Let &#x03C8;<!-- ψ --> be a linear operator on V , a vector space of dimension two. How d

boloman0z

boloman0z

Answered question

2022-06-21

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.

Answer & Explanation

Nola Rivera

Nola Rivera

Beginner2022-06-22Added 21 answers

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'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

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