Finding the Inverse of a Normal Matrix

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To show that the inverse of a normal matrix A is also normal, one must start with the definition of a normal matrix, which is AA* = A*A. The discussion suggests multiplying by A^(-1) to manipulate the equation, but initial attempts lead to confusion. A more effective approach involves manipulating the equation to reach the form (A^(-1))*(A^(-1)) = A^(-1)(A^(-1))*. The participants encourage experimentation with the properties of inverses and adjoints to find the solution. Ultimately, the key is to apply matrix operations correctly to demonstrate that A^(-1) retains the normality property.
chuy52506
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Homework Statement



Let A be a normal matrix in the complex field.

Homework Equations


Show that A-1 is normal.


The Attempt at a Solution


I know that a normal matrix is AA*=A*A
what would i multiply this to start?
 
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I would think multiplying by A^(-1) on the left would be a good start.
 
SO i would get A*=A-1A*A
I think i would just get stuck wouldn't I?
 
chuy52506 said:
SO i would get A*=A-1A*A
I think i would just get stuck wouldn't I?

If that's all the farther you've gotten then you are stuck already. Why don't you just try something. You want to get to (A^(-1))*A^(-1)=A^(-1)(A^(-1))*. Just mess around until you get there. Remember you can always take '*' or '^(-1)' of both sides in addition to multiplying both sides by stuff.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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