Past exam of the mathematical methods I course of the University of Oxford / 2015 / 13

A fixed point obeys $(I-M)X=c$. Thus the necessary and sufficient existence condition isIf $1$ is not an eigenvalue of $M$, then $I-M$ is an invertible matrix and the unique constant solution is $\boxed{X=(I-M{)}^{-1}c}$. If $I-M$ is singular but the equation is consistent, all constant solutions are $X_0+\ker (I-M)$ for any particular solution $X_0$. For an existing fixed point, $x_{k+1}-X=M(x_k-X)$; the successive difference in the question likewise obeys $\delta x_{k+1}=M\delta x_k$.

Repeated substitution in the affine iteration givesTherefore the required matrix polynomials areThese formulas do not require an inverse. If $I-M$ is an invertible matrix, multiplying the finite geometric sum by $I-M$ gives $g(M)=(I-M^{k-1})(I-M{)}^{-1}$.

For a diagonalizable matrix, write $M=P\mathrm{diag}({\lambda }_j)P^{-1}$. Relative to an existing fixed point,Convergence to the same fixed point for every initial vector therefore holds exactly whenHere $\rho (M)$ is the spectral radius. Necessity follows by choosing an initial displacement in each eigenspace; sufficiency follows because every displayed scalar power tends to zero. The condition also makes $I-M$ an invertible matrix, so that fixed point exists uniquely for every $c$.

If instead different initial vectors may converge to different constant solutions, an eigenvalue $1$ is allowed: all other eigenvalues must have modulus less than one, and the component of $c$ in the eigenspace for $1$ must vanish. That component of the initial vector then remains unchanged. No other unit-modulus eigenvalue is compatible with convergence for arbitrary initial vectors.

Because $A$ is a real symmetric matrix, differentiating the quadratic form givesThus the stationary vectors are precisely $\ker (A-\alpha I)$, and at every one of themThe zero vector is always stationary. Nonzero stationary vectors exist exactly when $\alpha$ is an eigenvalue of $A$, and are exactly its nonzero eigenvectors with that eigenvalue. There is no norm constraint here; introducing one would change the stationary-value problem.

The second gradient is ${\nabla }_{v}b=[A+B-(\alpha +\beta )I]v$. Combining its vanishing with $(A-\alpha I)v=0$ givesThe zero vector always works. A nonzero common stationary vector exists exactly when $A,B$ have a common eigenvector with respective eigenvalues $\alpha ,\beta$; both stationary values are then zero. Commuting symmetric matrices have a common orthonormal eigenbasis, so commutation is sufficient to find such vectors for the paired eigenvalues in that basis. It does not guarantee an arbitrary prescribed pair.

Commutation is also not necessary for one common eigenvector. For example,are real symmetric matrices sharing $e_1$ with eigenvalues $1,\beta$, but $AB\ne BA$. This separates the question's common-vector condition from the stronger simultaneous diagonalization of the entire space.