1. LetT :R3→R3 . Show that if
T ([c1 c2 c3]) = c1T ([1 0 0]) + c2T([0 1 0]) + c3T ([0 0 1]) for c1, c2, c3 ∈ R
then T satisfies
T(a⃗v1+b⃗v2)=aT(⃗v1)+bT(⃗v2), for a, b ∈R and⃗ v1, v2 ∈R .
To show that T satisfies T(a⃗v1+b⃗v2)=aT(⃗v1)+bT(⃗v2) for a, b ∈ R and ⃗v1, ⃗v2 ∈ R, we can start by rewriting the expression T(a⃗v1+b⃗v2) using the given property of T.
Let's expand T(a⃗v1+b⃗v2) using the linearity of T:
T(a⃗v1+b⃗v2) = T([a⃗v1 b⃗v2]) = aT([v1 0 0]) + bT([0 v2 0])
Here, we are using the fact that multiplying a vector by a scalar a is the same as multiplying each element of the vector by a. For example, [c1 c2 c3] can be written as c1[1 0 0] + c2[0 1 0] + c3[0 0 1].
Now, let's express T([v1 0 0]) and T([0 v2 0]) using the given property of T:
T([v1 0 0]) = v1T([1 0 0])
T([0 v2 0]) = v2T([0 1 0])
Substituting these expressions back into the previous equation, we have:
T(a⃗v1+b⃗v2) = a(v1T([1 0 0])) + b(v2T([0 1 0]))
Now we can use the distributive property of scalar multiplication:
T(a⃗v1+b⃗v2) = av1T([1 0 0]) + bv2T([0 1 0])
Finally, we can rearrange the terms to match the desired form:
T(a⃗v1+b⃗v2) = aT([v1 0 0]) + bT([0 v2 0])
This shows that T satisfies T(a⃗v1+b⃗v2) = aT(⃗v1) + bT(⃗v2) for a, b ∈ R and ⃗v1, ⃗v2 ∈ R.