Let ψ be a linear operator on V, a vector space of dimension two. How do I show, given ψ is not a scalar multiple of the identity, that there is a v such that v , ψ ( v ) forms a basis, and further how do I write the transformation as a matrix with respect to that matrix? For the matrix, the first column would be 0,1, but I don't know what the second column would be.

Question

Let   ψ be a linear operator on   V, a vector space of dimension two. How do I show, given   ψ is not a scalar multiple of the identity, that there is a   v such that       v    ,    ψ    (    v    )   forms a basis, and further how do I write the transformation as a matrix with respect to that matrix? For the matrix, the first column would be 0,1, but I don&#039;t know what the second column would be.

Nola Rivera · Accepted Answer

The key observation here seems to be: let   V be a vector space over a field   F [not necessarily finite-dimensional!], and let   ψ  :  V  →  V be a linear operator on   V. Then exactly one of the following occurs:(i) There is   λ  ∈  F such that   ψ  (  v  )  =  λ  v for all   v  ∈  V(ii) There is   v  ∈  V such that   ψ  (  v  ) is not in the span of   v.It is clear that the conditions are mutually exclusive. Seeking a contradiction we suppose that neither (i) nor (ii) holds. Then there are       λ    1    ≠      λ    2   in   F and       v    1    ,      v    2    ∈  V with   ψ  (      v    1    )  =      λ    1        v    1   and   ψ  (      v    2    )  =      λ    2        v    2  . Observe that       v    1   and       v    2   must be linearly independent. Let   v  =      v    1    +      v    2  . Then   ψ  (  v  )  =  ψ  (      v    1    )  +  ψ  (      v    2    )  =      λ    1        v    1    +      λ    2        v    2  . Now suppose there is       λ    3    ∈  F such that      λ    1        v    1    +      λ    2        v    2    =  ψ  (  v  )  =      λ    3    v  =      λ    3        v    1    +      λ    3        v    2  Since       v    1   and       v    2   are linearly independent, this equation forces       λ    1    =      λ    3    =      λ    2  . This is a contradiction.In other words, the only endomorphisms which preserve every one-dimensional subspace are the scalar linear transformations. The OP&#039;s first question is a special case of this.As for the second question: yes, the first column of the matrix is   (  0  ,  1  ). In fact the second column of the matrix could be anything whatsoever: since   ψ  (  v  ) is linearly indpendent from   v for any   w  ∈  V (here I am assuming, as the OP is, that   V is two-dimensional) there is a unique linear transformation   v such that   v  ↦  ψ  (  v  ) and   ψ  (  v  )  ↦  w

Let ψ be a linear operator on V , a vector space of dimension two. How d

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