Show that every square matrix can be written uniquely in the form $A= A_1 + jA_2$ with $A_1, A_2$ self adjoint.
In this question we have to show for both real and couples matrix case.
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Possible duplicate of Any n x n matrix A can be written as a sum A = B + C where B is symmetric and C is skew-symmetric. – Babelfish Sep 12 '18 at 06:25
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Take $A_1=\frac {A+A^{*}} 2, A_2= \frac {A-A^{*}} {2j}$. Uniqueness: suppose $A=A_1+jA_2$ where $A_1,A_2$ are self adjoint. Then $\frac {A+A^{*}} 2=\frac {A_1+jA_2+A_1-jA_2} 2=A_1$. Simialrly, $\frac {A-A^{*}} {2j}=A_2$.
Kavi Rama Murthy
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thanks. I just wanted to know 1 more thing, that $A_1,A_2$ being self adjoint was of no use in the solution? – Christy Sep 12 '18 at 06:24
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@mathamity Self adointness is needed for uniqueness. See my revised answer. – Kavi Rama Murthy Sep 12 '18 at 06:29
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In Sheldon Axler book on Linear Algebra done right, it says that self adjoint means $T=T^*$. Then by this definition, $A_1=A$ and $A_2=0$? Sorry I am new to this subject. – Christy Sep 12 '18 at 06:43
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$A$ is not given to be self adjoint. You are getting $A_1=A,A_2=0$ by assuming that $A$ is self adjoint. – Kavi Rama Murthy Sep 12 '18 at 06:45