Deciding which statements about matrix A is true where $A^3+A^2-3A+I=0$

Question

A is $2 \times 2$ matrix. $$A^3+A^2-3A+I=0$$ Decide which statements is true.

$\quad$ A) 1 is eigenvalue of A.
$\quad$ B) Det(A) is 1.
$\quad$ C) $A^{-1}$ exists.
$\quad$ D) If B is inverse of A, $B^3-3B^2+B+I=0$.

Choices are {A,B}, {A,C}, {C,D}, and {B,C,D}.

What I've did so far is,

A : If I multiply eigenvector v$_{(2 \times 1)}$ to given equation, It'll satisfy the equation if eigenvalue of A is 1.
$\quad$ But I'm not sure if it's enough to say statement A is true.

B: (?)

C: $A(-A^2-A+3)=I$, so it's true.

D: $B=-A^2-A+3. $
$\quad B^3-3B^2+B+I=0=B(B^2-3B+1)+I\;$ , If I substitute B then
$\quad =(-A^2-A+3)(A^4+2A^3-2A^2-3A+1)=-A^3-A^2+3A=I$.
$\quad$ So it's true.

Have I done correctly?
and How should I go for statements A and B?

What are the entries of $A$ : integer/real/ complex numbers? — Sarvesh Ravichandran Iyer, Sep 05 '18 at 05:30
Set $p(x) = x^3+x^2-3x+1$. Then $p(A) = 0$, which means that the minimal polynomial of $A$ divides $p$. What can you say about the eigenvalues of $A$? — amsmath, Sep 05 '18 at 05:31
I cannot follow what you did for D, plus the calculation is obviously wrong. — amsmath, Sep 05 '18 at 05:47
Actually, $D$ is very easy. The point is, that $A^{-1} = B$ ($A$ is invertible, you showed this), now you know $A^3 + A^2 - 3A + I = 0$, multiplying by $B^3$ gives $(AB)^3 + (AB)^2 B - 3(AB)B^2 + B^3 = 0$, and $AB = I$ implies $I + B - 3B^2 + B^3 = 0$. — Sarvesh Ravichandran Iyer, Sep 05 '18 at 06:04
@астон вілла олоф мэллбэрг It's not given. I believe it's real. — nik, Sep 05 '18 at 06:06
@amsmath I didn't think about characteristic polynomial. I get that it includes minimal polynomial. then how can I know which of the three roots is eigenvalue of A? — nik, Sep 05 '18 at 06:10

Theo Bendit · Accepted Answer · 2018-09-05T06:24:57.947

Your answers for (c) and (d) look good (though I admit I haven't crunched the algebra for your answer to part (d)).

For parts (a) and (b), you have a polynomial $p(x)$ such that $p(A) = 0$. The eigenvalues of $A$ must be roots of $p(x)$, but (importantly) not vice-versa. There are more roots than dimensions, meaning that at least one root is not an eigenvalue.

That said, we now have a list of possible eigenvalues that we can have:

$\lambda_1 = \lambda_2 = 1$
$\lambda_1 = 1$, $\lambda_2 = -1 + \sqrt{2}$
$\lambda_1 = 1$, $\lambda_2 = -1 - \sqrt{2}$
$\lambda_1 = -1 + \sqrt{2}$, $\lambda_2 = -1 + \sqrt{2}$
$\lambda_1 = -1 + \sqrt{2}$, $\lambda_2 = -1 - \sqrt{2}$
$\lambda_1 = -1 - \sqrt{2}$, $\lambda_2 = -1 - \sqrt{2}$

All 6 possibilities above are possible. In fact, put each of these pair of numbers into a diagonal matrix, and you'll have a matrix $A$ that satisfies the given polynomial equation, with the given pair of eigenvalues. Note that, in cases 4, 5, 6, $1$ is not an eigenvalue, so (a) is false. Also, in all cases except 1, the determinant is not $1$, so (b) is false.

@nik I told you 50 minutes ago: The minimal polynomial divides $p$. And you hopefully know that the roots of the minimal polynomial are the eigenvalues of $A$. So you could have easily got this answer by yourself. — amsmath, Sep 05 '18 at 06:25
I guess so. I didn't know which of roots can be eigenvalues. I see they are all possible. — nik, Sep 05 '18 at 06:30

score 1 · Answer 2 · answered Sep 05 '18 at 06:12

Explanations.

A. By assumption, $$ (\boldsymbol A - \boldsymbol I)(\boldsymbol A^2 + \boldsymbol A - \boldsymbol I) = \boldsymbol O = (\boldsymbol A -\boldsymbol I)(\boldsymbol A - (-1 + \sqrt 2)\boldsymbol I) (\boldsymbol A - (-1 - \sqrt 2)\boldsymbol I), $$ i.e. $p(x)=(x-1)(x + 1-\sqrt 2)(x + 1 + \sqrt 2)$ nullifies $\boldsymbol A$. Assume we are considering real matrices, then the minimal polynomial $m(x)$ divides $p(x)$. Clearly $m(x)$ could be $(x + 1 + \sqrt 2)$. Example $\boldsymbol A = (-1-\sqrt 2) \boldsymbol I_2$ satisfies the equation, but $\boldsymbol A$ has no eigenvalues equals $1$. Hence $\color{red}{\mathsf {False}}$.
B. Use the same counter example above, we know that $\det(\boldsymbol A) = (1 + \sqrt 2)^2 \neq 1$, so $\color{red}{\mathsf {False}}$.
C. You are right.
D. Seems fine.

I believe it says the same thing as the answer of Theo Bendit. Thanks for your help! — nik, Sep 05 '18 at 06:25

Sarvesh Ravichandran Iyer · Answer 3 · 2018-09-05T06:28:13.847

1

Let $\lambda$($\neq 0$) be a real root of $p(x) = x^3 + x^2 - 3x + 1 = 0$. Then, note that $\lambda I$ also satisfies the equation $p(x) = 0$,since: $$ p(\lambda I) = (\lambda^3 + \lambda^2 - 3 \lambda + 1) I = 0 $$

by how we chose $\lambda$. Note that the determinant of $\lambda I$ is $\lambda^2$, and all its eigenvalues are $\lambda$.

It is easy to see that the real roots of $p(x)$ are $1,-\sqrt 2 \pm 1$. Of these, two do not satisfy $\lambda^2 = 1$. Consequently, $\lambda I$, for $\lambda = -\sqrt 2 \pm 1$, serve as counterexamples of $a$ and $b$.

Your proof for $c$ is correct.

For $d$, if $B = A^{-1}$, simply multiply the equation $A^3 + A^2 -3A + I = 0$ by $B^3$ and use $AB = I$ to see what you get.

edited Sep 05 '18 at 06:28

answered Sep 05 '18 at 06:15

Sarvesh Ravichandran Iyer

76,105

Why is determinant of $\lambda I$ equal to ${\lambda}^3$? – nik Sep 05 '18 at 06:27
Correction needed. Correction made. – Sarvesh Ravichandran Iyer Sep 05 '18 at 06:27

score 0 · Answer 4 · answered Sep 05 '18 at 06:43

$$ A^3+A^2-3A+I = 0 \\ (A-I)(A^2+2A-I) = 0 \\ (A-I)((A+I)^2-2I) = 0 \\ (A-I)(A+(1-\sqrt{2})I)(A+(1+\sqrt{2})I)=0. $$

So, $A$ has possible eigenvalues $1,-1-\sqrt{2},-1+\sqrt{2}$. It is possible that $$ A = I, \mbox{ or } A=(-1+\sqrt{2})I, \mbox{ or } A=(-1-\sqrt{2})I. $$

(A) is not necessarily true because none of the above can be ruled out. So $1$ does not have to be an eigenvalue of $A$, even thought it could be.
(B) is not necessarily true. The determinant of $A$ could be $1$, but it does not have to be if, for example $A=(-I+\sqrt{2})I$.
(C) must be true because $$ A(-A^2-A+3I)=I \\ (-A^2-A+3I)A=I. $$
(D) is true because, if $B$ is the inverse of $A$, then

$$ (B^{-1})^3+(B^{-1})^2-3B^{-1}+I = 0 \\ I+B-3B^2+B^3 = 0. $$

Therfore, (C) and (D) must be true, but (A) and (B) may or may not be true.

Deciding which statements about matrix A is true where $A^3+A^2-3A+I=0$

4 Answers4