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Given the matrix

$$\begin{bmatrix} 0 & 0 & 0 & -8\\ 1 & 0 & 0& 16 \\ 0 & 1 & 0 & -14 \\ 0 & 0 & 1 & 6 \\ \end{bmatrix}$$

find its real canonical form.

Thanks to Wiki, I got the part where I finished Jordan normal form like below : \begin{bmatrix} 1-i & 0 & 0 & 0\\ 0 & 1+i & 0 & 0 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 2 \\ \end{bmatrix}

Now, I am stuck and have no clue how to convert this into "REAL" jordan form. Thank you.

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    $\begingroup$ Surely one of those $1 - i$'s should be $1 + i$? $\endgroup$ – Theo Bendit Apr 4 at 22:22
  • $\begingroup$ This answer of mine may be of some help. $\endgroup$ – Theo Bendit Apr 4 at 22:24
  • $\begingroup$ Sorry, I didn't notice there was a typo in it! $\endgroup$ – Complicated Apr 4 at 23:16
  • $\begingroup$ Do you mind explaining it over an example? ( doesn't have to be mine ) I tried to search over the internet for real jordan form, but i have trouble understanding those paragraphs... $\endgroup$ – Complicated Apr 4 at 23:17
  • $\begingroup$ I'll write up an answer, but I further want to point out that the matrix above is singular (since it has a column of $0$s), hence $0$ should be an eigenvalue. If you write out the matrix correctly, I'll use it as an example. $\endgroup$ – Theo Bendit Apr 5 at 0:25
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As always, when finding complex eigenvalues of a real matrix, our non-real eigenvalues come in conjugate pairs: $1 \pm i$. By my calculations, the eigenspaces for these eigenvalues are $1$-dimensional (each), which agrees with your JNF matrix. Moreover, we take non-zero complex eigenvectors $$v_1 = \begin{bmatrix} 4 + 4i \\ -8 - 4i \\ 5 + i \\ -1 \end{bmatrix} \quad v_2=\begin{bmatrix} 4 - 4i \\ -8 + 4i \\ 5 - i \\ -1 \end{bmatrix},$$ corresponding to eigenvalues $\lambda_1 = 1 - i$ and $\lambda_2 = 1 + i$ respectively, which must span their respective eigenspaces.

Note that the eigenvectors are "conjugates" of each other themselves (in the entry-wise sense). This is no accident. In general, we have that $$Av = \lambda v \implies \overline{Av} = \overline{\lambda} \overline{v} \implies A \overline{v} = \overline{\lambda} \overline{v},$$ which is to say, if $v$ is an eigenvector corresponding to $\lambda$, then $\overline{v}$ is an eigenvector corresponding ot $\overline{\lambda}$. This says something very trivial if $v$ and $\lambda$ are real, but something useful when they are complex.

So, we should be able to write our eigenvectors in the form $v \pm iw$, where $v$ and $w$ are real vectors. In fact, we have \begin{align*} v_1 &= \begin{bmatrix} 4 + 4i \\ -8 - 4i \\ 5 + i \\ -1 \end{bmatrix} = \begin{bmatrix} 4 \\ -8 \\ 5 \\ -1 \end{bmatrix} + i\begin{bmatrix} 4 \\ -4 \\ 1 \\ 0 \end{bmatrix} \\ v_2 &= \begin{bmatrix} 4 - 4i \\ -8 + 4i \\ 5 - i \\ -1 \end{bmatrix} = \begin{bmatrix} 4 \\ -8 \\ 5 \\ -1 \end{bmatrix} - i\begin{bmatrix} 4 \\ -4 \\ 1 \\ 0 \end{bmatrix}. \end{align*}

To compute the complex JNF matrix, you take the Jordan basis of complex eigenvectors, and compute the matrix with respect to the Jordan basis. To compute the real JNF matrix, replace the vectors $v_1$ and $v_2$ with their real and imaginary components, i.e. the vectors $[4, -8, 5, -1]^\top$ and $[4, -4, 1, 0]^\top$. As pointed out in the answer linked in the comments, this will replace the $\begin{bmatrix} 1 - i & 0 \\ 0 & 1 + i \end{bmatrix}$ block with $\begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}$.

Let's see it in action. We have a complex Jordan Basis: $$\left(\begin{bmatrix} 4 + 4i \\ -8 - 4i \\ 5 + i \\ -1 \end{bmatrix}, \begin{bmatrix} 4 - 4i \\ -8 + 4i \\ 5 - i \\ -1 \end{bmatrix}, \begin{bmatrix} -4 \\ 6 \\ -4 \\ 1 \end{bmatrix}, \begin{bmatrix} 2 \\ -2 \\ 1 \\ 0 \end{bmatrix}\right).$$ Changing to this basis produces the JNF you've written down.

As instructed, we will replace the first two vectors:

$$\mathcal{B} = \left(\begin{bmatrix} 4 \\ -8 \\ 5 \\ -1 \end{bmatrix}, \begin{bmatrix} 4 \\ -4 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -4 \\ 6 \\ -4 \\ 1 \end{bmatrix}, \begin{bmatrix} 2 \\ -2 \\ 1 \\ 0 \end{bmatrix}\right).$$

To transform the matrix into this basis, consider the action of this matrix on individual basis vectors. We don't need to check the last two vectors, as they remain unchanged from the complex Jordan basis. We have, $$\begin{bmatrix} 0 & 0 & 0 & -8 \\ 1 & 0 & 0 & 16 \\ 0 & 1 & 0 & -14 \\ 0 & 0 & 1 & 6 \end{bmatrix}\begin{bmatrix} 4 \\ -8 \\ 5 \\ -1 \end{bmatrix} = \begin{bmatrix} 8 \\ -12 \\ 6 \\ -1 \end{bmatrix} = \begin{bmatrix} 4 \\ -8 \\ 5 \\ -1 \end{bmatrix} + \begin{bmatrix} 4 \\ -4 \\ 1 \\ 0 \end{bmatrix} + 0\begin{bmatrix} -4 \\ 6 \\ -4 \\ 1 \end{bmatrix} + 0\begin{bmatrix} 2 \\ -2 \\ 1 \\ 0 \end{bmatrix},$$ hence the first column will be the coordinate vector $$\begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix}.$$ For the second vector, we get $$\begin{bmatrix} 0 & 0 & 0 & -8 \\ 1 & 0 & 0 & 16 \\ 0 & 1 & 0 & -14 \\ 0 & 0 & 1 & 6 \end{bmatrix}\begin{bmatrix} 4 \\ -4 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 4 \\ -4 \\ 1 \end{bmatrix} = -\begin{bmatrix} 4 \\ -8 \\ 5 \\ -1 \end{bmatrix} + \begin{bmatrix} 4 \\ -4 \\ 1 \\ 0 \end{bmatrix} + 0\begin{bmatrix} -4 \\ 6 \\ -4 \\ 1 \end{bmatrix} + 0\begin{bmatrix} 2 \\ -2 \\ 1 \\ 0 \end{bmatrix},$$ so the second column will be $$\begin{bmatrix} -1 \\ 1 \\ 0 \\ 0 \end{bmatrix}.$$ As stated before, the third and fourth columns remain unchanged. Thus, we obtain the matrix with respect to basis $\mathcal{B}$, $$\begin{bmatrix} 1 & -1 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 2 \end{bmatrix},$$ which is the real JNF.

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