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Let $f$ be a polynomial with real coefficients and $A$ a symmetric matrix of $n\times n$ with elements in $\mathbb{R}$. Prove that $f(A)$ is symmetric. Suppose $A$ is hermitian and that $f$ has complex coefficients.

Is hermitian the matrix of $f(A)$?.

I did this:

Let $f(x)=a_0x^n+a_1x^{n-1}+.......+a_n$, if A is symmetric then cA is symmetric where c $\in{}\mathbb{R}$ ,A$^n$ is symmetric too where n is an integer finally $f(A)=a_0A^n+a_1A^{n-1}+.......+a_n$ is symmetric

If A is Hermitian and n is an integer $A^n$ is symmetric but by hypothesis n can be a complex number then $A^n$ can not be hermitian.

Any suggestion?

Thanks for everything

Have a nice day

Pedro
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user63192
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2 Answers2

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For the first question, observe that positive integer powers of a symmetric matrix are symmetric: $(A^n)^t=(A\cdots A)^t=(A^t\cdots A^t)=(A\cdots A)=A^n$. Since a scalar times a symmetric matrix is clearly symmetric, then it follows that for a polynomial $f$ with real coefficients, $f(A)$ will be symmetric.

By mimicking the same proof you can see that if $A$ is Hermitian and $f$ has real coefficients the same proof will apply.

Added: If you allow complex coefficients you will have a problem since if $z$ is a complex number with nonzero imaginary part and $A$ is Hermitian, then $zA$ will not be Hermitian, since $(zA)^*=z^*A^*=z^*A$. Remember that the diagonal of a Hermitian matrix must be real-valued.

Lepidopterist
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  • Thanks a lot, but I have one question if A is hermitian f(A) isn't hermitian true? – user63192 Mar 16 '13 at 22:23
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    @user63192: If $A$ is self-adjoint (Hermitian) and $f$ is a complex polynomial, then $f(A)$ will be a normal operator with $f(A)^\ast = \bar{f}(A)$, e.g., $(iA)^\ast = -iA$. If $f$ is actually a real polynomial, however, then you're in business, for $\bar{f} = f$. – Branimir Ćaćić Mar 16 '13 at 22:33
  • @BranimirĆaćić thanks for you help :D Have a nice day – user63192 Mar 16 '13 at 22:37
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  • $f(x)\in\mathbb R[x]$ and $A\in Mat_{n\times n}(\mathbb R)$ is symmetric $\implies f(A)$ is symmetric.

Let $f(x)=a_0+a_1x+a_2x^2+...+a_nx^n\in\mathbb R[x].$ Then $f(A)=a_0I+a_1A+a_2A^2+...+a_nA^n.$

$(f(A))^t=(a_0I+a_1A+a_2A^2+...+a_nA^n)^t$$=a_0(I)^t+a_1(A)^t+a_2(A^2)^t+...+a_n(A^n)^t$$=a_0I+a_1A+a_2A^2+...+a_nA^n=f(A).$ (Recall: For symmetric matrix $A,(A^n)^t=(A.A...A)^t=(A^tA^t...A^t)=A^n~\forall~n\in\mathbb Z^+$)

  • Let $f(x)\in\mathbb C[x]$ and $A\in Mat_{n\times n}(\mathbb C)$ is hermitian. Is $f(A)$ hermitian?

Let $f(x)=a_0+a_1x+a_2x^2+...+a_nx^n\in\mathbb C[x].$ Then $f(A)=a_0I+a_1A+a_2A^2+...+a_nA^n.$

${\overline{f(A)}}^{~t}={\overline{a_0I+a_1A+a_2A^2+...+a_nA^n}}^{~t}$$=\overline{a_0I}^{~t}+\overline{a_1A}^{~t}+\overline{a_2A^2}^{~t}+...+\overline{a_nA^n}^{~t}$$=\overline{a_0}I+\overline{a_1}A+\overline{a_2}A^2+...+\overline{a_n}A^n$ (Since for a hermitian matrix $A,$ $\overline{A^n}^{~t}=\overline{A.A...A}^{~t}$$=\overline{A}^{~t}\overline{A}^{~t}...\overline{A}^{~t}=A^n$)

It's here where the problem begins since the coefficients turn into their corresponding conjugate. Keeping it in mind, let's try to construct a counterexample.

BTW if $f(x)\in\mathbb R[x],$ then hermitianness won't get affected by the transformation under $f$.

  • Counterexample: Take $A=\begin{pmatrix}1 & 2 & 3i\\2&0&4\\-3i&4&2\end{pmatrix}$ and $f(x)=ix.$
Sugata Adhya
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