# The Unapologetic Mathematician

## 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

## 3 Comments »

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! [...]

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