Shift matrix

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In mathematics, a shift matrix is a binary matrix with ones only on the superdiagonal or subdiagonal, and zeroes elsewhere. A shift matrix U with ones on the superdiagonal is an upper shift matrix. The alternative subdiagonal matrix L is unsurprisingly known as a lower shift matrix. The (i,j):th component of U and L are

 U_{ij} = \delta_{i+1,j}, \quad L_{ij} = \delta_{i,j+1},

where \delta_{ij} is the Kronecker delta symbol.

For example, the 5×5 shift matrices are


U_5=\begin{pmatrix}
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 0
\end{pmatrix} \quad
L_5=\begin{pmatrix}
0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0
\end{pmatrix}.

Clearly, the transpose of a lower shift matrix is an upper shift matrix and vice versa.

Premultiplying a matrix A by a lower shift matrix results in the elements of A being shifted downward by one position, with zeroes appearing in the top row. Postmultiplication by a lower shift matrix results in a shift left. Similar operations involving an upper shift matrix result in the opposite shift.

Clearly all shift matrices are nilpotent; an n by n shift matrix S becomes the null matrix when raised to the power of its dimension n.

Properties[edit]

Let L and U be the n by n lower and upper shift matrices, respectively. The following properties hold for both U and L. Let us therefore only list the properties for U:

p_U(\lambda) = (-1)^n\lambda^n.


The following properties show how U and L are related:

  • LT = U; UT = L
 N(U) = \operatorname{span}\{ (1,0,\ldots, 0)^T \},
 N(L) = \operatorname{span}\{ (0,\ldots, 0, 1)^T \}.
  • For LU and UL we have
UL = I - \operatorname{diag}(0,\ldots, 0,1),
LU = I - \operatorname{diag}(1,0,\ldots, 0).
These matrices are both idempotent, symmetric, and have the same rank as U and L
  • Ln-aUn-a + LaUa = Un-aLn-a + UaLa = I (the identity matrix), for any integer a between 0 and n inclusive.

Examples[edit]


S=\begin{pmatrix}
0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0
\end{pmatrix}; \quad A=\begin{pmatrix}
1 & 1 & 1 & 1 & 1 \\
1 & 2 & 2 & 2 & 1 \\
1 & 2 & 3 & 2 & 1 \\
1 & 2 & 2 & 2 & 1 \\
1 & 1 & 1 & 1 & 1
\end{pmatrix}.


Then 
SA=\begin{pmatrix}
0 & 0 & 0 & 0 & 0 \\
1 & 1 & 1 & 1 & 1 \\
1 & 2 & 2 & 2 & 1 \\
1 & 2 & 3 & 2 & 1 \\
1 & 2 & 2 & 2 & 1
\end{pmatrix}; \quad AS=\begin{pmatrix}
1 & 1 & 1 & 1 & 0 \\
2 & 2 & 2 & 1 & 0 \\
2 & 3 & 2 & 1 & 0 \\
2 & 2 & 2 & 1 & 0 \\
1 & 1 & 1 & 1 & 0
\end{pmatrix}.


Clearly there are many possible permutations. For example, S^{T}AS is equal to the matrix A shifted up and left along the main diagonal.



S^{T}AS=\begin{pmatrix}
2 & 2 & 2 & 1 & 0 \\
2 & 3 & 2 & 1 & 0 \\
2 & 2 & 2 & 1 & 0 \\
1 & 1 & 1 & 1 & 0 \\
0 & 0 & 0 & 0 & 0
\end{pmatrix}.

See also[edit]

References[edit]