Suppose we have some self adjoint operator, given by either a matrix $$\begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}$$ or a function $f \longmapsto xf$ on $L^2[-1,1]$. Is there a quick way of decomposing these self adjoint operators into the difference of positive operators?
Decomposition of self-adjoint operator
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real-analysis
linear-algebra
functional-analysis
hilbert-spaces
spectral-theory
1 Answers
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Compose the operator with the projections on the positive and negative part of the spectrum and take the difference. For your matrix $A$, you need to find the eigenvalues which are the roots of
$$ \lambda^2 - \operatorname{tr}(A)\lambda + \det(A) = \lambda^2 - 2\lambda -3 = (\lambda + 1)(\lambda - 3).$$
An eigenvector associated to $\lambda = 3$ is $v_1 = (1,1)^T$ and an eigenvector associated to $\lambda = -1$ is $v_2 = (-1,1)^T$. The orthogonal projection onto $\operatorname{span} \{ v_1 \}$ is given by
$$ P_1 = \begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix} $$
and the orthogonal projection onto $\operatorname{span} \{ v_2 \} = \operatorname{span} \{ v_1 \}^{\perp}$ is given by
$$ P_2 = I - P = \begin{pmatrix} \frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} \end{pmatrix}.$$
Finally, we have
$$ A = A(P_1 + P_2) = AP_1 - (-AP_2) = \begin{pmatrix} \frac{3}{2} & \frac{3}{2} \\ \frac{3}{2} & \frac{3}{2} \end{pmatrix} - \begin{pmatrix} \frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2}\end{pmatrix}$$
where $AP_1$ is positive (with spectrum $\{ 0, 3 \}$) and $-AP_2$ is positive (with spectrum $\{ 0, 1 \}$).
For the operator $(M(f))(x) = x f(x)$ on $L^2([-1,1])$ the situation is easier because it is already "diagonal". Just define
$$ g_1(x) = \begin{cases} x & x \geq 0, \\ 0 & x < 0 \end{cases} $$
and let $P_1(f)(x) := g_1(x) f(x)$. Then $P_1$ is positive (with spectrum $[0,1])$ and $M = P_1 - (P_1 - M)$ where $P_1 - M$ is also positive.