The Unapologetic Mathematician

Mathematics for the interested outsider

Decomposing Real Linear Transformations

Finally, we come to the analogue of Jordan normal form over the real numbers.

Given a linear transformation T:V\rightarrow V on a real vector space V of dimension d, we can find its characteristic polynomial. We can factor a real polynomial into the product of linear terms (X-\lambda_i) and irreducible quadratic terms (X^2-\tau_jX+\delta_j) with \tau_j^2<4\delta. These give us a list of eigenvalues and eigenpairs for T.

For each distinct eigenvalue \lambda_i we get a subspace U_i=\mathrm{Ker}\left((T-\lambda_iI_V)^d\right)\subseteq V of generalized eigenvectors, with m distinct eigenvalues in total. Similarly, for each distinct eigenpair (\tau_j,\delta_j) we get a subspace V_j=\mathrm{Ker}\left((T^2-\tau_jT+\delta_jI_V)^d\right)\subseteq V of generalized eigenvectors, with n distinct eigenpairs in total.

We know that these subspaces are mutually disjoint. We also know that the dimension of U_i is equal to the multiplicity of \lambda_i, which is the number of factors of (X-\lambda_i) in the characteristic polynomial. Similarly, the dimension of V_j is twice the multiplicity of (\tau_j,\delta_j), which is the number of factors of (X^2-\tau_jX+\delta) in the characteristic polynomial. Since each linear factor contributes {1} to the degree of the polynomial, while each irreducible quadratic contributes {2}, we can see that the sum of the dimensions of the U_i and V_j is equal to the degree of the characteristic polynomial, which is the dimension of V itself.

That is, we have a decomposition of V as a direct sum of invariant subspaces

\displaystyle V=\bigoplus\limits_{i=1}^mU_i\oplus\bigoplus\limits_{j=1}^nV_j

Further, we know that the restrictions (T-\lambda_iI_V)|_{U_i} and (T^2-\tau_jT+\delta_jI_V)|_{V_j} are nilpotent transformations.

I’ll leave it to you to work out what this last property implies for the matrix on the generalized eigenspace of an eigenpair, in analogy with a Jordan block for an eigenvalue.

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April 9, 2009 - Posted by | Algebra, Linear Algebra


  1. The formula after “For each distinct eigenvalue” does not parse.

    Comment by jr | April 9, 2009 | Reply

  2. Thanks. Not sure when the d in lambda disappeared.

    Comment by John Armstrong | April 9, 2009 | Reply

  3. [...] Real Inner Products Now that we’ve got bilinear forms, let’s focus in on when the base field is . We’ll also add the requirement that our bilinear forms be symmetric. As we saw, a bilinear form corresponds to a linear transformation . Since is symmetric, the matrix of must itself be symmetric with respect to any basis. So let’s try to put it into a canonical form! [...]

    Pingback by Real Inner Products « The Unapologetic Mathematician | April 15, 2009 | Reply

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