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Let $(V,<,>)$ be a real inner product space and $T \in \mathcal{L}(V,V)$ a normal operator. Assume that the minimal polynomial of $T$ is a real irreducible quadratic. Prove that there is a real linear polynomial $f(x)$ such that $T^{*}=f(T)$

There is a Lemma in the book : Let $T$ be a normal operator on $V$. Then for all vectors $v \in V$, $|| T(v) ||=||T^{*}(v)||$. Also there is another defintion of being normal, $TT^*=T^*T$

$T^{*}$ denote to the adjoint of an linear transformation $T$ between inner product spaces.

I believe a real irreducible quadratic something like, $X^2+1=0$

I do not know how would I get started.

Ben Grossmann
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ADAM
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  • Do we know whether $V$ is finite dimensional? Also, do you know about the spectral theorem for normal matrices? – Ben Grossmann Apr 27 '16 at 15:20
  • @Omnomnomnom The Spectral Theorem for normal matrices only applies in complex inner product spaces. Though there is a nice block diagonal representation for real normal matrices. – Ken Duna Apr 27 '16 at 15:35

2 Answers2

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Hint: The minimal polynomial has the form $m(x) = (x - \mu)(x - \overline{\mu})$ for some $\mu \in \Bbb C$. It suffices to find a polynomial $p$ for which $p(\mu) = \overline{\mu}$ and $p(\overline{\mu}) = \mu$.

Ben Grossmann
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Here is a theorem from Axler's book "Linear Algebra Done Right" that should be helpful.

Let $V$ be a finite dimensional real inner product space, and $T$ a linear operator on $V$. Then $T$ is normal if and only if there is an orthonormal basis of $V$ with respect to which the matrix of $T$ has block diagonal form with each block either $1\times 1$ or $2\times 2$. The $2 \times 2$ blocks are of the form $$\bigl(\begin{smallmatrix} a & -b\\ b & a \end{smallmatrix}\bigr)$$ with $b > 0$.

Since the minimal polynomial of $T$ is an irreducible quadratic, it has no real eigenvalues. Thus there are no $1\times 1$ blocks in the above decomposition. Using this, you can see exactly the form of $T$ and $T^*$.

The eigenvalues of the matrix $$\bigl(\begin{smallmatrix} a & -b\\ b & a \end{smallmatrix}\bigr)$$

are $a \pm bi$. So if there were more than one type of $2\times 2$ matrix showing up in the decomposition, then we would have more than two eigenvalues, and so our minimal polynomial would not be a quadratic. Therefore our decomposition is just a bunch of copies of $$\bigl(\begin{smallmatrix} a & -b\\ b & a \end{smallmatrix}\bigr)$$ along the diagonal.

So lets find a polynomial that will take the transpose of this $2\times 2$. Since taking the transpose will make the upper-right entry $b$, I will first multiply the matrix by $-1$. This gives $$\bigl(\begin{smallmatrix} -a & b\\ -b & -a \end{smallmatrix}\bigr).$$ But now the main diagonal is bad. We can fix it by adding $2a$ to the main diagonal. We can accomplish this by adding the matrix $2a I$.

So $$\bigl(\begin{smallmatrix} a & -b\\ b & a \end{smallmatrix}\bigr)^* = \bigl(\begin{smallmatrix} a & -b\\ b & a \end{smallmatrix}\bigr)^{tr} = -\bigl(\begin{smallmatrix} a & -b\\ b & a \end{smallmatrix}\bigr) + 2aI.$$

Therefore the polynomial we want is $f(x) = -x + 2a$.

You can now argue that this polynomial works for $T$.

Ken Duna
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  • Yes , I do have this Theorem in my book too. In the same section with this question. But, I did not get how would I write $f(T)$ – ADAM Apr 27 '16 at 17:14
  • I know that $T=T^{tr}$, $tr$ denote to the transpose. So, would I need to find the characteristics polynomial and the minimal polynomial for the above decomposition then it better to equal the $T^{}$ , is that sounds right ? – ADAM Apr 27 '16 at 18:00
  • @ADAM You can show that there is only one type of 2x2 block in the decompsition (otherwise you would get more eigenvalues and hence not have a quadratic minimal polynomial). Once you have that fact, it suffices to find a polynomial that will work on a 2x2 matrix as above. This goes relatively simply. – Ken Duna Apr 27 '16 at 18:55