# 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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