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How can i find the Jordan canonical form of this matrix $$A=\begin{pmatrix}1 &0 &0 &0 \\ 1& 2& 0& 0\\ 1 &0& 2& 0\\ 1 &1& 0& 2\end{pmatrix}.$$ In my book there are examples but all the matrices in these examples have only one eigenvalue repeated n times(for $n\times n$ matrices) but the matrix $A$ has eigenvalues $1$ and $2$(multiple of $3$). What is the way of finding $A$'s jordan canonical form? Thanks

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I'll answer to "What is the way of finding $A$'s jordan canonical form?".

You must first find how $A$ acts on the generalised eigenspaces. Since the characteristic polynomial of $A$ is clearly $(X-1)(X-2)^3$, the one for $\lambda=1$ is $1$-dimensional, and the one for $\lambda=2$ is $3$-dimensional. The former is just the eigenspace for $\lambda=1$, while the latter is the kernel of the linear map with matrix $(A-2I_4)^d$ for sufficiently large $d$. You can be sure that $d=3$ will suffice, but simple computation will tell you that in this case $(A-2I_4)^2$ already has rank $1$, and therefore gives a $3$-dimensional kernel which is the generalised eigenspace for $\lambda=2$. In fact it is just the space spanned by the last $3$ standard basis vectors, which subspace can be indeed seen to be stable under (the linear map with matrix) $A$, and the restriction of $A$ to it has, on the obvious basis, matrix $$ A'=\begin{pmatrix}2& 0& 0\\0& 2& 0\\1& 0& 2\end{pmatrix}. $$ You can see that $A'-2I_3\neq0$ and $(A'-2I_3)^2=0$ confirming the $d=2$ found above; this leaves for the Jordan canonical form of $A'$ as only possibility $$ \begin{pmatrix}2& 1& 0\\0& 2& 0\\0& 0& 2\end{pmatrix} $$ (indeed it suffices to permute the basis vectors, bringing the final one to the front, to obtain this form). The Jordan canonical form of $A$ then is $$ \begin{pmatrix}1 &0 &0 &0 \\ 0& 2& 1& 0\\ 0 &0& 2& 0\\ 0 &0& 0& 2\end{pmatrix}. $$ The basis on which this matrix is obtained is not difficult to give either: the first basis vector is an eignevector for $\lambda=1$, for instance $(-1,1,1,0)$, and for the remaing three, one can take the fourth, the second and the third standard basis vectors of $\Bbb C^4$ (assuming your field was $\Bbb C$).

This example is particularly simple. However, computing the ranks of the matrices $(A-\lambda I)^k$ for all eigenvalues $\lambda$ and sufficiently many $k$ (until the rank reaches the codimension of the generalised eigenspace) always gives you enough information to know the Jordan canonical form: when going from $k-1$ to $k$, this rank drops by the number of Jordan bocks for $\lambda$ that have size at least $k$.

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For your matrix right now, there are several Jordan structures possible. There will be a single trivial Jordan block corresponding to $1$. For the eigenvalue $2$ there will be several possibilities:

  1. We can have three trivial blocks in which case the entire sub-matrix corresponding to $2$ will look like $$\begin{pmatrix}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{pmatrix}$$ This is the case where the matrix is diagonaliable.

  2. There can be two Jordan blocks. In this case, one must be trivial and the other must be of size $2$. The sub-matrix will look like $$\begin{pmatrix}2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{pmatrix}$$

  3. There can be a single Jordan block of size $3$ $$\begin{pmatrix}2 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 2\end{pmatrix}$$

The three cases are distinguished by the geometric multiplicity of the eigenvalue $2$. Remember that the number of Jordan blocks corresponding to an eigenvalue is the geometric multiplicity of that eigenvalue.

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Can a 4x4 matrix have 3x3 Jordan canonical forms? i didn't know it.thank you. But what is the structure of finding them? –  aicha Nov 12 '12 at 7:42
    
Well no, a $4\times 4$ matrix will have a $4\times 4$ Jordan form. The above is only the $3\times 3$ portion which corresponds to the eigenvalue $2$. As for finding the Jordan form, most linear algebra textbooks will describe a procedure for finding the Jordan form in general. For small matrices, you can use a combination of factors to determine the structure including: algebraic multiplicity, geometric multiplicity and the minimal polynomial. –  EuYu Nov 12 '12 at 7:47
    
ok. My linear algebra books do not have this topic, differential equation book includes it, but it has examples having only one eigenvalue multiple of n, this is my midterm question and has 2 distinct eigenvalues,i'm confused, don't know what to do. –  aicha Nov 12 '12 at 8:23

The properties listed here can help you.
In this case you have three possibilities for the Jordan Canonical Form.These are: $$J_1=\begin{pmatrix}1 &0 &0 &0 \\ 0& 2& 0& 0\\ 0 &0& 2& 0\\ 0 &0& 0& 2\end{pmatrix} , \ J_2=\begin{pmatrix}1 &0 &0 &0 \\ 0& 2& 0& 0\\ 0 &0& 2& 1\\ 0 &0& 0& 2\end{pmatrix}, \ J_3=\begin{pmatrix}1 &0 &0 &0 \\ 0& 2& 1& 0\\ 0 &0& 2& 1\\ 0 &0& 0& 2 \end{pmatrix}.$$ If $(A-I)(A-2I)=0$ is $J_1$. If $(A-I)(A-2I)\neq 0$ and $(A-I)(A-2I)^2= 0$ is $J_2$. Else is $J_3.$

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this is my midterm question. will i write (A-I)(A-2I) ~= 0 , so J2=..... i think is not enogh for this question –  aicha Nov 12 '12 at 8:27
    
Here $(A-I)(A-2I) \neq 0$ and $(A-I)(A-2I)^2=0$ so the minimal polynomial of $A$ is $(x-1)(x-2)^2.$ Since similar matrices have the same minimal polynomial this mean $J_2$ is ... I believe this is enough. Of course it depends on what you have taught. –  P.. Nov 12 '12 at 9:02

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