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Show there is no matrix such that its square is equal to \begin{pmatrix} 0 & 1 \\ 0 & 0 \\ \end{pmatrix}

If you have an idea...

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closed as off-topic by Aqua, José Carlos Santos, GNUSupporter 8964民主女神 地下教會, MathOverview, Claude Leibovici Jan 13 '18 at 15:22

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In order to allow $A^2$ to have $0$ as eigenvalue with algebraic multiplicity $2$, $A$ itself must have $0$ as eigenvalue with algebraic multiplicity $2.$ But then the characteristic polynomial of $A$ is $\chi_A(t)=t^2$ and due to the Cayley-Hamilton theorem, $A^2=\chi_A(A)=0.$

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Let $\begin{pmatrix}a&b\\c&d\end{pmatrix}^2 = \begin{pmatrix}0&1\\0&0\end{pmatrix}$, i.e. $\begin{pmatrix}a^2+bc&ab+bd\\ac+cd&bc+d^2\end{pmatrix} = \begin{pmatrix}0&1\\0&0\end{pmatrix}$.

So, we obtain: $$\begin{cases} a^2+bc &=& 0 \\ ab+bd &=& 1 \\ ac+cd &=& 0 \\ bc+d^2 &=& 0 \end{cases}$$

From the second equation, $b(a+d) = 1$ and $c(a+d) = 0$. Either $c=0$ or $a+d=0$, but $b(a+d) = 1$, so $a+d \ne 0$, so $c=0$.

From the first equation, $a^2+bc = 0$, but since $c=0$ we have $a^2=0$, i.e. $a=0$.

From the last equation, $bc+d^2 = 0$, but since $c=0$ we have $d^2=0$, i.e. $d=0$.

Therefore, $a+d=0$, contradiction.

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  • $\begingroup$ Thanks. We can calculate directly, but there is not a more astute method? $\endgroup$ – user371663 Jan 6 '18 at 12:13
  • $\begingroup$ Why not just convince yourself by a calculation? If you do such a calculation yourself, then you will really remember this later on. $\endgroup$ – Dietrich Burde Jan 6 '18 at 15:52
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Such a matrix $A=\pmatrix{a&b\\c&d}$ would commute with $\pmatrix{0&1\\0&0}$. This means that $A=\pmatrix{a&b\\0&a}$. But then $A^2=\pmatrix{a^2&2ab\\0&a^2}$ which cannot equal $\pmatrix{0&1\\0&0}$.

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Hint: Suppose there is such a matrix, $A=\begin{pmatrix} a & b \\ c&d\end{pmatrix}$. Then $$A^2=\begin{pmatrix}a^2+bc & b(a+d) \\ c(a+d) & bc+d^2\end{pmatrix}=\begin{pmatrix} 0 & 1 \\ 0 & 0 \\ \end{pmatrix}.$$

What can you deduce?

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Assume $A^2=\begin{pmatrix}0&1\\0&0\end{pmatrix}$. Then the rank of $A$ is $1$ (it can not be $2$ or $0$). That is, $$A=\begin{pmatrix}a&b\\ak&bk\end{pmatrix},\;A^2=\begin{pmatrix}a^2+abk&ab+b^2k\\a^2k+abk&abk+b^2k^2\end{pmatrix}$$

Then $a^2+abk=0$ but $ab+b^2k=1$. Or $a(a+bk)=0$, $b(a+bk)=1$. So $a=0$ and $bk\neq 0$. But also $abk+b^2k^2=0$, a contradiction.

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Set $$A=\begin{pmatrix} 0 & 1 \\ 0 & 0 \\ \end{pmatrix}$$ $A$ has two zero eigenvalues and its characteristic equation is $\lambda^2=0$ and so any matrix $B$ such that $B^2=A$ has two 0 eigenvalues. Therefore $B$ has the same characteristic equation as of $A$ and according to Cayley-Hamilton theorem we must have $$B^2=0=A$$ which is a contradiction. So there exists no $B$ such that $B^2=A$.

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The matrix $$ B=\begin{pmatrix} 0 & 1 \\ 0 & 0 \\ \end{pmatrix} $$ is nilpotent with $B^2=0$. Hence $A^2=B$ says that $A$ is nilpotent, too, so that $A^2=0$. This is a contradiction, because $B\neq 0$.

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  • $\begingroup$ You use Cayley-Hamilton implicitly. $\endgroup$ – user371663 Jan 6 '18 at 14:11
  • $\begingroup$ @Lukas I use that a nilpotent matrix of size $n$ satisfies $A^n=0$. This can be proved without Cayley-Hamilton. $\endgroup$ – Dietrich Burde Jan 6 '18 at 15:29
  • $\begingroup$ Yes Indeed, the fact it is similar to a strict triangular is sufficient (so it seems there is reduction implicitly). Unless it can be proved without, I don't know. $\endgroup$ – user371663 Jan 6 '18 at 15:48
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If $A^2$ is as stated, then $(A^2)^2=0$ can be verified directly. Therefore, the minimum polynomial for $A$ must divide $\lambda^4$, and must be of order $0,1,2$. So $\lambda^2=0$ must hold regardless of which case is true, and this is a contradiction.

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