# Minimal polynomial

• Oct 7th 2009, 02:57 PM
Astrid
Minimal polynomial
Find the minimal polynomial given $\displaystyle T:\mathbb R^2\to\mathbb R^2$ where $\displaystyle T(x,y)=(x+y,x-y).$

I found the associated matrix, which is $\displaystyle A=\left[\begin{array}{rr}1&1\\1&-1\end{array}\right].$

Now I need to compute the characteristic polynomial, and this is $\displaystyle -\left( 1-\lambda ^{2} \right)-1=\lambda ^{2}-2=\big(\lambda-\sqrt2\big)\big(\lambda+\sqrt2\big).$

Now suppose I take $\displaystyle P(\lambda)=\lambda-\sqrt2$ then I make $\displaystyle P(A),$ if this is zero, then that polynomial is the minimal one right? If is not zero, then I consider the characteristic polynomial, and that's actually the minimal polynomial because of simple application of Cayley - Hamilton Theorem.

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Another question I have, suppose we're computing the characteristic polynomial, when is it equal to the minimal one?
• Oct 7th 2009, 03:48 PM
tonio
Quote:

Originally Posted by Astrid
Find the minimal polynomial given $\displaystyle T:\mathbb R^2\to\mathbb R^2$ where $\displaystyle T(x,y)=(x+y,x-y).$

I found the associated matrix, which is $\displaystyle A=\left[\begin{array}{rr}1&1\\1&-1\end{array}\right].$

Now I need to compute the characteristic polynomial, and this is $\displaystyle -\left( 1-\lambda ^{2} \right)-1=\lambda ^{2}-2=\big(\lambda-\sqrt2\big)\big(\lambda+\sqrt2\big).$

Now suppose I take $\displaystyle P(\lambda)=\lambda-\sqrt2$ then I make $\displaystyle P(A),$ if this is zero, then that polynomial is the minimal one right? If is not zero, then I consider the characteristic polynomial, and that's actually the minimal polynomial because of simple application of Cayley - Hamilton Theorem.

== Yup, but in fact it's easier: since the minimal and the char. polynomials have the same irreducible factors, you know the min. pol. = the char. pol. in this case!

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Another question I have, suppose we're computing the characteristic polynomial, when is it equal to the minimal one?

There's not easy way to say.

Tonio
• Oct 7th 2009, 05:06 PM
Astrid
Ohhh, so if the characteristic polynomial has no multiplicities then it'd be equal to the minimal one?
• Oct 7th 2009, 08:34 PM
tonio
Quote:

Originally Posted by Astrid
Ohhh, so if the characteristic polynomial has no multiplicities then it'd be equal to the minimal one?

Well, if you meant to say that if the char. pol. is the product of different irreducible factors then it equals the min. pol. the answer is yes.

Tonio