# Thread: Unique complement of subspace

1. ## Unique complement of subspace

Hi,

problem:

Can it happen that a non-trivial subspace of a vector space V (i.e., a subspace different from 0 and V) has a unique complement?

attempt:

My answer to this would be no in general, but I do not know if there exists some exotic subspace for which this holds.

Here's my attempt at some sort of a proof:

Let K and H be subspaces of V.
Let $\displaystyle k_1,\cdots,k_m$ be a basis of K.
Let $\displaystyle h_1,\cdots,h_n$ be a basis of H.

K and H are complementary subspaces if and only if the complete set,
$\displaystyle k_1,\cdots,k_m,h_1,\cdots,h_n$ is a basis of V.

Say now that I have K and want to find a complementary subspace. Since every set of linearly independent vectors may be used as a basis, H is not unique.

Any suggestions are greatly appreciated.

Thanks

2. Originally Posted by Mollier
Hi,

problem:

Can it happen that a non-trivial subspace of a vector space V (i.e., a subspace different from 0 and V) has a unique complement?

attempt:

My answer to this would be no in general, but I do not know if there exists some exotic subspace for which this holds.

Here's my attempt at some sort of a proof:

Let K and H be subspaces of V.
Let $\displaystyle k_1,\cdots,k_m$ be a basis of K.
Let $\displaystyle h_1,\cdots,h_n$ be a basis of H.

K and H are complementary subspaces if and only if the complete set,
$\displaystyle k_1,\cdots,k_m,h_1,\cdots,h_n$ is a basis of V.

Say now that I have K and want to find a complementary subspace. Since every set of linearly independent vectors may be used as a basis, H is not unique.

I can't understand how you may think that you've proved anything...What you now must prove is that no matter what H is (non-trivial, of course), complementing its basis to a basis of the whole space cannot be made in one unique way, and this is true ALWAYS...and here you may want to distinguish different fields: finite and infinite ones.

Tonio

Any suggestions are greatly appreciated.

Thanks
.

3. Originally Posted by Mollier
Hi,

problem:

Can it happen that a non-trivial subspace of a vector space V (i.e., a subspace different from 0 and V) has a unique complement?

attempt:

My answer to this would be no in general, but I do not know if there exists some exotic subspace for which this holds
Here's my attempt at some sort of a proof:

Let K and H be subspaces of V.
Let $\displaystyle k_1,\cdots,k_m$ be a basis of K.
Let $\displaystyle h_1,\cdots,h_n$ be a basis of H.

K and H are complementary subspaces if and only if the complete set,
$\displaystyle k_1,\cdots,k_m,h_1,\cdots,h_n$ is a basis of V.

Say now that I have K and want to find a complementary subspace. Since every set of linearly independent vectors may be used as a basis, H is not unique.

Any suggestions are greatly appreciated.

Thanks
No, you've shown that the basis is not unique. The problem was to show whether the subspace spanned by those other vectors is unique.

To prove that a statement like "the complementary subspace is unique" if false you only need to prove a counter example. And it doesn't have to be very "exotic".

Let V= $\displaystyle R^2$ and let K be {(x, y)|x= 0}, the x-axis. Let U be {(x,y)| y= x}. We can take {(1, 0)} as a basis for K and {(1 1)} as a basis for U. It's clear that those vectors are indpendent and so {(1, 0), (1, 1)} is a basis for $\displaystyle R^2$. By your definition, then, U is a "complementary" subspace to K.

Or, take W {(x,y)|y= 2x}. Now {(1, 2)} is a basis for U and, again, {(1, 0), (1, 2)} is a basis for V. But U and W are clearly NOT the same subspace.

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### a unique complement

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