Find all values of $a$ such that $A^3 = 2A$, where

$$A = \begin{bmatrix} -2 & 2 \\ -1 & a \end{bmatrix}.$$

The matrix I got for $A^3$ at the end didn't match up, but I probably made a multiplication mistake somewhere.

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    $\begingroup$ Please do not delete questions with good answers. Others have devoted effort to answer your question; deleting the question is disrespectful of their effort and prevents others from benefiting from your question and its answers. $\endgroup$ – robjohn Sep 29 '13 at 16:53


$$A^3 = \begin{bmatrix} -4 + 2 (2 - a) & -2 (-4 + 2 a) + 2 (-2 + a^2) \\-2 + (2 - a) a & 4 - 2 a + a (-2 + a^2) \end{bmatrix}$$

$$2A = \begin{bmatrix} -4 & 4 \\-2 & 2 a \end{bmatrix}$$

Can you take it from here?


Hover over the blue area for the result.

$a=2$ (Throw out the other solutions from equating the lower right items to solve for $a$).

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    $\begingroup$ What's that about the lower right? This is most easily seen by comparing the results in the upper left! $\endgroup$ – dfeuer Oct 6 '13 at 20:11
  • $\begingroup$ I could have used better terminology. If you equate $A_{2,2}^3$ to $2A_{2,2}$ and solve for $a$, there are 3 values. One works and the other two do not. All other positions only provide a single value. So we could have possibly had more than one solution. Regards $\endgroup$ – Amzoti Oct 6 '13 at 20:46
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    $\begingroup$ I think you're missing dfeuer's point. If you equate $A_{1,1}^3$ to $2A_{1,1}$ you get $-4 + 2(2-a) = -4$, which has one solution. Then it's just a case of checking it against the other three cells. $\endgroup$ – Peter Taylor Oct 7 '13 at 9:53
  • $\begingroup$ Yes, I looked at that when I originally solved it and all I am saying is that we needed a single solution, but that we also possibly had three, so I did not miss that. It happens that the $a$ that we find in all cases is coincident. Regards $\endgroup$ – Amzoti Oct 7 '13 at 12:50
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    $\begingroup$ No, you're still missing dfeuer's point. Because the solution has to work in all positions, once we see that the upper left has only one solution, that's the only possible solution to the problem, no matter what the other elements look like. The only thing that is left is to check whether $a=2$ works or not -- that there may be different numbers that work for some of the other elements is immaterial, because they will fail in the upper left position anyway. $\endgroup$ – Henning Makholm Oct 8 '13 at 0:59

Here's a different approach that avoids matrix multiplication in favor of determinants:

We want to find $a$ such that $A^3=2A$. This is equivalent to $A^3-2A=0$, and to $(A^2-2I)A=0$.

Thus either $A=0$, which is impossible, or $A^2-2I$ is a zero divisor.

Thus \begin{align}0&=\det (A^2-2I) \\ &= \det(A+\sqrt2I)\det(A-\sqrt2 I),\end{align} so one of these determinants must be $0$. That is, $(-2+\sqrt 2)(a+\sqrt 2)+2=0$ or $(-2-\sqrt 2)(a-\sqrt 2)+2=0$. These have the same sole solution: $a=2$.

Note that we get the value of $A^2$ too: plugging in $a=2$ gives $\det A = -2\ne 0$, so $A$ is not a zero divisor, so $A^2-2I=0$, so $A^2=2I$.

  • $\begingroup$ I don't follow. Since $x^3-2x$ annihilates the $2\times2$ matrix $A$, the minimal polynomial of $A$ could be $x,\ x-\sqrt{2},\ x+\sqrt{2},\ x(x-\sqrt{2}),\ x(x+\sqrt{2})$ or $x^2-2$. Among these six candidates, you only consider $x$ and $x^2-2$. Why? $\endgroup$ – user1551 Oct 7 '13 at 4:43
  • $\begingroup$ @user1551, I don't know what a "minimal polynomial" is. $(A^2-2I)A=0$ implies that either $A=0$ or $\det(A^2-2I)=0$. $A$ obviously cannot be $0$, because it has non-zero constant entries. $\endgroup$ – dfeuer Oct 7 '13 at 4:56
  • $\begingroup$ I see. So, you are using abstract algebra terminology ('zero divisor') to refer to what is usually called 'noninvertible/singular matrix' in linear algebra. Thanks. +1 $\endgroup$ – user1551 Oct 7 '13 at 5:26
  • $\begingroup$ @user1551, would you mind telling me what a minimal polynomial is? $\endgroup$ – dfeuer Oct 7 '13 at 6:37
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    $\begingroup$ See Wikipedia. A simple and routine solution (without tedious calculation of $A^3$) to the OP's problem would employ minimal polynomial, but your solution is admittedly even simpler. $\endgroup$ – user1551 Oct 7 '13 at 6:42

Any element of the set $\left\{\lambda\right\}$ of $A$ eigenvalues satisfy the equations

$$ \lambda^{2} = \left(a - 2\right)\lambda + 2\left(a - 1\right)\,, \quad \lambda^{3} = 2\lambda\,, \quad \sum_{\lambda}\lambda = a - 2 $$

Then, $\sum_{\lambda}\lambda^{3} = 2\left(a - 2\right)$. Also, \begin{align} \sum_{\lambda}\lambda^{3} &= \sum_{\lambda}\lambda\left[\left(a - 2\right)\lambda + 2\left(a - 1\right)\right] = \left(a - 2\right)\sum_{\lambda}\lambda^{2} + 2\left(a - 1\right)\left(a - 2\right) \\[3mm]&= \left(a - 2\right)^{3} + 6\left(a - 1\right)\left(a - 2\right) \end{align} We get $$ \left(a - 2\right)\left[\left(a - 2\right)^{2} + 6a - 8\right] = 0 $$

$$ \color{#ff0000}{\large% a = 2\,,\qquad\qquad a = -1 \pm \sqrt{5\,}} $$

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    $\begingroup$ How does a $2\times2$ matrix have $3$ eigenvalues? $\endgroup$ – Marc van Leeuwen Oct 9 '13 at 22:00
  • $\begingroup$ @MarcvanLeeuwen That´s true. I have to think another way. Thanks. $\endgroup$ – Felix Marin Oct 9 '13 at 22:02

The matrix $A$ is clearly never a multiple of the identity, so its minimal polynomial is of degree$~2$, and equal to its characteristic polynomial, which is $P=X^2-(a-2)X+2-2a$. Now the polynomial $X^3-2X$ annihilates $A$ if and only if it is divisible by the minimal polynomial $P$. Euclidean division of $X^3-2X$ by $P$ gives quotient $X+a-2$ and remainder $(a^2-2a)X-(a-2)(2-2a)$. This remainder is clearly$~0$ if $a=2$ and for no other values of $a$, so $A^3=2A$ holds precisely for that value.

You can in fact avoid doing the Euclidean division by observing, since $X$ divides $X^3-2X$, that for $P$ to divide $X^3-2X$, unless $P$ itself has a factor $X$ (which happens for $a=1$ but divisibility then fails anyway), $P$ must divide (and in fact be equal to) $(X^3-2X)/X=X^2-2$; this happens for $a=2$.


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