what will be the determinant of the matrix A

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November 24th 2011, 01:42 AM
sorv1986

what will be the determinant of the matrix A

Let A be 3 by 3 matrix over real numbers satisfying A⁻¹= I -2A. Then what will be the

determinant of the matrix A?

thanks in advance. regards.(Nod)
November 24th 2011, 11:14 AM
FernandoRevilla

Re: what will be the determinant of the matrix A

Quote:

Originally Posted by sorv1986

Let A be 3 by 3 matrix over real numbers satisfying A⁻¹= I -2A. Then what will be the determinant of the matrix A?

Perhaps (perhaps) that is not the exact formulation of the problem. Could you review it?
November 24th 2011, 01:47 PM
HallsofIvy

Re: what will be the determinant of the matrix A

There are two possible answers. First, the determinant of a matrix is the same as the product of all of its eigenvalues. Second, if A satisfies that equation, its eigenvalues must also. You can manipulate the equation to a quadratic that has two solutions. Since A is 3 by 3, it has three eigenvalues. Since only two numbers sastisfy that equation, one of them must be a duplicate eigenvalue. The determinant depends upon which of them is the duplicate.
November 24th 2011, 04:48 PM
sorv1986

Re: what will be the determinant of the matrix A

Quote:

Originally Posted by FernandoRevilla

Perhaps (perhaps) that is not the exact formulation of the problem. Could you review it?

i m sry .but that was the only information given about the matriX A.
November 24th 2011, 05:34 PM
CaramelCardinal

Re: what will be the determinant of the matrix A

det(AB) = det(A)det(B)

Determinant is non-zero if it is invertible

det(A) = 1/(det(A^-1))
November 24th 2011, 11:36 PM
FernandoRevilla

Re: what will be the determinant of the matrix A

Quote:

Originally Posted by HallsofIvy

The determinant depends upon which of them is the duplicate.

But that is not possible. We have $A^{-1}=I-2A\Leftrightarrow 2A^2-A+I=0$ . That is , $p(\lambda)=2(\lambda-\lambda_1)(\lambda-\lambda_2)$ is an annihilator polynomial of $A$ being $\lambda_1=(1+\sqrt{7}i)/4$ and $\lambda_2=(1-\sqrt{7}i)/4$ . As $A\in\mathbb{R}^{3\times 3}$ , $A-\lambda_1I\neq 0$ and $A-\lambda_2I\neq 0$ . This means that the minimal polynomial of $A$ is just $(1/2)p(\lambda)$ and $\lambda_1,\lambda_2$ are eigenvalues of $A$ as you said. The third one $\lambda_3$ should be necessarily real and $A$ is diagonalizable in $\mathbb{C}$ . So, $A$ is similar (in $\mathbb{C}$ ) to $D=\textrm{diag}\;(\lambda_1,\lambda_2,\lambda_3)$ and as a consequence:

$2D^2-D+I\Rightarrow 2\lambda_i^2-\lambda_i+1=0\;(i=1,2,3)$

Again, as you said $\lambda_3=\lambda_1$ or $\lambda_3=\lambda_2$ which implies $\det A=\det D=\lambda_1\lambda_2\lambda_3=(1/2)\lambda_3\not\in\mathbb{R}$ (contradiction).
November 25th 2011, 02:40 AM
FernandoRevilla

Re: what will be the determinant of the matrix A

Quote:

Originally Posted by CaramelCardinal

Determinant is non-zero if it is invertible det(A) = 1/(det(A^-1))

Rigorously true.
November 25th 2011, 04:39 AM
sorv1986

Re: what will be the determinant of the matrix A

Quote:

Originally Posted by FernandoRevilla

But that is not possible. We have $A^{-1}=I-2A\Leftrightarrow 2A^2-A+I=0$ . That is , $p(\lambda)=2(\lambda-\lambda_1)(\lambda-\lambda_2)$ is an annihilator polynomial of $A$ being $\lambda_1=(1+\sqrt{7}i)/4$ and $\lambda_2=(1-\sqrt{7}i)/4$ . As $A\in\mathbb{R}^{3\times 3}$ , $A-\lambda_1I\neq 0$ and $A-\lambda_2I\neq 0$ . This means that the minimal polynomial of $A$ is just $(1/2)p(\lambda)$ and $\lambda_1,\lambda_2$ are eigenvalues of $A$ as you said. The third one $\lambda_3$ should be necessarily real and $A$ is diagonalizable in $\mathbb{C}$ . So, $A$ is similar (in $\mathbb{C}$ ) to $D=\textrm{diag}\;(\lambda_1,\lambda_2,\lambda_3)$ and as a consequence:

$2D^2-D+I\Rightarrow 2\lambda_i^2-\lambda_i+1=0\;(i=1,2,3)$

Again, as you said $\lambda_3=\lambda_1$ or $\lambda_3=\lambda_2$ which implies $\det A=\det D=\lambda_1\lambda_2\lambda_3=(1/2)\lambda_3\not\in\mathbb{R}$ (contradiction).

how [COLOR=rgb(0, 0, 0)]lambda3=lambda1 or lambda3= lambda2[/COLOR] is possible? lambda_3 must be real as the matrix is 3 by 3.Is that problem is wrong or something? Still i got no clue. 2A^2-A+I=0 cant be the minimal polynomial as every annihilating polynomial is multiple of the minimal polynomial. Please help me
November 25th 2011, 04:44 AM
sorv1986

Re: what will be the determinant of the matrix A

Quote:

Originally Posted by CaramelCardinal

det(AB) = det(A)det(B)

Determinant is non-zero if it is invertible

det(A) = 1/(det(A^-1))

how the det of A^-1 i.e. det (I-2A) would be calculated?
November 25th 2011, 06:00 AM
FernandoRevilla

Re: what will be the determinant of the matrix A

Quote:

Originally Posted by sorv1986

Is that problem is wrong or something?

Plainly: $A$ does not exist.

Quote:

Still i got no clue. 2A^2-A+I=0 cant be the minimal polynomial as every annihilating polynomial is multiple of the minimal polynomial. Please help me

The minimal polynomial $\mu (\lambda)$ of $A$ divides to any annihilating polynomial of $A$ so $\mu(\lambda)=\lambda-\lambda_1$ or $\mu(\lambda)=\lambda-\lambda_2$ or $\mu(\lambda)=(\lambda-\lambda_1)(\lambda-\lambda_2)$ . As $A-\lambda_1I\neq 0$ and $A-\lambda_2I\neq 0$ we conclude that $\mu(\lambda)=(\lambda-\lambda_1)(\lambda-\lambda_2)$ is the minimal polynomial of $A$ .