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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.

Empty
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Mücahit Meral
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9 Answers9

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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 $A^{3}=2I$.

Lythia
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  • 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 ?? – Freelancer Mar 02 '16 at 17:38
  • 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 $$\left(\alpha A^2+\beta A +\gamma I\right)B=I$$ – Freelancer Mar 02 '16 at 17:46
  • 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 $QB=I$. – Lythia Mar 03 '16 at 04:36
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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!

David Zhang
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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.

ajotatxe
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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.

Dario
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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 $A^{-1}=\frac{1}{2}A^2$.

Ramiro
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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.

kensaii
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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.

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

Ben Grossmann
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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
$$B^{-1}=(A^2+3A+4I)/10$$

Mehran
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