Logarithm of A Matrix - The Logarithm of A Non-diagonalizable Matrix

The Logarithm of A Non-diagonalizable Matrix

The algorithm illustrated above does not work for non-diagonalizable matrices, such as

For such matrices one needs to find its Jordan decomposition and, rather than computing the logarithm of diagonal entries as above, one would calculate the logarithm of the Jordan blocks.

The latter is accomplished by noticing that one can write a Jordan block as

$B=\begin{pmatrix} \lambda & 1 & 0 & 0 & \cdots & 0 \\ 0 & \lambda & 1 & 0 & \cdots & 0 \\ 0 & 0 & \lambda & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & \lambda & 1 \\ 0 & 0 & 0 & 0 & 0 & \lambda \\\end{pmatrix} = \lambda \begin{pmatrix} 1 & \lambda^{-1} & 0 & 0 & \cdots & 0 \\ 0 & 1 & \lambda^{-1} & 0 & \cdots & 0 \\ 0 & 0 & 1 & \lambda^{-1} & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & 1 & \lambda^{-1} \\ 0 & 0 & 0 & 0 & 0 & 1 \\\end{pmatrix}=\lambda(I+K)$

where K is a matrix with zeros on and under the main diagonal. (The number λ is nonzero by the assumption that the matrix whose logarithm one attempts to take is invertible.)

Then, by the Mercator series

one gets

This series in general does not converge for every matrix K, as it would not for any real number with absolute value greater than unity, however, this particular K is a nilpotent matrix, so the series actually has a finite number of terms (Km is zero if m is the dimension of K).

Using this approach one finds

$\ln \begin{bmatrix}1 & 1\\ 0 & 1\end{bmatrix} =\begin{bmatrix}0 & 1\\ 0 & 0\end{bmatrix}.$

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