# Matrix whose square is in Jordan normal form

Let $$A = \begin{bmatrix}J_0^2 \\ & J_0^2 & \\ && J_{1/4}^3\end{bmatrix}\in M_7(\mathbb{Q})$$ Find, with proof, a matrix $$B$$ so that $$B^2 = A$$.

I'm not sure how to find this matrix. Clearly, $$A$$ is not invertible as it is contains rows solely consisting of zeros. I know how to convert a matrix into Jordan form, but I'm not sure how this can be applied here. Also, I know that for any invertible matrix $$A$$ in $$M_n(\mathbb{C})$$, there exists a matrix $$B$$ so that $$B^2 = A$$. I can find a matrix $$B$$ so that $$B^2 = J_{1/4}^3$$, where

$$B = \begin{bmatrix}1/2 & 1 & -1 \\ & 1/2 & 1\\ & & 1/2\end{bmatrix} = \frac{1}2 \left(I + \frac{1}2N + {1/2\choose 2}N^2 \right)$$

where $$N^k$$ is the matrix where $$N_{i, i+k} = 4^k$$ and $$N_{i,j} = 0$$ otherwise. I think I might need to solve some system of equations to find the matrix $$B$$.

Clarification: $$A := J_k^m$$ is the $$m\times m$$ matrix with entries given by $$A_{m,m}= k$$ and for $$1\leq i\leq m-1, A_{i,i} = k$$ and $$A_{i,i+1} = 1$$ and $$A_{i,j}=0$$ otherwise.

• What are $J_0$ and $J_{1/4}$? Jul 28 '20 at 15:48
• Is $J_0^2$ means order $2$ block for eigenvalue $0$? Or there are two blocks of order $1$ ? conform it. Jul 28 '20 at 15:55
• Sorry for the confusion. I'll update my question
– user763400
Jul 28 '20 at 16:02
• @Subhajit yes $J_0^2$ means order $2$ block for eigenvalue $0$.
– user763400
Jul 28 '20 at 16:05
• $M_7(\Bbb Q)$ isn’t the space of $7\times7$ matrixes with rational entries, is it? Because $A$ looks $3\times3$ right? I just haven’t studied this stuff before, but I’d like to read some more about it. Jul 28 '20 at 16:30

Think of $$A$$ as a block diagonal matrix with blocks $$\begin{bmatrix} J_0^2 & 0 \\ 0 & J_0^2 \end{bmatrix}$$ and $$\begin{bmatrix} J_{1/4}^3 \end{bmatrix}$$. If you find matrices that square to each of these blocks then form a block diagonal matrix out of them and the result will square to $$A$$. You've already figured out a matrix for the second block, for the first block use $$\begin{bmatrix} 0 & J_0^2 \\ I & 0 \end{bmatrix}$$.
Take the matrix, $$B= \begin{pmatrix} 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1/2 & 1 & -1 \\ 0 & 0 & 0 & 0 & 0 & 1/2 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1/2 \\ \end{pmatrix}$$ , which fulfills your requirement.