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True/False Question:

Let $V$ be the vector space of consisting polynomials of $\mathbb{R}\left [ t \right ]$ of deg$\leq 2$. The map $T:V\rightarrow V$ sending $f\left ( t \right )$ to $f\left ( t \right )+{f}'\left ( t \right )$ is invertible.

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R[t] deg≤2 means function  line or point function.So it easy to represent it in martix form.

Let V, W be vector spaces over a field F. Let T : V → W be a linear transformation. Suppose T is one-to-one and onto. Then there exists a linear transformation S : W → V that is the inverse of T. Proof. We first have to somehow define a linear transformation S : W → V that inverts T. Given any w ∈ W, since T is bijective, there exists a unique v ∈ V such that w = T(v). So, define S(w) := v. (∗) Since v uniquely depends on w, the map S : W → V defined in this way is well-defined. We now show that S is linear. Let w, w0 ∈ W. Since T is bijective, there exist unique v, v0 ∈ V such that T(v) = w and T(v 0 ) = w 0 . In particular, by the definition (∗), S(w) = v and S(w 0 ) = v 0 . Since T is linear, T(v + v 0 ) = w + w 0 . So, by the definition (∗), we have S(w + w 0 ) = v + v 0 = S(w) + S(w 0 ). Now, let α ∈ F. Since T(v) = w and T is linear, T(αv) = αT(v) = αw. By the definition (∗), S(αw) = αv. Since v = S(w), we therefore have S(αw) = αS(w), as desired. So, S is linear. 13 It remains to show that S inverts T. Applying T to both sides of (∗), note that T S(w) = T(v) = w, so T S = IW . Also, substituting w = T(v) into (∗), we get S(T(v)) = v, so that ST = IV , as desired. Combining Lemmas 6.4 and 6.6, we see that a linear transformation T : V → W is invertible if and only if T is one-to-one and onto. Invertible linear transformations are also known as isomorphisms.
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