Question

Show that every square matrix can be uniquely expressed as the sum of a Hermitian and a skew-Hermitian matrix.

Answer 1

Answer:

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Step-by-step explanation:

Answer 2

ANSWER

$Consider F(x)=f(x)+f(−x)...(1)</p><p>F(−x)=f(−x)+f(x)=F(x)∴Even G(x)=f(x)−f(−x)...(2)</p><p>G(−x)=f(−x)−f(x)=−G(x)∴Odd.</p><p>Now F(x)+G(x)=2f(x)by(1)and(2) ∴f(x)= </p><p>2</p><p>1</p><p> </p><p> [F(x)+G(x)]...(3) where F(x) is even G(x) is odd function of x.</p><p>For uniqueness : Let,if possible, there exist F </p><p>1</p><p> </p><p> (x)(even)andG </p><p>1</p><p> </p><p> (x)(odd) function of x such that f(x)= </p><p>2</p><p>1</p><p> </p><p> [F </p><p>1</p><p> </p><p> (x)+G </p><p>1</p><p> </p><p> (x)]...(4)</p><p>Subtracting (3) and (4),we get 0= </p><p>2</p><p>1</p><p> </p><p> [{F(x)−F </p><p>1</p><p> </p><p> (x)}+{G(x)−G </p><p>1</p><p> </p><p> (x)}]</p><p>∴F(x)−F </p><p>1</p><p> </p><p> (x)=0andG(x)−G </p><p>1</p><p> </p><p> (x)=0</p><p>∴F </p><p>1</p><p> </p><p> (x)=F(x)andG </p><p>1</p><p> </p><p> (x)=G(x)$

Hence the expression is unique.

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