Spectral Theorem

Prove that a normal operator on a complex inner product space is self-adjoint if and only if all its eigenvalues are real.

Suppose $\mathbf{F} = \mathbb {C}$. Suppose $T \in \mathcal{L}(V)$ is normal and has only one eigenvalue. Prove that $T$ is a scalar multiple of the identity operator.

Suppose $\mathbf{F} = \mathbb {C}$ and $T \in \mathcal{L}(V)$ is normal. Prove that the set of eigenvalues of $T$ is contained in $\left\{ 0,1\right\}$ if and only if there is a subspace $U$ of $V$ such that $T = P_{U}$.

Prove that a normal operator on a complex inner product space is skew (meaning it equals the negative of its adjoint) if and only if all its eigenvalues are purely imaginary (meaning that they have real part equal to $0$).

Prove or give a counterexample: If $T \in \mathcal{L}(\mathbb {C}^{3})$ is a diagonalizable operator, then $T$ is normal (with respect to the usual inner product).

Suppose $V$ is a complex inner product space and $T \in \mathcal{L}(V)$ is a normal operator such that $T^{9} = T^{8}$. Prove that $T$ is self-adjoint and $T^{2} = T$.

Give an example of an operator $T$ on a complex vector space such that $T^{9} = T^{8}$ but $T^{2} \neq T$.

Suppose $\mathbf{F} = \mathbb {C}$ and $T \in \mathcal{L}(V)$. Prove that $T$ is normal if and only if every eigenvector of $T$ is also an eigenvector of $T^{*}$.

Suppose $\mathbf{F} = \mathbb {C}$ and $T \in \mathcal{L}(V)$. Prove that $T$ is normal if and only if there exists a polynomial $p \in \mathcal{P}(\mathbb {C})$ such that $T^{*} = p(T)$.

Suppose $V$ is a complex inner product space. Prove that every normal operator on $V$ has a square root.

Prove that every self-adjoint operator on $V$ has a cube root.

Suppose $V$ is a complex vector space and $T \in \mathcal{L}(V)$ is normal. Prove that if $S$ is an operator on $V$ that commutes with $T$, then $S$ commutes with $T^{*}$.

Without using the complex spectral theorem, use the version of Schur's theorem that applies to two commuting operators (take $\mathcal{E} = \left\{ T, T^{*}\right\}$ in Exercise 20 in Section 6B) to give a different proof that if $\mathbf{F} = \mathbb {C}$ and $T \in \mathcal{L}(V)$ is normal, then $T$ has a diagonal matrix with respect to some orthonormal basis of $V$.

Exercise 20 in Section 6B (Exercise 6B.20): if $\mathbf{F} = \mathbb {C}$, $V$ is finite-dimensional, and $\mathcal{E} \subseteq \mathcal{L}(V)$ is such that $S T = T S$ for all $S, T \in \mathcal{E}$, then there is an orthonormal basis of $V$ with respect to which every element of $\mathcal{E}$ has an upper-triangular matrix.

Suppose $\mathbf{F} = \mathbb {R}$ and $T \in \mathcal{L}(V)$. Prove that $T$ is self-adjoint if and only if all pairs of eigenvectors corresponding to distinct eigenvalues of $T$ are orthogonal and $V = E(\lambda_{1}, T) \oplus \cdots \oplus E(\lambda_{m}, T)$, where $\lambda_{1}, \ldots , \lambda_{m}$ denote the distinct eigenvalues of $T$.

Suppose $\mathbf{F} = \mathbb {C}$ and $T \in \mathcal{L}(V)$. Prove that $T$ is normal if and only if all pairs of eigenvectors corresponding to distinct eigenvalues of $T$ are orthogonal and $V = E(\lambda_{1}, T) \oplus \cdots \oplus E(\lambda_{m}, T)$, where $\lambda_{1}, \ldots , \lambda_{m}$ denote the distinct eigenvalues of $T$.

Suppose $\mathbf{F} = \mathbb {C}$ and $\mathcal{E} \subseteq \mathcal{L}(V)$. Prove that there is an orthonormal basis of $V$ with respect to which every element of $\mathcal{E}$ has a diagonal matrix if and only if $S$ and $T$ are commuting normal operators for all $S, T \in \mathcal{E}$.

Suppose $\mathbf{F} = \mathbb {R}$ and $\mathcal{E} \subseteq \mathcal{L}(V)$. Prove that there is an orthonormal basis of $V$ with respect to which every element of $\mathcal{E}$ has a diagonal matrix if and only if $S$ and $T$ are commuting self-adjoint operators for all $S, T \in \mathcal{E}$.

Give an example of a real inner product space $V$, an operator $T \in \mathcal{L}(V)$, and real numbers $b, c$ with $b^{2}<4 c$ such that

is not invertible.

Suppose $T \in \mathcal{L}(V)$ is self-adjoint and $U$ is a subspace of $V$ that is invariant under $T$.

Suppose $T \in \mathcal{L}(V)$ is normal and $U$ is a subspace of $V$ that is invariant under $T$.

Suppose that $T$ is a self-adjoint operator on a finite-dimensional inner product space and that $2$ and $3$ are the only eigenvalues of $T$. Prove that

Give an example of an operator $T \in \mathcal{L}(\mathbb {C}^{3})$ such that $2$ and $3$ are the only eigenvalues of $T$ and $T^{2}-5 T+6 I \neq 0$.

Suppose $T \in \mathcal{L}(V)$ is self-adjoint, $\lambda \in \mathbf{F}$, and $\epsilon >0$. Suppose there exists $v \in V$ such that $\left\| v\right\| = 1$ and

Prove that $T$ has an eigenvalue $\lambda^{\prime }$ such that $\left|\lambda -\lambda^{\prime }\right|<\epsilon$.

Suppose $U$ is a finite-dimensional vector space and $T \in \mathcal{L}(U)$.

Suppose that $T \in \mathcal{L}(V)$ and there is an orthonormal basis $e_{1}, \ldots , e_{n}$ of $V$ consisting of eigenvectors of $T$, with corresponding eigenvalues $\lambda_{1}, \ldots , \lambda_{n}$. Show that if $k \in \left\{ 1, \ldots , n\right\}$, then the pseudoinverse $T^{\dagger }$ satisfies the equation