# Determinant of matrix which satisfy certain conditions

Let $A$ be a $3\times3$ matrix over real numbers such that $A^{-1} =I-2A$. Find the determinant of $A$.

1. $\dfrac12$
2. $-\dfrac12$
3. $1$
4. $2$

Since the above equation results in a two-degree polynomial, $2x^2-x+1$, I got two eigenvalues as $\dfrac{1 + i\sqrt{7}}4$ and its conjugate. Its product give me $\dfrac12$. But what about the other eigenvalue? We should get a 3-degree polynomial to proceed further, right? I do not know whether this helps me.

Thankyou

• Hint:Trace is sum of Eigen values Apr 9, 2018 at 3:49
• I don't believe any such real matrix can exist. After all, the characteristic polynomial must be divisible by the minimal polynomial $m(x)=2x^2-x+1$ and can't have any other irreducible factors. Apr 9, 2018 at 3:52
• Me neither. I'm very suspicious about the existence of such a matrix! Apr 9, 2018 at 4:00
• I'm guessing it was supposed to be a $2\times 2$ matrix in the first place. Apr 9, 2018 at 4:18
• Yea... May be question was wrong.. Apr 9, 2018 at 4:52

No such matrix exists.

For, we have:

$A^{-1} = I - 2A; \tag 1$

multiplying by $A$:

$I = A - 2A^2; \tag 2$

re-arranging:

$2A^2 - A + I = 0. \tag 3$

We now make use of the following

Fact: If a matrix $A$ satisfies a polynomial $p(x)$, then every eigenvalue $\lambda$ of $A$ also satisfies $p(x)$, i.e. $p(\lambda) = 0$; futhermore, if the roots of $p(x)$ are distinct, and $p(x)$ is of minimal degree among polynomials satisfied by $A$, then every zero of $p(x)$ is an eigenvalue of $A$.

Proof of Fact: If

$p(A) = 0, \tag 4$

and

$A \vec v = \lambda \vec v, \tag 5$

then writing

$p(x) = \displaystyle \sum_0^{\deg p} p_i x^i, \tag 6$

we find

$0 = p(A) \vec v = \left (\displaystyle \sum_0^{\deg p} p_i A^i \right )\vec v = \left ( \displaystyle \sum_0^{\deg p} p_i \lambda^i \right ) \vec v = p(\lambda) \vec v, \tag 7$

for some $\vec v \ne 0$, since

$A^i \vec v = \lambda^i \vec v, \tag 8$

as may easily be seen by repeated application of $A$ to (5), as in

$A^2 \vec v = AA\vec v = A(\lambda \vec v) = \lambda A \vec v = \lambda^2 \vec v, \tag 9$

etc. Since $\vec v \ne 0$ we see that

$p(\lambda) = 0; \tag{10}$

now if the roots $\lambda_1, \lambda_2, \ldots, \lambda_{\deg p}$ of $p(x)$ are distinct, we may write

$p(x) = \displaystyle \prod_1^{\deg p} (x - \lambda_i); \tag{11}$

then

$\displaystyle \prod_1^{\deg p} (A - \lambda_i) = p(A) = 0; \tag{12}$

and since $p(x)$ is minimal for $A$, we must have

$\displaystyle \prod_2^{\deg p} (A - \lambda_i) \ne 0, \tag{13}$

so there is some vector $\vec w$ such that

$\vec v = \displaystyle \prod_2^{\deg p} (A - \lambda_i) \vec w \ne 0, \tag{14}$

thus from (12),

$(A - \lambda_1) \vec v = (A - \lambda_1) \displaystyle \prod_2^{\deg p} (A - \lambda_i) \vec w = 0, \tag{15}$

which shows that $\lambda_1$ is indeed an eigenvalue of $A$; clearly the same argument may be repeated whilst swapping the roles of $\lambda_1$ and $\lambda_i$, $2 \le i \le \deg p(x)$, which shows that every $\lambda_j$, $1 \le j \le \deg p(x)$ is an eigenvalue of $A$. End of Proof of Fact.

Now suppose that $2x^2 - x + 1$ were not of minimal degree for $A$; the for some $\alpha$, $\beta$ we must have

$\alpha A + \beta I = 0, \tag{16}$

whence

$A = -\dfrac{\beta}{\alpha}I = \gamma I; \tag{17}$

since $A$ is real, so $\gamma \in \Bbb R$; but this is impossible since we easily see that $\gamma$ must obey

$2\gamma^2 - \gamma + 1 = 0, \tag{18}$

whose roots are known to be

$\dfrac{1 \pm i \sqrt 7}{4}; \tag{19}$

thus $2x^2 - x + 1$ is minimal for $A$ and our Fact applies to allow us to conclude that

$\lambda = \dfrac{1 \pm i \sqrt 7}{4} \tag{20}$

are all the eigenvalues of $A$; but $A$ is of size $3$, and every real matrix of odd size has at least one real eigenvalue, since its characteristic polynomial is of odd degree. This contradicts our proven assertion that the totality of eigenvalues of $A$ is (17); thus no such $A$ can exist.

There are such $2 \times 2$ $A$, however; for example

$A = \begin{bmatrix} -1/4 & -1 \\ 7/16 & -1/4 \end{bmatrix} \tag{21}$

does the job. We have

$\det(A) = \dfrac{1}{2}, \tag{22}$

and the characteristic polynomial of $A$ is

$x^2 -\dfrac{1}{2} x + \dfrac{1}{2}. \tag{23}$