Solving for Inverse of a Complex Matrix with Cayley-Hamilton Theorem

[Delta what]

Homework Statement

I have matrix F=(1_1,1, -i_1,2, i_2,1, 1_2,2)
I calculated the characteristic polynomial to be x²-2x
The question is to find the inverse of F

Homework Equations

I am having a hard time trying to get the inverse out of this. I am used to dealing with real matrices so this may be the source of my error. I have done these with the usual case being that there was a constant term in the polynomial which one could get on the other side of the equality and get the inverse easily.

The Attempt at a Solution

I am ending up with something like X²=2x and then multiplying by x^-1 and ending up with x=2. What am I missing here?

Side note (new guy question) could someone point me into the direction of an explanation of how to do math notation of this forum such as matrices and similar?

 

[Ray Vickson]

Homework Statement

I have matrix F=(1_1,1, -i_1,2, i_2,1, 1_2,2)
I calculated the characteristic polynomial to be x²-2x
The question is to find the inverse of F

Homework Equations

I am having a hard time trying to get the inverse out of this. I am used to dealing with real matrices so this may be the source of my error. I have done these with the usual case being that there was a constant term in the polynomial which one could get on the other side of the equality and get the inverse easily.

The Attempt at a Solution

I am ending up with something like X²=2x and then multiplying by x^-1 and ending up with x=2. What am I missing here?

Side note (new guy question) could someone point me into the direction of an explanation of how to do math notation of this forum such as matrices and similar?

If the characteristic polynomial is $c(x)= x^2 - 2x$ the eigenvalues are $\lambda_1 = 0$ and $\lambda_2 = 2$. Since one of the eigenvalues is zero the matrix does not have an inverse.

 

[Delta what]

If the characteristic polynomial is $c(x)= x^2 - 2x$ the eigenvalues are $\lambda_1 = 0$ and $\lambda_2 = 2$. Since one of the eigenvalues is zero the matrix does not have an inverse.

Thank you for the help!

 

[Delta what]

Also on the side note I found that LaTeX is used to write in various notations.

 

[vela]

The Attempt at a Solution

I am ending up with something like X²=2x and then multiplying by x^-1 and ending up with x=2. What am I missing here?

Side note (new guy question) could someone point me into the direction of an explanation of how to do math notation of this forum such as matrices and similar?

Here's a LaTeX FAQ: https://www.physicsforums.com/help/latexhelp/

The Cayley-Hamilton theorem tells you $F^2 = 2F$. When you multiplied by $F^{-1}$, you assumed the inverse exists, and this assumption leads to the conclusion that $F=2I$, which clearly contradicts what you started with. Therefore, you would conclude that the assumption was wrong. $F^{-1}$ doesn't exist.

 

[Delta what]

Here's a LaTeX FAQ: https://www.physicsforums.com/help/latexhelp/

The Cayley-Hamilton theorem tells you $F^2 = 2F$. When you multiplied by $F^{-1}$, you assumed the inverse exists, and this assumption leads to the conclusion that $F=2I$, which clearly contradicts what you started with. Therefore, you would conclude that the assumption was wrong. $F^{-1}$ doesn't exist.

Thank you for your help vela!

 

[Ray Vickson]

Thank you for the help!

OK. An easy way to see it is to note that the product of the eigenvalues equals the determinant; see, eg.,
http://math.stackexchange.com/quest...-a-is-equal-to-the-product-of-its-eigenvalues
for a simple proof.

So, if 0 is an eigenvalue the determinant of the matrix equals 0 ==> not invertible.

 

[Delta what]

OK. An easy way to see it is to note that the product of the eigenvalues equals the determinant; see, eg.,
http://math.stackexchange.com/quest...-a-is-equal-to-the-product-of-its-eigenvalues
for a simple proof.

So, if 0 is an eigenvalue the determinant of the matrix equals 0 ==> not invertible.

Little bit of a different question but pertains to the same matrices. If i have a eigenvalue of zero can i still use that as a value in my diagonal matrix?

 

[vela]

Little bit of a different question but pertains to the same matrices. If i have a eigenvalue of zero can i still use that as a value in my diagonal matrix?

Yes.

 

[epenguin]

More elementary, if I understand your notation, then if you multiply the second row by i, and then add that to the first row you see that the matrix is singular.