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Show that ๐ฑฬƒ = ๐€ ๐“ (๐€๐€ ๐“ ) โˆ’๐Ÿ ๐ฌ + ๐null๐šฬƒ Given that: 1) ๐ฑฬƒ = argmin ๐ฑ โ€–๐ฑโ€–1 s.t. ๐ฌ = ๐€๐ฑ ๐€: ๐‘š ร— ๐‘› , ๐‘› > ๐‘š dim(ker(๐€)) = ๐‘› โˆ’ ๐‘š ๐ฑ โˆˆ ๐‘… ๐‘› , ๐š๐ง๐ฎ๐ฅ๐ฅ โˆˆ ๐‘… ๐‘›โˆ’๐‘š 2) ๐šฬƒ = arg min ๐šnull โ€–๐ฑLS + ๐null๐šnullโ€–1 ๏‚ท The columns of ๐null spans ker๐€ and ๐null: ๐‘› ร— ๐‘› โˆ’ ๐‘š ๏‚ท ๐ฑLS is the least squares solution of ๐ฌ = ๐€๐ฑ ๐ฑLS = argmin ๐ฑ โ€–๐ฑโ€–2 2 s. t. ๐ฌ = ๐€๐ฑ โ†’ ๐ฑLS = ๐€ ๐“ (๐€๐€ ๐“ ) โˆ’๐Ÿ ๐ฌ

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Least Squares Optimization
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