i\neq j$
  2. P_1+\dots+ P_k=I_V
then $V=V_1\oplus\dots\oplus V_k$ for $V_i=\text{Ran}P_i$.
Hence, direct sum decompositions correspond to families of projections satisfying the two properties.

Concepts

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Hypothesis

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Results

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Proof

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If $V$ is a vector space and $P_1, \dots, P_k\in\mathcal{L}(V)$ are projections satisfying
  1. $P_iP_j=0$ if $i\neq j$
  2. P_1+\dots+ P_k=I_V
then $V=V_1\oplus\dots\oplus V_k$ for $V_i=\text{Ran}P_i$.
Hence, direct sum decompositions correspond to families of projections satisfying the two properties.

Concepts

Coming soon

Hypothesis

Coming soon

Results

Coming soon

Proof

Coming soon
FullPage
result
concepts
hypothesis
implications
proof
FullPage
result
concepts
hypothesis
implications
proof
If $V$ is a vector space and $P_1, \dots, P_k\in\mathcal{L}(V)$ are projections satisfying
  1. $P_iP_j=0$ if $i\neq j$
  2. P_1+\dots+ P_k=I_V
then $V=V_1\oplus\dots\oplus V_k$ for $V_i=\text{Ran}P_i$.
Hence, direct sum decompositions correspond to families of projections satisfying the two properties.

Concepts

Coming soon

Hypothesis

Coming soon

Results

Coming soon

Proof

Coming soon
If $V$ is a vector space and $P_1, \dots, P_k\in\mathcal{L}(V)$ are projections satisfying
  1. $P_iP_j=0$ if $i\neq j$
  2. P_1+\dots+ P_k=I_V
then $V=V_1\oplus\dots\oplus V_k$ for $V_i=\text{Ran}P_i$.
Hence, direct sum decompositions correspond to families of projections satisfying the two properties.

Concepts

Coming soon

Hypothesis

Coming soon

Results

Coming soon

Proof

Coming soon
FullPage
result
concepts
hypothesis
implications
proof