# The Unapologetic Mathematician

## Jordan Normal Form

Okay, let’s put everything together now. Start with a linear endomorphism $T:V\rightarrow V$ on a vector space $V$ of finite dimension $d$ over an algebraically closed field $\mathbb{F}$. If you want to be specific, use the complex numbers $\mathbb{C}$.

Now we can calculate the characteristic polynomial of $T$, whose roots are the eigenvalues of $T$. For each eigenvalue $\lambda$, we can define the generalized eigenspace as the kernel $U_\lambda=\mathrm{Ker}\left((T-\lambda1_V)^d\right)$, since if some power of $T-\lambda1_V$ kills a vector then the $d$th power will.

These generalized eigenspaces do not overlap, and each one is invariant under $T$. The dimension of the generalized eigenspace $U_\lambda$ associated to $\lambda$ is the multiplicity of $\lambda$, which is the number of times $\lambda$ shows up on the diagonal of an upper-triangular matrix for $T$. Since the total number of diagonal entries is $d$, we see that the dimensions of all the generalized eigenspaces add up to the dimension of the entire space $V$. Thus, we have a decomposition

$\displaystyle V=\bigoplus\limits_{\lambda}U_\lambda$

of $V$ as the direct sum of these generalized eigenspaces, where $\lambda$ runs over the roots of the characteristic polynomial.

If we restrict $T$ to the generalized eigenspace $U_\lambda$ with eigenvalue $\lambda$, the transformation $T\vert_{U_\lambda}-\lambda1_{U_\lambda}$ is nilpotent. Thus we can find a Jordan basis for $U_\lambda$, which puts $T\vert_{U_\lambda}-\lambda1_{U_\lambda}$ into the block-diagonal form

$\displaystyle\begin{pmatrix}A_1&&{0}\\&\ddots&\\{0}&&A_n\end{pmatrix}$

where each block has the form

$\displaystyle\begin{pmatrix}{0}&1&&&{0}\\&{0}&1&&\\&&\ddots&\ddots&\\&&&{0}&1\\{0}&&&&{0}\end{pmatrix}$

We can now add back in the eigenvalue $\lambda$ times the identity transformation to the restriction of $T$. Now we still have the block-diagonal form, but the blocks themselves now have the form

$\displaystyle\begin{pmatrix}\lambda&1&&&{0}\\&\lambda&1&&\\&&\ddots&\ddots&\\&&&\lambda&1\\{0}&&&&\lambda\end{pmatrix}$

where, of course, a block could be a single $1\times1$ matrix whose only entry is $\lambda$.

Putting these together for all the different eigenvalues, we have a Jordan basis for $V$. This puts the matrix $T$ into “Jordan normal form”. That is, the matrix of $T$ with respect to a Jordan basis is block-diagonal, with each block consisting of one eigenvalue $\lambda$ down its diagonal, and ${1}$s just above the diagonal.

Unfortunately, if the base field $\mathbb{F}$ is not algebraically closed, we may not have any upper-triangular matrix for $T$, and so we can only put the portion of $T$ captured by generalized eigenspaces into Jordan normal form. There may still be another direct summand which contains no generalized eigenvectors at all. Over an arbitrary field, this sort of thing gets complicated quickly, but it will be useful for us to consider what happens over the real numbers $\mathbb{R}$. We’ll come back to this.

March 4, 2009 - Posted by | Algebra, Linear Algebra

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4. Hello!
Very Interesting post! Thank you for such interesting resource!
PS: Sorry for my bad english, I’v just started to learn this language ;)
See you!
Your, Raiul Baztepo

Comment by RaiulBaztepo | March 29, 2009 | Reply

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