SupposeU={(x,y,x+y,x−y,2x)∈F5,x,y∈F}&nbsp;. Find a subspace&nbsp;Wof&nbsp;F5&nbsp;such that&nbsp;F5=U⊕W.

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SupposeU={(x,y,x+y,x−y,2x)∈F5,x,y∈F}&amp;nbsp;. Find a subspace&amp;nbsp;Wof&amp;nbsp;F5&amp;nbsp;such that&amp;nbsp;F5=U⊕W.

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We are given a set U which is defined as:U={(x,y,x+y,x−y,2x)∈F5:x,y∈F}We need to find a subspace W of F5.To determine W, we need to check if U satisfies the three conditions for being a subspace:1. The zero vector 0→ is in U.2. U is closed under vector addition.3. U is closed under scalar multiplication.First, let&#039;s verify if 0→ is in U. We have:0→=(0,0,0,0,0)Substituting x=0 and y=0 in the definition of U, we get:(0,0,0,0,0)∈UTherefore, condition 1 is satisfied.Next, we need to check if U is closed under vector addition. Let u→=(x1,y1,x1+y1,x1−y1,2x1) and v→=(x2,y2,x2+y2,x2−y2,2x2) be two arbitrary vectors in U. Then, their sum u→+v→ is:u→+v→=(x1+x2,y1+y2,x1+x2+y1+y2,x1−x2+y1−y2,2x1+2x2)=(x1,y1,x1+y1,x1−y1,2x1)+(x2,y2,x2+y2,x2−y2,2x2)=u→+v→Since u→+v→ is also of the form (x,y,x+y,x−y,2x), it follows that u→+v→∈U. Therefore, U is closed under vector addition.Finally, we need to check if U is closed under scalar multiplication. Let u→=(x,y,x+y,x−y,2x) be an arbitrary vector in U, and let c be an arbitrary scalar in F. Then, their product cu→ is:cu→=(cx,cy,c(x+y),c(x−y),2cx)Since cx and cy are in F (as F is a field), it follows that cu→∈U. Therefore, U is closed under scalar multiplication.Since all three conditions are satisfied, U is a subspace of F5. Thus, we can take W=U as the subspace of F5 that we were looking for.

SupposeU={(x,y,x+y,x−y,2x)∈F5,x,y∈F

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