Bases and Subspaces Worksheet Advanced Linear Algebra, Spring 2013 The vector space for this worksheet is P2; the set of polynomials of degree less that or equal to 2. We are given a basis for this...

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Bases and Subspaces Worksheet Advanced Linear Algebra, Spring 2013 The vector space for this worksheet is P2; the set of polynomials of degree less that or equal to 2. We are given a basis for this vector space S =  1 + t + t 2 ; 1 + t; 1 + t 2 and a subspace W = span  1 + 2t + 3t 2 ; 2 + t; 1 + t + t 2 : Additionally we are given v 2 P2; but in terms of its coordinates: [v]S = 2 4 1 0 1 3 5 : 1. Does v 2 W? 2. Find a basis for W 3. Extend the previous basis to a basis of P2; and call this basis S 0 : 4. Characterize w 2 W in terms of [w]S0 5. If we know [w]S ; how do we Önd [w]S0?




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Bases and Subspaces Worksheet Advanced Linear Algebra, Spring 2013 The vector space for this worksheet is P ; the set of polynomials of degree 2 less that or equal to 2. We are given a basis for this vector space  2 2 S = 1+t+t ;1+t;1+t and a subspace  2 2 W =span 1+2t+3t ;2+t;1+t+t : 2 3 1 4 5 Additionally we are given v2P ; but in terms of its coordinates: [v] = 0 : 2 S 1 1. Does v2W? 2. Find a basis for W 0 3. Extend the previous basis to a basis ofP ; and call this basis S : 2 4. Characterize w2W in terms of [w] 0 S 5. If we know [w] ; how do we ?nd [w] ? 0 S S 1






Bases and Subspaces Worksheet Advanced Linear Algebra, Spring 2013 The vector space for this worksheet is P2, the set of polynomials of degree less that or equal to 2. We are given a basis for this vector space S = { 1 + t+ t2, 1 + t, 1 + t2 } and a subspace W = span { 1 + 2t+ 3t2, 2 + t, 1 + t+ t2 } . Additionally we are given v ∈ P2, but in terms of its coordinates: [v]S = 10 1  . 1. Does v ∈W? 2. Find a basis for W 3. Extend the previous basis to a basis of P2, and call this basis S′. 4. Characterize w ∈W in terms of [w]S′ 5. If we know [w]S , how do we find [w]S′? 1
Answered Same DayDec 22, 2021

Answer To: Bases and Subspaces Worksheet Advanced Linear Algebra, Spring 2013 The vector space for this...

Robert answered on Dec 22 2021
123 Votes
1. Suppose v ∈ W then we can written as linear combination of spanning
element of W .
We have give
n [v]S =
 10
1
 This gives v = 1.(1 + t + t2) + 0.(1 + t) +
1(1 + t2)
v = 2 + t + 2t2
Now suppose v ∈W , there must exists a, b, c scalars, such that
2 + t + 2t2 = a(1 + 2t + 3t2) + b(2 + t) + c(1 + t + t2)
2 + t + 2t2 = (a + 2b + c) + (2a + b + c)t + (3a + c)t2
Hence we have
3a + c = 2 2a + b + c = 1 a + 2b + c = 2
We have c = 2− 3a, putting this in remaining equation we get
2a + b + 2− 3a = 1 a + 2b + 2− 3a = 2
that is we have
−a + b = −1 2b− 2a = 0
Which is absurd and do not have solution. Hence we can not find...
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