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4

Let denote $\lambda$ the eigenvalue with geometric multiplicity $n-1$ and since the algebraic multiplicity of $\lambda$ is greater than or equal to the geometric multiplicity then there's two cases the algebraic multiplicity of $\lambda$ is $n$ hence the characteristic polynomial of $S$ is $$\chi_S(x)=(x-\lambda)^n$$ so $S-\lambda \operatorname{id}$ is ...


4

Take $$ A= \left( \begin{matrix} 1 \\ 1 \end{matrix} \right) \ \ \ \ \ \ \ \ \mbox{ and } \ \ \ \ \ \ \ \ B= ( \begin{matrix} 1 & 1 \end{matrix} ) $$


3

I think you're correct in your initial post- you need to know that the given eigenvalue has geometric multiplicity equal to its algebraic multiplicity (which is $n-1$). Then the matrix must have another eigenvalue, with multiplicity equal to 1, which implies that $S$ is diagonal. If you don't use this assumption, you can get something like: $$S = ...


2

If $$ A = \left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{matrix} \right) \quad B = \left( \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix} \right) $$ then $$ A B = \left( \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{matrix} \right) \quad B A = \left( ...


2

Using the definition of matrix multiplication, you can just check directly that $De_n$ ($e_n$ is the unit vector with $1$ at the $n$-th position) is $\kappa e_n$.


1

See if you can extend your operator to some $L^2$ space containing your functions. Maybe your operator is even self adjoint and still compact. Use the result for compact self adjoint operators on Hilbert spaces to find a basis of eigenvectors in that $L^2$ space. Show that in fact your eigenfunctions are already in your original space. Show that any ...


1

I think you are thinking about it backwards. Every Banach space is a vector space. Every vector space has a basis. If your eigenfunctions are: (1) linearly independent under the inner product (2) have the same dimension as the space. They must be an equivalent basis. However, neither (1) nor (2) are necessary conditions of eigenfunctions as far as I know. ...


1

Compact operators on Banach spaces can in general be upper-triangularized---generalizing the Jordan canonical form. This means that the space is Schauder-spanned by the generalized eigenvectors. So what you would need to show is that, given any non-zero $\lambda \in \sigma(L)$, the finite dimensional subspace $\mbox{ker}(L - \lambda)$ is spanned by ...


1

This isn't actually correct. One can have $A = \left[ \begin{array}{cc} 1 & -1 \\ 3 & -2 \end{array} \right]$, so that $A^3 = I$, but the characteristic polynomial of $A$ is $\lambda^2 + \lambda + 1$, whose roots are complex. What is guaranteed is that the eigenvalues will be roots of the polynomial $\lambda^3 - 1,$ one root of which is $\lambda = ...



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