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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.

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    $\begingroup$ What are $J_0$ and $J_{1/4}$? $\endgroup$ – primes.against.humanity Jul 28 at 15:48
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    $\begingroup$ Is $ J_0^2 $ means order $ 2 $ block for eigenvalue $ 0 $? Or there are two blocks of order $1$ ? conform it. $\endgroup$ – A learner Jul 28 at 15:55
  • $\begingroup$ Sorry for the confusion. I'll update my question $\endgroup$ – user763400 Jul 28 at 16:02
  • $\begingroup$ @Subhajit yes $J_0^2$ means order $2$ block for eigenvalue $0$. $\endgroup$ – user763400 Jul 28 at 16:05
  • $\begingroup$ $M_7(\Bbb Q)$ 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. $\endgroup$ – gen-ℤ ready to perish Jul 28 at 16:30
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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 $\begin{bmatrix} 0 & J_0^2 \\ I & 0 \end{bmatrix}$.

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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.

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  • $\begingroup$ Just try to creat such matrix that the square of the matrix is the matrix made by first two blocks what you have given. It can be done by a simple try. $\endgroup$ – A learner Jul 28 at 16:48

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