- #1
elo
- 2
- 0
I have a linear transformation, T, from P3 (polynomials of degree ≤ 3) to R4 (4-dimensional real number space). I have a second linear transformation, U, from R4 back to P3.
In the first step of this four-step problem, I have shown that the composition TU from R4 to R4 is the identity linear transformation by sending a vector x through U, then sending the product of that through T, and ending up with x again.
What I don't understand is the next step: show that T is an isomorphism. The problem gives the hint that I should first show that the image of T spans R4: okay, fair enough. After some misguided attempts to prove linear independence of the entries of the column vector of T(f), I skipped ahead to the solutions, and was given this:
"Given a vector x in R4, we have T(U(x)) = x, by the previous part, and so x is a value of T, and im(T) = R4."
I do not understand this answer. I did T(U(x)) = x, so I get that. But what does it mean that "x is a value of T?" And how does that mean that the image of T spans R4?
In the first step of this four-step problem, I have shown that the composition TU from R4 to R4 is the identity linear transformation by sending a vector x through U, then sending the product of that through T, and ending up with x again.
What I don't understand is the next step: show that T is an isomorphism. The problem gives the hint that I should first show that the image of T spans R4: okay, fair enough. After some misguided attempts to prove linear independence of the entries of the column vector of T(f), I skipped ahead to the solutions, and was given this:
"Given a vector x in R4, we have T(U(x)) = x, by the previous part, and so x is a value of T, and im(T) = R4."
I do not understand this answer. I did T(U(x)) = x, so I get that. But what does it mean that "x is a value of T?" And how does that mean that the image of T spans R4?