Linear Algebra-Changing Bases-2, guaranteed and verified 100% Pass

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Changing Bases


Suppose 𝑉 is a finite dimensional vector space with two different ordered bases
𝐵1 = {𝑤1 , … , 𝑤𝑛 } and 𝐵2 = {𝑣1 , … , 𝑣𝑛 }. Let’s call 𝐵1 the old basis and 𝐵2 the new
basis. Given a vector 𝑣 ∈ 𝑉 which is expressed as
𝑣 = 𝑏1 𝑣1 + ⋯ + 𝑏𝑛 𝑣𝑛
in the new basis, how do we express 𝑣 in the old basis 𝑣 = 𝑎1 𝑤1 + ⋯ + 𝑎𝑛 𝑤𝑛 ?
That is, if we know 𝑏1 , … , 𝑏𝑛 , how do we find 𝑎1 , … , 𝑎𝑛 ?
Clearly, if we can express each new basis vector 𝑣𝑖 , 1 ≤ 𝑖 ≤ 𝑛, in terms of
𝑤1 , … , 𝑤𝑛 , ie
𝑣𝑖 = 𝑐𝑖1 𝑤1 + ⋯ + 𝑐𝑖𝑛 𝑤𝑛
we can express 𝑣 in terms of 𝑤1 , … , 𝑤𝑛 .


Notice that we can think of changing bases from 𝐵2 to 𝐵1 as a linear
transformation, 𝐼: 𝑉 → 𝑉, where 𝐼 is just the identity map, 𝐼(𝑣) = 𝑣, but we are
using the basis 𝐵2 = {𝑣1 , … , 𝑣𝑛 } for the domain space and 𝐵1 = {𝑤1 , … , 𝑤𝑛 } as
the basis for the range of 𝐼. It’s also worth noting that when we represent 𝐼 in
1 ⋯ 0
𝐵1
matrix form, [𝐼]𝐵2 , it does not look like [ ⋮ ⋱ ⋮ ] (it only looks this way if
0 ⋯ 1
𝐵1 = 𝐵2 ).


Ex. Let 𝑤1 =< 1,2 > , 𝑤2 =< 3,5 > be the old basis for ℝ2 and 𝑣1 =< 1, −1 >,
𝑣2 =< 1, −2 > be the new basis for ℝ2 . Express 𝑣1 and 𝑣2 in terms of 𝑤1 and 𝑤2 .


𝑣1 =< 1, −1 >= 𝑐11 𝑤1 + 𝑐12 𝑤2
= 𝑐11 < 1,2 > +𝑐12 < 3,5 >
< 1, −1 >=< 𝑐11 + 𝑐12 , 2𝑐11 + 5𝑐12 >.

, 2


So we must solve a linear system of equations:
1 = 𝑐11 + 3𝑐12
−1 = 2𝑐11 + 5𝑐12


⟹ 𝑐11 = −8, 𝑐12 = 3. So we have:


𝑣1 =< 1, −1 >= −8 < 1,2 > +3 < 3,5 >= −8𝑤1 + 3𝑤2 .



𝑣2 =< 1, −2 >= 𝑐21 𝑤1 + 𝑐22 𝑤2
= 𝑐21 < 1,2 > +𝑐22 < 3,5 >
< 1, −2 >=< 𝑐21 + 3𝑐22 , 2𝑐21 + 5𝑐22 >.



So we must solve a linear sytem of equations:
1 = 𝑐21 + 3𝑐22
−2 = 2𝑐21 + 5𝑐22


⟹ 𝑐21 = −11, 𝑐22 = 4. So we have:


𝑣2 =< 1, −2 >= −11 < 1,2 > +4 < 3,5 >= −11𝑤1 + 4𝑤2 .

Notice that with respect to the (new) basis 𝐵2 = {𝑣1 , 𝑣2 }
𝑣1 =< 1,0 >𝐵2 = 1𝑣1 + 0𝑣2
𝑣2 =< 0,1 >𝐵2 = 0𝑣1 + 1𝑣2 .