Lean, informalized

Nielsen & Chuang · Chapter 2, Introduction to quantum mechanics · p. 69

Exercise 2.12 - No change of basis turns the matrix with ones on and below the diagonal into a diagonal one

The shear of a qubit has no diagonal form.

Hover over a blue or green term in the text, or a node in the graph, to find it in the other. Select it to read its declaration both in Lean and in words.

The Exercise

Exercise 2.12 asks for a proof that one operator on a qubit is not diagonalizable: the shear, which leaves |1⟩|1\rangle fixed and sends |0⟩|0\rangle to |0⟩+|1⟩|0\rangle + |1\rangle.

The Shear Matrix

The shear matrix MM is the complex 2×22 \times 2 matrix M=(1011).M = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}.(1)✦ In the computational basis {|0⟩,|1⟩}\{|0\rangle, |1\rangle\} of the qubit, the shear has the matrix MM.

The Claim

Theorem 1. The shear matrix MM is not diagonalizable: there is no invertible complex 2×22 \times 2 matrix PP and no pair of complex numbers d0,d1d_0, d_1 with M=P(d000d1)P−1.M = P \begin{pmatrix} d_0 & 0 \\ 0 & d_1 \end{pmatrix} P^{-1}.(2)✦

What Is Checked

The claim is stated over a single object, the shear matrix (1).