Given that the the Jordan normal form of a matrix is,
$J=\begin{bmatrix}2&1&0&0\\0&2&0&0\\0&0&1-i&0\\0&0&0&1+i\end{bmatrix}$
How do you find the 'real' canonical form of the matrix?
[Math] How to find the ‘real’ jordan canonical form of a matrix
jordan-normal-formlinear algebramatrices
Related Solutions
We have:
$$A = \begin{bmatrix}1&1&1\\0&2&0 \\ 0&0&2\end{bmatrix}$$
We find the eigenvalues of $|A - \lambda I| = 0$, hence:
$$\lambda_1 = 1, \lambda_{2,3} = 2$$
That is, we have a single root and a double root eigenvalue, algebraic multiplicity.
To find the eigenvectors, we set up and solve $[A - \lambda_i I]v_i = 0$.
For $\lambda_1 = 1$, we get the eigenvector:
$$v_1 = (1,0,0)$$
For $\lambda_{2,3} = 2$, we get the eigenvectors (normally, we do not get two linearly independent eigenvectors):
$$v_2 = (1,0,1), v_3 = (1,1,0)$$
We now can write $P$ using the eigenvectors as columns. We have,
$$P = [v_1 | v_2 | v_3 ] = \begin{bmatrix}1&1&1\\0&0&1 \\ 0&1&0\end{bmatrix}$$
We can write the Jordan Normal Form (notice that we do not have any Jordan blocks), $J$, using the corresponding eigenvalues:
$$J = P^{-1} A P$$
However, we can also write this straight off from the eigenvalues and knowing we do not need any Jordan blocks.
$$J = \begin{bmatrix}1&0&0\\0&2&0 \\ 0&0&2\end{bmatrix}$$
Lastly, we should verify:
$$A = P J P^{-1}$$
I purposely left things so you can fill in the details of the calculations.
Firstly, we have that $\dim\ker(\mathbf A-4\mathbf I) = 1$. That means the eigenvalue $\lambda = 4$ is not semi - simple *, thus we have to add a unit above the main diagonal on the Jordan block that corresponds to the eigenvalue $\lambda = 4$, i.e. $$\mathbf J = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & 1 \\ 0 & 0 & 4 \end{bmatrix}.$$
*An eigenvalue $\lambda$ is called semi - simple, when: $$\operatorname{algebraic\, mult}\, \lambda =\operatorname{geometric\, mult} \lambda.$$
The matrix $\mathbf P$ will contain (as its columns) the eigenvectors that correspond to the eigenvalues. It is easy to find $2$ eigenvectors that correspond to $\lambda_1=1$ and $\lambda_2 = 4$, say $\mathbf u$ and $\mathbf v_1$ respectively.
However, we need to find an extra generalized eigenvector $\mathbf v_2$ that corresponds to the eigenvalue $\lambda_2 =4$. To do so, we may solve the equation: $$(\mathbf A-\lambda_2 \mathbf I)\cdot \mathbf v_2 = \mathbf v_1.\tag{1}$$
We can check that $\mathbf u = (1,0,0)^\top$ is an eigenvector that corresponds to $\lambda_ 1 = 1$. Also, we can check that $\mathbf v_1 = (1, \frac{3}{2}, 0)^\top$ is an eigenvector that corresponds to the eigenvalue $\lambda_2 = 4$.
For $\mathbf v_2 $ it must hold: $$(\mathbf A - 4\mathbf I) \cdot \mathbf v_2 \neq \mathbf 0 \text{ and }(\mathbf A - 4\mathbf I)^2 \cdot \mathbf v_2 = \mathbf 0. $$
By $(1)$ we can compute $\mathbf v_2 = (1,\frac {31}{20},\frac{3}{10})^\top$. That means our matrix $\mathbf P$ is: $$\mathbf P = \begin{bmatrix} \mathbf u &\mid &\mathbf v_1 & \mid &\mathbf v_2 \end{bmatrix}.$$
If we do the math, we will see that $$\mathbf P^{-1} \cdot \mathbf A \cdot \mathbf P = \mathbf J.$$
Best Answer
Wubbish. There clearly is such a thing. Just replace the complex pair by the $2\times 2$ block below.
$$J=\left[\begin{array}{rrrr}2&1&0&0\\0&2&0&0\\0&0&1&-1\\0&0&1&1\end{array}\right]$$
You can compare the block to the matrix:
$$\left[\begin{array}{rr}a&-b\\b&a\\\end{array}\right]$$ which can be used to represent arbitrary complex numbers $a+bi$ or $a-bi$ as long as you are consistent on which you are using for each new number / block.