If A is an nxn matrix with real eigenvalues show that A=B+C where B is symmetric and C is nilpotent. Plz help!?

Question

If A is an nxn matrix with real eigenvalues, show that A=B+C where B is symmetric and C is nilpotent? (Hint: Use the fact that an nxn upper triangular matrix U with zeros on the main diagonal satisfies Uⁿ=0)

I know that:
A has real eigenvalues, therefore P^TAP is upper triangular
B is symmetric, orthogonally diagonizable, and has an orthonormal set of eigenvectors.
and because C is nilpotent, Cⁿ=0


I'm not sure how to solve this or how to use the hint because I'm not sure whether A has zeros on the main diagonal. I was hoping that I could take A-B=C and have A-B be an upper triangular matrix with zeros on the main diagonal but I don't think that is necessarily the case. How can I solve this and where does the hint come in?

Anonymous · Accepted Answer

The "real eigenvalues"  assumption is not needed. It works for any matrix A.

Define B as a matrix which is the same as A on the diagonal
and everything below the diagonal, and its entries are filled out
above the diagonal so as to be symmetric.  Then define C = A -B, 
and note that this is an upper triangular matrix with zeros on the diagonal.

idiart · Answer

a + b + c = 0 --> c = -(a + b) c^4 = (-a million)^4 * (a + b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4 2a^4 + 2c^4 + 2b^4 = 4a^4 + 8a^3b + 12a^2b^2 + 8ab^3 + 4b^4 = 4[a^4 + 2a^3b + 3a^2b^2 + 2ab^3 + b^4] = 4[a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4 - 2a^3b - 4a^2b^2 - 2ab^3 + a^2b^2] = 4[(a + b)^4 - 2ab(a^2 + 2ab + b^2) + a^2b^2] = 4[(a + b)^4 - 2ab(a + b)^2 + (ab)^2] = 4{[(a + b)^2 - ab] * [(a + b)^2 - ab} = 4[(a + b)^2 - ab]^2 --> 2a^4 + 2b^4 + 2c^4 = {2[(a + b)^2 - ab]}^2 so enable n = 2[(a + b)^2 - ab]. replace: I only found out that this answer could be wiped sparkling up somewhat in view that a + b = -c. n = 2[c^2 - ab]

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If A is an nxn matrix with real eigenvalues show that A=B+C where B is symmetric and C is nilpotent. Plz help!?

2 Answers