- #1

joshmccraney

Gold Member

- 2,185

- 124

hi pf!

I am reading a text and am stuck at a part. this is what is being said:

If ##g## is a graph we have ##L(g) + L(\bar{g}) = nI - J## where ##J## is the matrix of ones. Let ##f^1,...f^n## be an orthogonal system of eigenvectors to ##L(g) : f^1 = \mathbb{1}## and ##L(g)f^i = \lambda_i f^i##. Evidently ##L(g) + L(\bar{g}) = nI - J## implies ##L(\bar{g}) f^1 = 0##.

Now for ##2 \leq i \leq n## we have ##L(\bar{g}) f^i = (n-\lambda_i)f^i##

The text continues by saying ##\lambda_1(\bar{g}) = 0##

Thus we have that ##\lambda_i(\bar{g}) = n - \lambda_{n-i+2}(g) : i \geq 2##. This is the part I don't understand. If eigenvalues of ##L(g)## are eigenvalues to ##L(\bar{g})## then I see that ##\lambda_i (\bar{g})f^i = (n-\lambda_i)f^i \implies \lambda_i (\bar{g}) = n-\lambda_i(g)## but the index is now wrong...please help me out!

Thanks so much!

Josh

I am reading a text and am stuck at a part. this is what is being said:

If ##g## is a graph we have ##L(g) + L(\bar{g}) = nI - J## where ##J## is the matrix of ones. Let ##f^1,...f^n## be an orthogonal system of eigenvectors to ##L(g) : f^1 = \mathbb{1}## and ##L(g)f^i = \lambda_i f^i##. Evidently ##L(g) + L(\bar{g}) = nI - J## implies ##L(\bar{g}) f^1 = 0##.

Now for ##2 \leq i \leq n## we have ##L(\bar{g}) f^i = (n-\lambda_i)f^i##

The text continues by saying ##\lambda_1(\bar{g}) = 0##

Thus we have that ##\lambda_i(\bar{g}) = n - \lambda_{n-i+2}(g) : i \geq 2##. This is the part I don't understand. If eigenvalues of ##L(g)## are eigenvalues to ##L(\bar{g})## then I see that ##\lambda_i (\bar{g})f^i = (n-\lambda_i)f^i \implies \lambda_i (\bar{g}) = n-\lambda_i(g)## but the index is now wrong...please help me out!

Thanks so much!

Josh

Last edited: