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linear algebra – Low rank approximation proof: norm with Euclidean vector Answer

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linear algebra – Low rank approximation proof: norm with Euclidean vector

I learn High-dimensional knowledge evaluation with low-dimensional fashions: rules, computation and purposes, the place one of many duties is to show this theorem:

SENTENCE $ 4.5 $ (Best approximation with low rank). Permit $ boldsymbol {Y} in mathbb {R} ^ {n_ {1} occasions n_ {2}} $, and deem the next optimization downside commence {equation *} commence {array} {ll} min & | varvec {X} – varvec {Y} | _ {F} textual content {matter to} & operatorname {rank} ( varvec {X}) leq r. aim {array} aim {equation *} Every optimum resolution $ hat { varvec {X}} $ to the above downside has the figure $ hat { boldsymbol {X}} = sum_ {i = 1} ^ {r} sigma_ {i} boldsymbol {u} _ {i} boldsymbol {v} _ {i} ^ {*} $ Where
$ boldsymbol {Y} = sum_ {i = 1} ^ { min left (n_ {1}, n_ {2} privilege)} sigma_ {i} boldsymbol {u} _ {i} boldsymbol {v} _ {i} ^ {*} $ is a (full) eccentric worth decomposition of $ Bold attribute {Y} $.

Part of the rehearse:

$ 4.3 $ (Best rank$ r $ Approximation). We show Theorem 4.5.

First deem the particular illustration through which $ boldsymbol {Y} = boldsymbol { Sigma} = operatorname {diag} left ( sigma_ {1}, ldots, sigma_ {n} privilege) $ with $ sigma_ {1}> sigma_ {2}> cdots> sigma_ {n} $. Any rank r matrix $ boldsymbol {X} $ can breathe expressed as $ varvec {X} = varvec {F} varvec {G} ^ {*} $ with $ boldsymbol {F} in mathbb {R} ^ {n_ {1} occasions r} $, $ varvec {F} ^ {*} varvec {F} = varvec {I} $ and $ boldsymbol {G} in mathbb {R} ^ {n_ {2} occasions r}. $

  1. Permit $ varvec {P} = varvec {I} – varvec {FF} ^ {*} $, and write $ nu_ {i} = left | varvec {P} varvec {e} _ {i} privilege | _ {2} ^ {2}. $ Deny that $ sum_ {i = 1} ^ {n} nu_ {i} = n_ {1} -r $ and $ nu_ {i} in[0,1] . $ cease it
    commence {Equation *} | P Sigma | _ {F} ^ {2} = sum_ {i = 1} ^ {n_ {1}} sigma ^ {2} nu_ {i} geq sum_ {i = r + 1} ^ {n_ {1}} sigma_ {i} ^ {2} aim {equation *}
    with equality if and provided that $ nu_ {1} = nu_ {2} = cdots = nu_ {r} = 0 $ and $ nu_ {r + 1} = cdots = nu_ {n} $. Conclude that theorem $ 4.5 $ applies in particular circumstances $ Y = Sigma $.

My downside

My downside is exhibiting that $ sum_ {i = 1} ^ {n} nu_ {i} = n_ {1} -r $ and $ nu_ {i} in[0,1] . $

My ideas:

$$ | Pe_i | ^ 2_2 = | (I – FF ^ T) e_i | _2 ^ 2 = | e_i – FF ^ Te_i | _2 ^ 2 = 1 – | FF ^ Te_i | _2 ^ 2 $$

I can grasp how that $ n_1 $ within the sum takes place, however I do not behold how I ought to get it $ r $. Any steering would breathe enormously appreciated.

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