Square root of a matrix 3 without diagonalization

Question

I have two questions.

I want to know how I can say that this matrix admits an infinite number of square roots :

$\begin{pmatrix} 0 & 0 & 1\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix}$

And that this matrix doesn't admit square root.

$\begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix}$

I am a beginner, anyone can help me ? Thank you in advance.

score 2 · Accepted Answer · answered Apr 09 '17 at 17:54

2

For the first problem, you can just write down square roots: $$\begin{pmatrix} 0 & a & 0 \\ 0 & 0 & a^{-1} \\ 0 & 0 & 0 \end{pmatrix}$$ will work for any $a \ne 0.$

For the second problem, a square root $A$ of $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ would have to be nilpotent ($A^4 = 0$); but then the Cayley-Hamilton theorem implies that $A^2 = 0$, contradiction.

answered Apr 09 '17 at 17:54

user399601

1,711

Thank you so much, but why a square root of the second matrix must be nilpotent of order 4 ? – Mélanie De la Cheminée Apr 09 '17 at 18:08
$\begin{pmatrix} 0 & 1 \ 0 & 0 \end{pmatrix}^2 = 0$ – user399601 Apr 09 '17 at 18:28

Square root of a matrix 3 without diagonalization

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