Linear Map T: Proving Existence of KerT=U if dimU≥dimV-dimW

[kostas230]

Suppose that V and W are finite dimensional and that U is a subspace of V. If dimU≥dimV-dimW prove that there exists a linear map T from V to W such that kerT=U.

My answer is this:

Consider the following linear map:

T|u>=|u> if |u> belongs to V-U and T|u>=|0> if |u> belongs to U​

Therefore kerT=U
Is this correct?

 

[tiny-tim]

hi kostas230!

Consider the following linear map:

T|u>=|u> if |u> belongs to V-U and T|u>=|0> if |u> belongs to U​

Therefore kerT=U
Is this correct?

nooo …

i] that's a map from V to V, not to W
ii] what do you mean by V-U ? (eg what is R³ - R² ?)

 

[kostas230]

i] that's a map from V to V, not to W

Silly me, I overlooked it xD

ii] what do you mean by V-U ? (eg what is R³ - R² ?)

I mean the elements of V that do not belong in U.

 

[tiny-tim]

I mean the elements of V that do not belong in U.

then R³ - R² would include all elements with z ≠ 0

 

[kostas230]

Well, the correct answer would be: ℝ^3-ℝ^2={(0,0,z):z is real}

Complement (set theory) - Wikipedia, the free encyclopedia

 

[tiny-tim]

Well, the correct answer would be: ℝ^3-ℝ^2={(0,0,z):z is real

that's a line

add a line to R² and you don't get R³

 

[HallsofIvy]

R² direct sum a line gives R³ but that is not at all what you said!R³- R² is all (x, y, z) such that $z\ne 0$ which is not a subspace.
I think you need to review your definitions!