# How to find a matrix $B$ and an invertible matrix $P$ such that this matrix $A$ is in Jordan Canonical Form?

I am working on the following exercise:

Find a matrix B and an invertible matrix P such that

$$A = \begin{bmatrix} 1 & -2 & 1 & 0 \\ 1 & -2 & 1 & 0 \\ 1 & -2 & 1 & 0 \\ 1 & -1 & 0 & 0 \\ \end{bmatrix}$$

is in Jordan Canonical Form.

This is my working so far:

I calculated the characteristic polynomial of $A$ to be $c_A(x) = -x^4$ (which implies that the only eigenvalue of $A$ is $\lambda = 0$) and the minimal polynomial to be $\mu_A(x) = x^2$.

Since $\text{nullity}((A - 0I_4)^2) - \text{nullity}((A - 0I_4)^1) = 4 - 2 = 2$, the JCF of $A$ is composed of two Jordan blocks, each of degree 2. The JCF is

$$B = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}.$$

To find $P$, I then proceeded to find two Jordan chains, $v_1, v_2$ and $v_3, v_4$, such that $A^2v_2 = 0$, but $Av_2 \neq 0$ (similarly for $v_4$), and $v_1 = Av_2 - 0v_2$, $v_3 = Av_4 - 0v_4$.

I chose $v_2 = \begin{bmatrix} 0 & 0 & 1 & 0 \\ \end{bmatrix}^T$ (then, $v_1 = \begin{bmatrix} 1 & 1 & 1 & 0 \\ \end{bmatrix}^T$), $v_4 = \begin{bmatrix} 0 & 1 & 0 & 0 \\ \end{bmatrix}^T$ (then, $v_3 = \begin{bmatrix} -2 & -2 & -2 & -1 \\ \end{bmatrix}^T$).

I then put

$$P = \begin{bmatrix} 1 & 0 & -2 & 0 \\ 1 & 0 & -2 & 1 \\ 1 & 1 & -2 & 0 \\ 0 & 0 & -1 & 0 \\ \end{bmatrix}$$

and calculated its inverse to be

$$P^{-1} = \frac{1}{\det(P)}\text{adj}(P) = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ \end{bmatrix}.$$

However, when I calculated $P^{-1}AP$ this gave me the matrix

$$\begin{bmatrix} 0 & 1 & 0 & -2 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 0 & -1 \\ 0 & 1 & 0 & -2 \\ \end{bmatrix}.$$

That matrix is clearly not equal to $B$.

Could I have some help to complete this exercise, please? Perhaps I have made an error in calculation or understanding.

Notice that $A^2 = 0$, so $A^2v=0$ for all $v$. Clearly starting with this condition does not help much.
You need to start with the eigenvectors associated with $\lambda=0$. Suppose $v_1$ is one such an eigenvector, then by the Jordan form you have calculated we must have some $v_{11}$ which is not a scalar multiple of $v_1$ satisfying $$Av_{11} = v_1.$$ (So just solve $Av_1=0$ and then solve $Av_{11}=v_1$). Similar equation follows for the second eigenvector $v_2$. It is easiest to see this by letting $$P = [v_1, v_{11},v_2,v_{21}]$$ and of course we must have $v_{11}$ is not a multiple of $v_1$ and $v_{21}$ is not a multiple of $v_2$, and we must have $$AP=PJ.$$ (Note: $J=B$, using the convention to denote the matrix in Jordan form as $J$)
Since $A^2$ is the zero matrix and $0I = 0$, you can just denote that the generalized eigenspace for $E_0$ is simply $Ker(A)$. Now, you require two vectors such that $A^2x = 0$ and $x \in Img(A)$ (this creates the chain $Ay = 0$, i.e. the chain is $y, x$ since $Ay = AAx = 0$, but you require that $y$ must be reachable from $x$ in the first place...)
If you row reduce $A$, you get that the linearly independent column vectors are: $$\begin{pmatrix}1\\1\\1\\1\\\end{pmatrix}, \begin{pmatrix}-2\\-2\\-2\\-1\\\end{pmatrix} = imgA$$ You chose the vector $[0, 0, 1, 0]$ which is NOT in $img(A)$ (verify this by putting the matrix $Ax = [0, 0, 1, 0]$ in augmented form and row-reduce. For example, $v_1 = \begin{bmatrix} 1 & 1 & 1 & 0 \\ \end{bmatrix}$, but: $$Av_1 \neq v_2$$ So even though (didn't check but assuming you are right about this), that $Av_2 = 0$, it is not reachable in the chain. So this cannot be a basis eigenvector for the Jordan canonical form.