Let A be a square matrix such that $A^3 = 2I$

Question

Let $A$ be a square matrix such that $A^3 = 2I$

i) Prove that $A - I$ is invertible and find its inverse

ii) Prove that $A + 2I$ is invertible and find its inverse

iii) Using (i) and (ii) or otherwise, prove that $A^2 - 2A + 2I$ is invertible and find its inverse as a polynomial in $A$

$I$ refers to identity matrix.

Am already stucked at part i). Was going along the line of showing that $(A-I)([...]) = I$ by manipulating the equation to $A^3 - I = I$ and I got stuck... :(

Answer 1

We have $A^3-2I=0$ which is same as $A^3-I=I$

Recall the formula $a^3-b^3=(a-b)(***)$

Answer 2

If $A^3 = 2 I$, then you know that if $v$ is an eigenvector of $A$ with eigenvalue $\lambda$, then $A^3 v = A (A (A v) = \lambda^3 v$. But, since $A^3 = 2 I$, you have $\lambda^3 v = 2 v$ or $\lambda^3 = 2$.

There are 3 cube roots of $2$, none of which is $1$ or $-2$. Thus, $A-I$ and $A+2I$ are invertible (they have trivial nullspaces, since 1 and -2 are not eigenvalues).

Answer 3

For finding the inverse of $B_a=A-aI$ (provided it exists), write $A=B_a+aI$; then $$ (B_a+aI)^3=B_a^3+3aB_a^2+3a^2B_a+a^3I $$ so you have $$ B_a^3+3aB_a^2+3a^2B_a=(2-a^3)I $$ and multiplying by $B_a^{-1}$ gives $$ B_a^{-1}=\frac{1}{2-a^3}(B_a^2+3aB_a+3a^2I)= \frac{1}{2-a^3}(A^2+aA+a^2I) $$ You see that the inverse exists provided $a^3\ne2$.

Can you do the last part?

Hint: \begin{align} A^2-2A+2I&=A^2-2A+A^3\\ &=A(A^2+A-2I)\\ &=A(A-I)(A+2I) \end{align}

Answer 4

For parts (i) and (ii): \begin{align} (A-I)^{-1} & = A^2+A+I \\ (A+2 I)^{-1} & = \frac{1}{10}(A^2-2A+4). \end{align} For (iii): $$ A^2-2A+2I=A^2-2A+A^3=A(A+2I)(A-I) $$ The inverse of $A$ is $\frac{1}{2}A^2$. So, \begin{align} (A^2-2A+2I)^{-1}&=(A-I)^{-1}(A+2I)^{-1}A^{-1}\\ &=\frac{1}{20}(A^2+A+I)(A^2-2A+4)A^2 \end{align}