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Show that the cyclic shift operator is unitary and determine its diagonalization: $$A=\begin{bmatrix} 0&1 \\[0.3em] &0&1 \\[0.3em] & & \ddots \\ &&&.&1\\ 1&&&&0 \end{bmatrix}.$$

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    What have you tried so far? As a hint, notice that there is a permutation which takes it to the identity.2012-11-29
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    An easy way to prove that the matrix is an isometry is to recall the fact that a matrix is unitary iff the map $x \to Ax$ is an isometry i.e. prove that $$\Vert Ax \Vert = \Vert x \Vert$$2012-11-29
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    I'm not really sure how cyclic shift operators work to begin with, this is the first time I encounter them.2012-11-29
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    Ok, so letting $A^t$ be the transpose, then this matrix will have $1$s on the positions immediately below the diagonal, another $1$ on the top rightmost corner and zeros everywhere else, so then $A^tA=I$ so $A$ is unitary. How about the diagonalization part?2012-11-29
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    Can you compute the eigenvalues, or where are your problems?2012-11-29
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    Is there a fast way of computing the eigenvalues for this particular matrix, or should I go the brute force way?2012-11-29
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    Taking the last line to develop should yield the result easily.2012-11-29

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