IMC 2017 Problem 1

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math90

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math90

#1 h Aug 2, 2017, 1:36 PM • 5 Y

Y by GoJensenOrGoHome, Davi-8191, Adventure10, Mango247, mathematicsy

Determine all complex numbers $\lambda$ for which there exists a positive integer $n$ and a real $n\times n$ matrix $A$ such that $A^2=A^T$ and $\lambda$ is an eigenvalue of $A$ .

This post has been edited 1 time. Last edited by math90, Aug 2, 2017, 3:28 PM

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ISHO95

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ISHO95

#3 h Aug 2, 2017, 2:58 PM • 4 Y

Y by GoJensenOrGoHome, MSTang, Adventure10, Mango247

$A^4=(A^2)^2=(A^T)^2=(A^2)^T=A$ . Hence, all eigenvalues $\alpha$ of $A$ is a root of the polynomial $x^4-x=0$ . For $\alpha =0$ , $A=O_n$ , zero matrix satisfies the conditions. For $(\alpha)^3=1$ ,
$A= \left( \begin{matrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 1 & 0 & 0 \end{matrix} \right)$ is a matrix which is asked.

This post has been edited 3 times. Last edited by ISHO95, Aug 2, 2017, 3:12 PM

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loup blanc

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loup blanc

#4 h Aug 4, 2017, 9:49 AM • 2 Y

Y by Adventure10, Mango247

More generally, we can solve the equations
(1) $A^2=A^*$ in $M_n(\mathbb{C})$ and (2) $A^2=A^T$ in $M_n(\mathbb{R})$ .
for (1). Since $AA^*=A^*A$ , $A$ is normal and, consequently, is diagonalizable in an orthonormal basis. Then we may assume that $A=diag((\lambda_i))$ where, for every $i$ , $\lambda_i^2=\overline{\lambda_i}$ . Thus $\lambda_i\in\{0,1,j,j^2\}$ where $j=\exp(2i\pi/3)$ .
Conclusion. $A$ is unitarily similar to $diag(0I_p,I_q,jI_r,j^2I_s)$ where $p+q+r+s=n$ .
for (2). It suffices to find the real solutions of (1). $A$ is unitarily similar to $diag(0I_p,I_q,V_1,\cdots,V_r)$ where $V_i=diag(j,j^2)$ . It remains to find a $2\times 2$ real matrix that is unitarily similar to $V_i$ .

This post has been edited 4 times. Last edited by loup blanc, Aug 4, 2017, 9:14 PM

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loup blanc

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loup blanc

#5 h Aug 4, 2017, 9:07 PM • 2 Y

Y by Adventure10, Mango247

Part 2. Let $R=Rot(2\pi/3)$ . Since $spectrum(R)=\{j,j^2\}$ , $R^2=R^{-1}=R^T$ , $R$ is unitarily similar to $diag(j,j^2)$ .
Conclusion. The solutions of (2) are those that are orthogonally similar to $diag(0I_p,I_q,U_1,\cdots,U_r)$ , where $U_j=R$ and $p+q+2r=n$ .
In particular, the solution $\begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix}$ given by ISHO95, is orthogonally similar to $diag(1,R)$ .

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Tyrone03

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Tyrone03

#6 h Jun 16, 2022, 3:08 PM

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ISHO95 wrote:

. Hence, all eigenvalues $\alpha$ of $A$ is a root of the polynomial $x^4-x=0$ .

why exactly is this step true? Is there some matrix property that im not aware of

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Adamsawyer

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Adamsawyer

#7 h Jun 16, 2022, 4:16 PM • 2 Y

Y by Tyrone03, Mango247

This is the consequence of Cayley hamilton theorem https://en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem#:~:text=In%20linear%20algebra%2C%20the%20Cayley,satisfies%20its%20own%20characteristic%20equation.

you can see that matrix satisfies its characteristics polynomial. ( sorry for bad precise language).
thank you @acridian9. I was thinking that if I saw matrix then there will be corresponding characteristics equation which is not true(is so?)
Can you please tell then what helps us to conclude that the minimal polynomial divides $x^4-x$ . Thank you

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Acridian9

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Acridian9

#8 h Jun 16, 2022, 5:58 PM

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Adamsawyer wrote:

This is the consequence of Cayley hamilton theorem https://en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem#:~:text=In%20linear%20algebra%2C%20the%20Cayley,satisfies%20its%20own%20characteristic%20equation.

you can see that matrix satisfies its characteristics polynomial. ( sorry for bad precise language).

No, it is not. We don't actually know what the characteristic polynomial is in this case.

We just know that the minimal polynomial divides $x^4-x$ , which concludes, but actually it's more a direct computation:
suppose $v$ is an eigenvector of $A$ for the eigenvalue $\lambda$ , then
$0=(A^4-A)v=(\lambda^4-\lambda)v$ which implies $\lambda^4-\lambda=0$ .

@GianDR: thanks for the correction!

This post has been edited 1 time. Last edited by Acridian9, Jun 16, 2022, 7:38 PM

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407420

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407420

#9 h Jun 16, 2022, 6:39 PM

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Acridian9 wrote:

Adamsawyer wrote:

This is the consequence of Cayley hamilton theorem https://en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem#:~:text=In%20linear%20algebra%2C%20the%20Cayley,satisfies%20its%20own%20characteristic%20equation.

you can see that matrix satisfies its characteristics polynomial. ( sorry for bad precise language).

No, it is not. We don't actually know what the characteristic polynomial is in this case.

We just know that the minimal polynomial divides $x^4-x$ , which concludes, but actually it's more a direct computation:
suppose $v$ is an eigenvalue of $A$ for the eigenvalue $\lambda$ , then
$0=(A^4-A)v=(\lambda^4-\lambda)v$ which implies $\lambda^4-\lambda=0$ .

Hi!
Small typo, you mean $v$ is an eigenvector, not an eigenvalue.

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BlochStokes

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BlochStokes

#10 h Jul 31, 2022, 3:58 PM

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As a new user, I'm not allowed to use LaTeX in my text so I encourage you to edit my text a bit and put it in the compiler.
What one can also use is the properties of #A^*#. Since #A# is real #A^T=A^*# and
# \lambda\|x\|^2=\langle Ax,x \rangle = \langle x, A^*x\rangle = \langle x, A^2x\rangle = \langle x, \lambda^2x\rangle = \overline{\langle \lambda^2x,x\rangle}=\overline{\lambda}^2\|x\|^2 \quad \Rightarrow \quad \lambda=\overline{\lambda}^2 #
where #x# is an eigenvector. In form #\lambda=a+bi# we get a system of equations:
#\left\{ \begin{matrix}
a^2-b^2&=a
-2ab &=b
\end{matrix}\right.#
which trivially gives us all the solutions. Remark that examples of such matrices must be provided in this way of solving too.

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recoco

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recoco

#12 h Jul 31, 2022, 4:05 PM • 1 Y

Y by BlochStokes

BlochStokes wrote:

As a new user, I'm not allowed to use LaTeX in my text so I encourage you to edit my text a bit and put it in the compiler.
What one can also use is the properties of $A^*$ . Since $A$ is real $A^T=A^*$ and
$\lambda\|x\|^2=\langle Ax,x \rangle = \langle x, A^*x\rangle = \langle x, A^2x\rangle = \langle x, \lambda^2x\rangle = \overline{\langle \lambda^2x,x\rangle}=\overline{\lambda}^2\|x\|^2 \quad \Rightarrow \quad \lambda=\overline{\lambda}^2$
where $x$ is an eigenvector. In form $\lambda=a+bi$ we get a system of equations:
$\left\{ \begin{matrix} a^2-b^2&=a \\ -2ab &=b \end{matrix}\right.$
which trivially gives us all the solutions. Remark that examples of such matrices must be provided in this way of solving too.

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VicKmath7

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VicKmath7

#13 h Aug 31, 2024, 8:57 PM

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Solution

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