Solving quadratic matrix equation for X \(X^2 = \begin{bmatrix}1&a\\0&1\\\end{bmatrix}\) where a\in\frac{\mathbb{R}}{0}. Solve



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Solving quadratic matrix equation for X
where aR0. Solve for matrix X.

Answer & Explanation

Kyler Jacobson

Kyler Jacobson

Beginner2022-01-31Added 8 answers

Step 1
Note M(a) has one eigenvalue: 1, and is not diagonalisable. This means that, if X2=M(a), then X has eigenvalue 1 or -1, but not both (if it had both, then X would be diagonalisable, and so would X2=M(a))
We also note that any eigenvectors of X will be an eigenvector for X2, and X2 has only the eigenvector
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
on the right reveals the first column of a 2×2 matrix. Therefore, the first column of X is
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:
where the three ±s are all the same. That is,
for some b. Note that (±M(b))2=M(b)2, so we are looking for b such that
which was solved by Angel in their answer: b=a2. This gives us precisely two solutions:


Beginner2022-02-01Added 10 answers

Step 1
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,
the answer is trivially obtained from that.

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