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Let $$A=\begin{pmatrix} 0&0&1\\0&1&0\\1&0&0 \end{pmatrix}$$and $$B=\begin{pmatrix} 0&1&0\\0&0&1\\1&0&0 \end{pmatrix}$$ Find one solution to each of the following matrix equation over real numbers which is not diagonal.

1)$X^{2}=-A$

2)$X^{3}=A$

3)$X^{3}=B$

4)$X^{2}=B$

My attempts are as follows:

  1. Write the matrics as linear transformations and try to figure out a solution.

$$B:\begin{pmatrix} x\\y\\z\end{pmatrix}\mapsto\begin{pmatrix} y\\z\\x\end{pmatrix}$$

So that $$X:\begin{pmatrix} x\\y\\z\end{pmatrix}\mapsto\begin{pmatrix} z\\x\\y\end{pmatrix}$$ is a solution to 4). However, the rest are quite hard to observe.

  1. Diagonalize $A,B$. However, they cannot be diagonalized in real numbers.
Isomorphism
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1 Answers1

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If you know about eigenvalues and diagonalization then you can diagonalize the matrices which allow for it and do the operations on the eigenvalues.

As an example for the first one

$$-A = SDS^{-1}, D = \left[\begin{array}{lll}-1&0&0\\0&-1&0\\0&0&1\end{array}\right], S = \left[\begin{array}{lll}0&\sqrt{1/2}&-\sqrt{1/2}\\-1&0&0\\0&\sqrt{1/2}&\sqrt{1/2}\end{array}\right]$$

then solve the scalar equation $X^2 = D_{i,i}$, As you will realize, this gives us two solutions for each such scalar. $\pm i$ for the two first, and $\pm 1$ for the last. Now what is left to do is to check if any of these lead to a real valued $X$.

I doubt that any of them will, because it must be diagonalizable and have complex eigenvalues to satisfy the equation.

So the lesson here is that we sometimes need a richer scalar field to be able to diagonalize.

mathreadler
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  • $E=\begin{pmatrix} 0&1&0\-1&0&0\0&0&1\end{pmatrix}\in\mathbb{R}^{3\times 3}$. Then $E^{2}=D$ – Isomorphism Jun 16 '20 at 14:22
  • I just realized that the square root of a diagonal matrix need not to be a diagonal one. – Isomorphism Jun 16 '20 at 14:24
  • @dilong, you are correct. I made the assumption at the start that we are working with matrices which are diagonalizable over our field. If that assumption is not true, then we don't have to search for $E$s which are also diagonalizable but can search for non-diagonal ones like the one you found. Either way, diagonalization provided us a way to simplify the problem even if we do not put restriction on $E$ to be diagonal. – mathreadler Jun 16 '20 at 14:30
  • In fact the solution you found is a diagonal matrix where upper left $2\times 2$ matrix corresponds to the complex conjugate pair $\pm i$. – mathreadler Jun 16 '20 at 14:35
  • You can play around with matrices of the form $\begin{bmatrix}a&b\-b&a\end{bmatrix}$, you may find that they behave just like $a+bi$, cool huh? – mathreadler Jun 16 '20 at 14:41
  • It is actually a rotation and is similar to my first attempt. Thanks a lot! – Isomorphism Jun 16 '20 at 14:52
  • @dilong Yep multiplication by $\pm i$ rotates 90 degrees for complex numbers, and it is a 90 degree rotation matrix as well. – mathreadler Jun 16 '20 at 15:11