A book I am reading keeps referring to the canonical form of the matrices when considering e.g. $\mathrm O (n)$ and its Lie algebra.

Although this is done in another, quicker, way (by considering the exponent of a matrix $A \in \mathrm O (n)$) it claims that the Lie algebra of the group can also be found by considering the canonical forms of the elements in $\mathrm O (n)$.

Also, for the general linear group $\textrm{GL}(\mathbb R, n)$ it is first shown that any $n\times n$ matrix is in the tangent room of the unit element $T_e$. Then the author considers the canonical form of a general $n \times n$ matrix and from these considerations draws conclusions about the one-parameter group.

I fail to see why this is the case. In the case of $\mathrm O (n)$ it is shown that you can always find a $B \in \mathrm{SO}(n)$ so that $B^{-1} A B$ is in canonical form. In this case I can accept that this is equivalent to choosing a different basis for the tangent space. And, I guess (?) this means that the conclusions you can draw from looking at the canonical form should hold in general. However, in the case of the Lie algebra of $\textrm{GL}(\mathbb R, n)$ it is not clearly stated if there are any special conditions on the matrices that put the tangent vector in canonical form, hence I am not sure if going over to this form is equivalent to a change of basis.

I might have misunderstood or confused some concepts, so please comment if there is anything wrong with my question and I will try to fix and clarify it.

  • $\begingroup$ Maybe the canonical form is only used as a tool to compute the matrix exponential. For example you readily see that for a trace-free upper triangular matrix the exponential has determinant one, and via canonical forms this extends to all tracefree matrices. $\endgroup$ Nov 27, 2016 at 9:15
  • $\begingroup$ Ok, this might be it. I have been thinking about this for some time, but I need to thinks some more. Thank you very much for the input. $\endgroup$ Nov 27, 2016 at 21:42


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