Question

Show that every square matrix can be uniquely expressed as the sum of hermitian and skew hermitian maatrix

Answer 1

Let, A be a given square matrix.

We can write A as-

A=1/2(A+A^θ)+1/2(A−A^θ)= say,

P+Q where P=1/2(A+A^θ) and Q= 1/2(A−A^θ)

Now, Pθ =[1/2(A+A^θ)]^θ= 1/2(A^θ+(A^θ)^θ)=1/2(A^θ+A)…(∵(A^θ)^θ)

P^θ=1/2(A^θ+A)=1/2(A+A^θ)=P

Hence, P is a Hermitian Matrix.…(By Definition)

Also, Q^θ=[1/2(A−A^θ)]^θ=1/2(A^θ−(A^θ)^θ)=1/2(A^θ−A)…(∵(Aθ)θ)

Q^θ= −1/2(A−A^θ) = −Q

Hence,Q is a skew−Hermitian Matrix.…(ByDefinition)

For uniqueness:

Let, A=R+S, where R is a Hermitian and S is a skew-Hermitian matrix, be another representation of A.

Now, A^θ= (R+S)^θ= R^θ+S^θ= R−S…(∵R^θ=R & S^θ=−S…By Definition)

∴ 1/2(A+A^θ)= 1/2[(R+S)+(R−S)]=R.

But 1/2(A+A^θ)=P.

∴R=P

Also, 1/2(A−A^θ) = 1/2[(R+S)−(R−S)]=S.

But 1/2(A−A^θ)=Q.

∴S=Q

Hence, every square matrix can be uniquely expressed as a sum of Hermitian and skew-Hermitian Matrix.

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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