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\begin{center}
{\Huge 189-251B: Algebra 2} \\ \vskip 0.1in
{\Huge Assignment 9} \\ \vskip 0.1in
{\Large Due: Wednesday, March 19 }
{(Extended to Wednesday March 26.)}
\end{center}
\sk\noindent
1. Prove or give a counterexample: the
product of any two self-adjoint operators on an
inner product space is self adjoint.
\sk\noindent
2. Let $T\in \cL(V)$ be an idempotent linear transformation,
(i.e., a transformation satisfying $T^2=T$) on a finite-dimensional
inner product space.
Show that $T$ is the orthogonal projection onto its image
if and only if $T$ is self-adjoint.
\sk\noindent
3. Show that a normal operator on an inner product space is self-adjoint if and only if all its eigenvalues are real.
\sk\noindent
4. Let $V$ be the real vector space of infinitely
differentiable
$\R$-valued functions $f:[0,1]\lra \R$ satisfying
$$f(0)=f(1) = f'(0) = f'(1) = \cdots = f^{(j)}(0) = f^{(j)}(1) = \cdots = 0.$$
Equip $V$ with the
standard inner product $\langle f,g\rangle = \int_0^1 f(t) g(t) dt.$
Let $T:V\lra V$ be the linear transformation given by $T(f)=f'$.
Show that $T$ is normal.
\sk\noindent
5. Suppose $V$ is a (real or complex) inner product
space, and that $T:V\lra V$ is self-adjoint. Suppose that there is a vector
$v$ with $||v||=1$, a scalar $\lambda\in F$, and a real
$\epsilon>0$ such that
$$ || T(v) - \lambda v|| < \epsilon.$$
Show that $T$ has an eigenvalue $\lambda'$ such that
$|\lambda-\lambda'|<\epsilon.$
Discuss the practical significance of this result.
\sk\noindent
6. Prove that if $T$ is a normal operator on a finite-dimensional
inner product space, then it has the same image as its adjoint.
\sk\noindent
7. Prove that there does not exist a self-adjoint operator $T:\R^3\lra \R^3$
(where $\R^3$ is equipped with the standard dot product) satisfying
$$ T(1,2,3)=(0,1,0), \quad T(2,5,7) = (1,1,1).$$
\sk\noindent
8. Let $T$ be a linear transformation on a finite dimensional real vector
space $V$. Show that $T$ is diagonalisable if and only if there exists an
inner product on $V$ relative to which $T$ is self-adjoint.
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