$$X^2 = \begin{bmatrix}1&a\\0&1\\\end{bmatrix}$$ where $a \in \Bbb R \setminus \{0\}$. Solve for matrix $X$.
I was practicing for matrix equations and this is the first one where it has squared matrix and another number, in this case, $a$. I would be grateful if you could help. If you can please suggest a book that has matrix equations to practice. Thank you!
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Let $$M(a) = \begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix}.$$ Note $M(a)$ has one eigenvalue: $1$, and is not diagonalisable. This means that, if $X^2 = M(a)$, then $X$ has eigenvalue $1$ or $-1$, but not both (if it had both, then $X$ would be diagonalisable, and so would $X^2 = M(a)$). We also note that any eigenvectors of $X$ will be an eigenvector for $X^2$, and $X^2$ has only the eigenvector $\begin{bmatrix} 1 \\ 0\end{bmatrix}$. Thus, $X$ must have only this eigenvector as well. (Of course, when I say a matrix has only one eigenvector, I mean that there is one eigenspace, and it is one-dimensional, so we can only find one linearly independent eigenvector.)
Now, multiplication by $\begin{bmatrix} 1 \\ 0\end{bmatrix}$ on the right reveals the first column of a $2\times 2$ matrix. Therefore, the first column of $X$ is $$X \begin{bmatrix} 1 \\ 0\end{bmatrix} = \pm 1 \cdot \begin{bmatrix} 1 \\ 0\end{bmatrix} = \begin{bmatrix} \pm1 \\ 0\end{bmatrix},$$ depending on whether the unique eigenvalue is $1$ or $-1$. This makes the matrix $X$ upper-triangular (due to the $0$ in the bottom left corner), so the eigenvalues of $X$ lie on the diagonal. That is, $X$ takes the form: $$\begin{bmatrix} \pm 1 & * \\ 0 & \pm 1\end{bmatrix} = \pm\begin{bmatrix} 1 & * \\ 0 & 1\end{bmatrix},$$ where the three $\pm$s are all the same. That is, $$X = \pm M(b)$$ for some $b$. Note that $(\pm M(b))^2 = M(b)^2$, so we are looking for $b$ such that $$\begin{bmatrix}1 & b \\ 0 & 1 \end{bmatrix}^2 = \begin{bmatrix}1 & a \\ 0 & 1 \end{bmatrix},$$ which was solved by Angel in their answer: $b = \frac{a}{2}$. This gives us precisely two solutions: $$X = \begin{bmatrix}1 & \frac{a}{2} \\ 0 & 1 \end{bmatrix}, -\begin{bmatrix}1 & \frac{a}{2} \\ 0 & 1 \end{bmatrix}.$$
HINT: Let $X = (I + N)$ where $I$ is the identy matrix and $N$ is the nilpotent matrix with all the diagonal entries and the lower left coner being zero. Take the second power of $X$, $$X^2 = I + N + N + N^2 = I + 2N $$ the answer is trivially obtained from that.