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HomeMatlabWhat Is a Nilpotent Matrix? – Nick Higham

# What Is a Nilpotent Matrix? – Nick Higham

An $ntimes n$ matrix $A$ is nilpotent if $A^k =0$ for some optimistic integer $k$. A nonzero nilpotent matrix will need to have each optimistic and unfavorable entries to ensure that cancellation to happen within the matrix powers. The smallest $k$ for which $A^k =0$ known as the index of nilpotency. The index doesn’t exceed $n$, as we are going to see beneath.

Listed here are some examples of nilpotent matrices.

notag begin{aligned} A_1 &= begin{bmatrix}0 & 1 0 & 0 end{bmatrix}, quad A_1^2=0, A_2 &= begin{bmatrix}0 & 1 & 1 0 & 0 & 1 0 & 0 & 0 end{bmatrix}, quad A_2^3 = 0, A_3 &= begin{bmatrix}1 & -1 1 & -1 end{bmatrix}, quad A_3^2 = 0, A_4 &= begin{bmatrix}2 & 2 & 4 -4 & -4 & -8 1 & 1 & 2 end{bmatrix}, quad A_4^2 = 0. end{aligned}

Matrix $A_1$ is the $2times 2$ occasion of the higher bidiagonal $ptimes p$ matrix

$notag N = begin{bmatrix} 0 & 1 & & & 0 & ddots & & & ddots & 1 & & & 0 end{bmatrix}, qquad (1)$

for which

$notag N^2 = begin{bmatrix} 0 & 0 & 1 & & & 0 & ddots & ddots & & & & ddots & 1 & & & ddots & 0 & & & & 0 end{bmatrix}, quad dots, quad N^{p-1} = begin{bmatrix} 0 & 0 & dots & 0 & 1 & 0 & ddots & & 0 & & ddots & ddots & vdots & & & 0 & 0 & & & & 0 end{bmatrix}$

and $N^p = 0$. The superdiagonal of ones strikes as much as the fitting with every improve within the index of the facility till it disappears off the highest proper nook of the matrix.

Matrix $A_4$ has rank $1$ and was constructed utilizing a basic formulation: if $A = xy^T$ with $y^Tx = 0$ then $A^2 = xy^T xy^T = (y^Tx) xy^T = 0$. We merely took orthogonal vectors $x =[2, -4, 1]^T$ and $y = [1, 1, 2]^T$.

If $A$ is nilpotent then each eigenvalue is zero, since $Ax = lambda x$ with $xne 0$ implies $0 = A^nx = lambda^n x$ or $lambda = 0$. Consequently, the hint and determinant of a nilpotent matrix are each zero.

If $A$ is nilpotent and Hermitian or symmetric, or extra typically regular ($A^*A = AA^*$), then $A = 0$, since such a matrix has a spectral decomposition $A = Q mathrm{diag}(lambda_i)Q^*$ and the matrix $mathrm{diag}(lambda_i)$ is zero. It is just for nonnormal matrices that nilpotency is a nontrivial property, and one of the best ways to grasp it’s with the Jordan canonical kind (JCF). The JCF of a matrix with solely zero eigenvalues has the shape $A = XJX^{-1}$, the place $J = mathrm{diag}(J_{m_1},J_{m_2}, dots, J_{m_p})$, the place $J_{m_i}$ is of the shape (1) and therefore $J_{m_i}^{m_i} = 0$. It follows that the index of nilpotency is $k = max{,m_i : i=1colon p,} le n$.

What’s the rank of an $ntimes n$ nilpotent matrix $A$? The minimal doable rank is $0$, attained for the zero matrix. The utmost doable rank is $n-1$, attained when the JCF of $A$ has only one Jordan block of measurement $n$. Any rank between $0$ and $n-1$ is feasible: rank $j$ is attained when there’s a Jordan block of measurement $j+1$ and all different blocks are $1times 1$.

Lastly, whereas a nilpotent matrix is clearly not invertible, like each matrix it has a Moore–Penrose pseudoinverse. The pseudoinverse of a Jordan block with eigenvalue zero is simply the transpose of the block: $N^+ = N^T$ for $N$ in (1).

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