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A real $2 \times 2 $ matrix $M$ such that $$M^2 = \begin{pmatrix} -1&0 \\ 0&-1-\epsilon \\ \end{pmatrix}$$

(a) exists for all $\epsilon > 0$.

(b) does not exist for any $\epsilon > 0$.

(c) exists for some $\epsilon > 0$.

(d) None of the above is true.

Attempt: I couldn't think of any theory base to prove / disprove the existence of such a matrix.

Could someone please give me a hint on how to go about this problem?

Thank you very much for your help in this regard.

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no matter if $\epsilon \neq 0$ is , there is no real matrix $M.$

for $\epsilon = 0, M = \pmatrix{0&b\\-\frac 1b&0}, b\neq 0$

pick a matrix $$M = \pmatrix{a&b\\c&d}, M^2 =\pmatrix{a^2 + bc&(a+d)b\\(a+d)c&bc+d^2}=\pmatrix{-1&0\\0&-1-\epsilon} $$ we have the constraints $$a^2 + bc = -1, bc+d^2 = -1-\epsilon \to a^2 - d^2 = \epsilon \tag 1$$

that means if $\epsilon \neq 0,$ then $a+d \neq0\implies b = c = 0.$ from $(1), a^2 = -1, d^2 = -1 - \epsilon$ there are no real solutions as $a^2 = -1.$

if $\epsilon = 0, a^2 - d^2 = 0$ it forces $a = -d, b = c = 0$ but then (1) forces $d^2 = -1$ which cannot be satisfied. the other case is $a = d.$ the constraints are $$a^2 + bc = -1, 2ab = 0 = 2ac, bc + d^2 = -1 $$ two choices: (i) $a = 0$ gives $bc = -1, d = 0$

(ii) $a \neq 0$ which gives $b = 0, c = 0, a^2 = -1$ not possible.

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  • $\begingroup$ @Travis, thanks for spotting the error. hopefully it is fixed now. $\endgroup$ – abel Apr 26 '15 at 16:48
  • $\begingroup$ You're welcome, and (+1) for the nice proof, by the way. $\endgroup$ – Travis Willse Apr 26 '15 at 16:51
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(I assume $M$ is supposed to be real; if it is allowed to be complex, then the exercise is trivial.)

Hint

  1. What are the possible eigenvalues of $M$?
  2. What can one say about the eigenvalues of real $2 \times 2$ matrices...?

Continued hint ...in particular what can one say about the eigenvalues of real $2 \times 2$ matrices when those eigenvalues are nonreal?

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    $\begingroup$ does $A^2$ diagonalizable imply that $A$ is too? $\endgroup$ – abel Apr 26 '15 at 16:36
  • $\begingroup$ @abel In general no, consider, e.g., $A = \begin{pmatrix}0 & 1\\0 & 0\end{pmatrix}$. $\endgroup$ – Travis Willse Apr 26 '15 at 16:41
  • $\begingroup$ Perhaps the downvoter could explain their objection? $\endgroup$ – Travis Willse Apr 27 '15 at 8:28
  • $\begingroup$ i got pinged but i did not down vote. in fact i upvoted you. $\endgroup$ – abel Apr 27 '15 at 11:31
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Let $M=\left(\begin{matrix}a&b\\c&d\end{matrix}\right)$. Then we want $M^2=\left(\begin{matrix}a^2+bc&b(a+d)\\c(a+d)&d^2+bc\end{matrix}\right)=\left(\begin{matrix}-1&0\\0&-1-\epsilon\end{matrix}\right)$.

Now we can't have $b=0$ or $c=0$ (why?). Therefore $a=-d$ and $$M^2=\left(\begin{matrix}a^2+bc&0\\0&a^2+bc\end{matrix}\right)=(a^2+bc)I$$ Then ?

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Assume $v$ is an eigenvector of $M$ with eigenvalue $\lambda$. Then $v$ is eigenvector of $M^2$ wih eigenvalue $\lambda^2\ge0$. Since $M^2$ has only negative eigenvalues $-1$ and $-1-\epsilon$, $M$ has no eigenvectors. Thus the matrix $A$ that maps $e_1\mapsto e_1$ and $e_2\mapsto Me_1$ is invertible and since $M$ maps $Me_1$ to $-e_1$ we find that $$M=A\begin{pmatrix}0&-1\\1&0\end{pmatrix}A^{-1}$$ so that $$ M^2=A\begin{pmatrix}0&-1\\1&0\end{pmatrix}^2A^{-1}=A\begin{pmatrix}-1&0\\0&-1\end{pmatrix}A^{-1}=\begin{pmatrix}0&-1\\1&0\end{pmatrix}.$$

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    $\begingroup$ In the last line isn't it supposed to be $A(-I_2)A^{-1} = -I_2$? $\endgroup$ – Winther Apr 26 '15 at 16:35

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