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... :(

$A^3-I=(A-I)(A^2+A+I)$. But $A^3-I=I$. So $A^2+A+I$ is the inverse of $A-I$. Now it's your turn. — André Nicolas, Feb 12 '16 at 15:56
Have you tried a search on this website before asking your question? One can easily find: this for the first question and this for the third question (looked at Dario's answer in particular!) — user37238, Feb 12 '16 at 15:56
If you know about minimal polynomials, it suffices to note that the minimal polynomial of $A$ divides $x^3 - 2$. So, $A - \lambda I$ will necessarily be invertible if $\lambda^3 \neq 2$. — Ben Grossmann, Feb 12 '16 at 15:59

score 8 · Answer 1 · answered Feb 12 '16 at 15:56

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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).

answered Feb 12 '16 at 15:56

Batman

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This is gem of an answer assuming that the person who asked this knows what eigenvalues are.. – Feb 12 '16 at 16:04
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How do we know that v exists? – sav Feb 12 '16 at 16:05
Eigenvalues was not taught in class.. but really appreciate the help! :D – noobnerd Feb 12 '16 at 16:09
@SharmaineSwee : You can appreciate by upvoting this answer... There are two arrows to the left of this answer.. One which points above one which points below.. If you like this, you click the one which points above if you do not like this you click the one which points below.. – Feb 12 '16 at 16:14
@sav - fair point; if we're working over C, we're fine. – Batman Feb 12 '16 at 16:39

score 2 · Answer 2 · answered Feb 12 '16 at 15:55

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

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

answered Feb 12 '16 at 15:55

Hello, please correct me if i'm wrong~ we can use $a^3 - B^3$ to factorize because identity matrix I can be treated as the value 1, which is the same as factorizing $A^3 - 1$ (can be done using the long division method) ?? Which derives the equation $(A-1)(A^2 + A + I)$ – noobnerd Feb 12 '16 at 16:02
Yes Yes.. You are correct.. – Feb 12 '16 at 16:03

egreg · Answer 3 · 2016-02-12T16:33:13.403

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}

Disintegrating By Parts · Answer 4 · 2016-02-13T02:34:28.733

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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}

edited Feb 13 '16 at 02:34

answered Feb 12 '16 at 21:52

Disintegrating By Parts

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

4 Answers4