Signed-off-by: szdytom <szdytom@qq.com>
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方而静 2025-08-11 20:35:02 +08:00
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Signed by: szTom
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@ -460,7 +460,7 @@
$ dim range S T <= min{dim range S, dim range T} $
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首先证明 $dim range S T <= dim range S$。设 $u in U$,则 $S T u = S (T u) in range S$,故 $range S T subset.eq range S$,即$dim range S T <= dim range S$
首先证明 $dim range S T <= dim range S$。设 $u in U$,则 $S T u = S (T u) in range S$,故 $range S T subset.eq range S$,即 $dim range S T <= dim range S$
#let SI = $restricted(S, I)$
#tab 现在证明 $dim range S T <= dim range T$。令 $I = range T$,则 $I$ $V$ 的子空间。根据线性映射引理原书3.4),存在 $SI in LinearMap(I, W)$,使得对于任意 $v in I$,有 $SI v = S v$。设 $u in U$,则 $T u in I$,因此 $S T u = SI (T u)$。故 $range S T = range SI T subset.eq range SI$,即 $dim range S T <= dim range SI$