Question

a) Prove or disprove: The linear sum of two subspaces of a vector space is also a

subspace.

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Answer 1

Answer:

it is right question of your are sure

Answer 2

The linear sum of two subspaces of a vector space is also a subspace.

Proof:

Let V be any vector space over a field M. Also, let us assume  $S_{1}$ and $S_{2}$ are the subspaces of the vector space V.

We need to prove the following subspace criteria:

1. The zero vector 0 of V is in  $S_{1} and S_{2}$ .
2. For any u, v ∈ $S_{1} + S_{2}$, we have u + v ∈ $S_{1} + S_{2}$.
3. For any y ∈ $S_{1} + S_{2}$ and p ∈ M, we have py ∈ $S_{1} + S_{2}$.

∵  $S_{1} and S_{2}$ are subspaces of V, the zero vector 0 of V  is in both  $S_{1} and S_{2}$.

Thus we have,

                 $0 = 0 + 0$  ∈ $S_{1} + S_{2}$

So, condition 1 is met.

Next, let x, y ∈ $S_{1} + S_{2}$

∵  x ∈ $S_{1} + S_{2}$, we can write

                   $x = u + v$

For some u ∈ $S_{1}$ and v ∈ $S_{2}$

                  $y = u^{`} + v^{`}$

For some $u^{`}$ ∈ $S_{1}$ and $v^{`}$ ∈ $S_{2}$

Then we have,

                  $x + y = ( u+ v ) + ( u^{`} + v^{`} )$

                            $= ( u + u^{`} ) + ( v + v^{`} )$

∵ u and $u^{`}$ are both in the vector space $S_{1}$, their sum $u + u^{`}$ is also in $S_{1}$.

Similarly, we have $v + v^{`}$ ∈ $S_{2}$ since  $v , v^{`}$ ∈ $S_{2}$ .

Thus from the expression above, we see that

                           $x + y$ ∈ $S_{1} + S_{2}$

Hence condition 2 is met.

Finally, let y ∈ $S_{1} + S_{2}$  and p ∈ M.

Then there exists u ∈ $S_{1}$ and v ∈ $S_{2}$ such that,

                               $y = u + v$

Since $S_{1}$ is a subspace, it is closed under scalar multiplication.

Hence we have $pu$ ∈ $S_{1}$

Similarly, we have $pv$ ∈ $S_{2}$

It follows from this observation that

                            $py = p ( u + v )$

                                 $= pu + pv$  ∈ $S_{1} + S_{2}$

Thus condition 3 is met.

∴ by the subspace criteria  $S_{1} + S_{2}$  is a subspace of V.