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If $A$ is $40\times 40$ matrix such that $A^3=2I$ show that $B$ is invertible where $B=A^2-2A+2I$.

I tried to evaluate $B(A-I)$ , $B(A+I)$ , $B(A-2I)$ ... but I couldn't find anything.

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You can find the inverse directly by solving $$\left(\alpha A^2+\beta A +\gamma I\right)B=I.$$ If you do, you will find that $\alpha = 1/10$, $\beta = 3/10$, and $\gamma = 4/10$. Hence, $$B^{-1} = \tfrac{1}{10}A^2+\tfrac{3}{10}A+\tfrac{4}{10}I.$$ This can be easily checked by multiplying out and using the relation $A^{3}=2I$.

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  • $\begingroup$ This method is really good ...but it looks kind of circular to me as the question was asking to prove that $B$ is invertible ...so doesn't that mean we have to somehow find another matrix say $Q$ such that $BQ=I$ but you have just assumed in the first step itself that $B$ is an invertible matrix so that $B^{-1}B=I$ ...sure you have indeed said in the end we can cross-check it by multiplying out...but is this method correct mathematically speaking ?? $\endgroup$
    – Freelancer
    Mar 2 '16 at 17:38
  • $\begingroup$ Because the Question was to prove $B$ is invertible and you first used $BB^{-1}=I$ which is applicable when $B$ was invertible and then you say we have found the value of $B^{-1}$ and we have also checked it ...thus $B$ must be invertible ....but you had assumed this(that $B$ is invertible) right from the start !!! Otherwise one couldn't have said that $B^{-1}B=I$ or $$\left(\alpha A^2+\beta A +\gamma I\right)B=I$$ $\endgroup$
    – Freelancer
    Mar 2 '16 at 17:46
  • $\begingroup$ Hi Freelancer. There is no guarantee that my initial equation will have a solution. If it did not, then we would have wasted some time. But if it does have a solution, then the matrix that is found (call it $Q$) is the inverse because it has the required property that $QB=I$. $\endgroup$
    – Lythia
    Mar 3 '16 at 4:36
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Here's something I call the "miracle method" for this type of problem. Suspend your disbelief for a moment and suppose $A$ and $B$ were scalars, not matrices. Then, we would simply be looking for $$ \frac{1}{B} = \frac{1}{A^2 - 2A - 2} = \frac{1}{2}+\frac{A}{2}+\frac{A^2}{4}-\frac{A^4}{8}-\frac{A^5}{8} + \cdots$$ where in the power series expansion, the coefficient of $A^n$ is $$ c_n = \frac{1+i}{2^{n+2}} \left((1-i)^n-i (1+i)^n\right). $$ But we know that $A^3 = 2$, so this becomes $$ \frac{1}{2}+\frac{A}{2}+\frac{A^2}{4}-\frac{A^4}{8}-\frac{A^5}{8} + \cdots = \frac{1}{2}+\frac{A}{2}+\frac{A^2}{4}-\frac{A}{4}-\frac{A^2}{4} + \cdots $$ and by summing the resulting coefficients on $1$, $A$, and $A^2$, we find that $$ \frac{1}{B} = \frac{2}{5} + \frac{3}{10}A + \frac{1}{10}A^2. $$ Now, what we've just done should be total nonsense if $A$ and $B$ are really matrices, not scalars. But try setting $B^{-1} = \frac{2}{5}I + \frac{3}{10}A + \frac{1}{10}A^2$, and you'll find that, miraculously, this answer works!

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  • $\begingroup$ What the hell hahaha, why does this work? I'm astounded! $\endgroup$ Aug 16 '15 at 18:19
  • $\begingroup$ @YoTengoUnLCD To be honest, I'm really not sure! I discovered this technique some time ago by accident, and it seems to produce meaningful answers for problems of this type. I've asked another question here if anyone has ideas as to why this works. $\endgroup$ Aug 16 '15 at 23:09
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The eigenvalues of $A$ are the cubic roots of $2$. Then $$B=(A-(1+i)I)(A-(1-i)I)$$ is a regular matrix.

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You have $$B=A^3+A^2-2A=A(A-I)(A+2I)\ .$$ $A$ is invertible, and since $$A^3-I=I\qquad\Rightarrow\qquad(A-I)(A^2+A+I)=I$$ and $$A^3+8I=10I\qquad\Rightarrow\qquad(A+2I)(A^2-2A+4I)=10I$$ also $(A-I)$ and $(A+2I)$ are invertible.

Now, since $B$ is the product of invertible matrices, it is invertible.

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  • $\begingroup$ I think it is correct: $B=A^2-2A+2I$ and $2I=A^3$ $\endgroup$
    – Dario
    Jul 29 '15 at 21:41
  • $\begingroup$ Ohh......!!!!!. $\endgroup$
    – Empty
    Jul 30 '15 at 3:21
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One of the easiest ways to prove $B$ is invertible is the following;
$$B=A^2-2A+2I$$ Using that $A^3=2I$, we can write $$B=A^2-2A+2I= A^2-2A + A^3 = A(A^2+A-2I)=A(A+2I)(A-I)$$ Now let us prove $A$, $A+2I$ and $A-I$ are invertible.

  1. Supose there is $v$ such that $Av=0$, then we have $0=A^20=A^3v=2Iv=2v$. So we have $v=0$. So $A$ is invertible.

  2. Supose there is $v$ such that $(A+2I)v=0$, then we have $Av=-2v$ and we have $A^3v=-8v$. So we get $-8v=A^3v=2Iv=2v$. So we have $v=0$. So $(A+2I)$ is invertible.

  3. Supose there is $v$ such that $(A-I)v=0$, then we have $Av=v$ and we have $A^3v=v$. So we get $v=A^3v=2Iv=2v$. So we have $v=0$. So $(A-I)$ is invertible.

Since B is the product of three invertible matrices, B itself is invertible.

Remark: For item 1 above, we also simply present the inverse of $A$. From $A^3=2I$, we have $A^{-1}=\frac{1}{2}A^2$.

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  • $\begingroup$ Nice. Best solution. $\endgroup$ Jul 29 '15 at 17:32
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A few manipulations:

$$B=A^2-2A+2I$$ $$A^3 = 2I$$

then,

$$AB = A^3 -2A^2+2A \\= 2I -2A^2+2A \\= -2(A^2-2A+2I)-2A+6I \\= -2B -2A+6I$$

Now, the fact that $A^3 = 2I$ means $A$ is invertible, because $\det(A) \neq 0$

Thus,

$$B = -2A^{-1}B-2I+6A^{-1} $$

Take $B$ to one side:

$$B(I+2A^{-1}) = -2I+6A^{-1}$$ Multiply both side with $A$: $$B(A+2I) = -2A+6I$$

I suppose those quantities will be similar.

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$B=(A-I)^2+I$. Now claim that $1\pm i$ are NOT eigen values of $A$. Otherwise , $(1\pm i)^3$ are eigen values of $A^3$ , which is NOT possible. So , $\pm i$ are NOT eigen values of $A-I$. Then , $(\pm i)^2=-1$ is NOT an eigen value of $(A-I)^2$ .

Then $0$ is NOT an eigen value of $(A-I)^2+I$ and consequently $B$ is invertible.

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Hint: $B$ is invertible if and only if $0$ is not and eigenvalue of $B$.

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$B=A^2-2A+2I$. Now multiplying the sides respectively by $A$ and $A^2$, and using the assumption $A^3=2I$ yields:
$$AB=BA=-2A^2+2A+2I$$
$$A^2B=BA^2=2A^2+2A-4I$$
Therefore
$$AB+A^2B=4A-2I$$
$$AB+2B=-2A+6I$$
By straightforward algebraic manipulation, it comes that:
$$A^2B+3AB+4B=BA^2+3BA+4B=10I$$
$$(A^2+3A+4I)B=B(A^2+3A+4I)=10I$$
Therefore
$$B^{-1}=(A^2+3A+4I)/10$$

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