Mixing
Polar Decomposition proof in Axler
$begingroup$
I'm having trouble understanding the proof for existence of isometry between null(T) (same as null(|T|) and (range T)⊥. They do have the same dimension, but how can you guarantee the existence of a linear map that maps orthonormal vectors
e1 to f1 (with no dependency on f2, f3, ...),
e2 to f2 (with no dependency on f1, f3, ...),
e3 to f3 (with no dependency on f1, f2, ...),
...
Shouldn't e1 be expressed as a linear combination of f1, f2, f3, ...
e1 = <e1.f1>f1 + <e1.f2>f2 + .....
excerpt from Linear Algebra Done Right
linear-algebra
asked Jan 12 at 20:20
SonnySonny
11
$endgroup$
add a comment |
$begingroup$
I'm having trouble understanding the proof for existence of isometry between null(T) (same as null(|T|) and (range T)⊥. They do have the same dimension, but how can you guarantee the existence of a linear map that maps orthonormal vectors
e1 to f1 (with no dependency on f2, f3, ...),
e2 to f2 (with no dependency on f1, f3, ...),
e3 to f3 (with no dependency on f1, f2, ...),
...
Shouldn't e1 be expressed as a linear combination of f1, f2, f3, ...
e1 = <e1.f1>f1 + <e1.f2>f2 + .....
excerpt from Linear Algebra Done Right
linear-algebra
asked Jan 12 at 20:20
SonnySonny
11
$endgroup$
add a comment |
$begingroup$
I'm having trouble understanding the proof for existence of isometry between null(T) (same as null(|T|) and (range T)⊥. They do have the same dimension, but how can you guarantee the existence of a linear map that maps orthonormal vectors
e1 to f1 (with no dependency on f2, f3, ...),
e2 to f2 (with no dependency on f1, f3, ...),
e3 to f3 (with no dependency on f1, f2, ...),
...
Shouldn't e1 be expressed as a linear combination of f1, f2, f3, ...
e1 = <e1.f1>f1 + <e1.f2>f2 + .....
excerpt from Linear Algebra Done Right
linear-algebra
asked Jan 12 at 20:20
SonnySonny
11
$endgroup$
I'm having trouble understanding the proof for existence of isometry between null(T) (same as null(|T|) and (range T)⊥. They do have the same dimension, but how can you guarantee the existence of a linear map that maps orthonormal vectors
e1 to f1 (with no dependency on f2, f3, ...),
e2 to f2 (with no dependency on f1, f3, ...),
e3 to f3 (with no dependency on f1, f2, ...),
...
Shouldn't e1 be expressed as a linear combination of f1, f2, f3, ...
e1 = <e1.f1>f1 + <e1.f2>f2 + .....
excerpt from Linear Algebra Done Right
linear-algebra
linear-algebra
asked Jan 12 at 20:20
SonnySonny
11
asked Jan 12 at 20:20
SonnySonny
11
asked Jan 12 at 20:20
SonnySonny
11
asked Jan 12 at 20:20
SonnySonny
11
asked Jan 12 at 20:20
SonnySonny
11
11
add a comment |
add a comment |
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