Some straightforward properties of eigenvalues and eigenvectors are as follows.

• If $\hat{A}\ket{\psi} = a\ket{\psi}$ then $\hat{A}^{n}\ket{\psi} = a^n \ket{\psi}$.

The case $n=1$ is true by definition, and we can prove the general case by induction. Suppose that $\hat{A}^{n}\ket{\psi} = a^n \ket{\psi}$. Then \[\hat{A}^{n+1}\ket{\psi} = \hat{A}\hat{A}^n \ket{\psi} = \hat{A}a^n\ket{\psi} = a^n \hat{A}\ket{\psi} = a^n a\ket{\psi} = a^{n+1} \ket{\psi}.\]

• If $\hat{A}\ket{\psi} = a\ket{\psi}$ then, for any function $f$, $f(\hat{A})\ket{\psi} = f(a)\ket{\psi}$.

This follows from the fact that functions are represented by their power series, the previous result, and linearity. Suppose the power series representation of $f$ is \[f(z) = \sum_{n=0}^{\infty} c_n z^n.\] Then \begin{align*} f(\hat{A})\ket{\psi} & = \left ( \sum_{n=0}^{\infty} c_n \hat{A}^n \right ) \ket{\psi} \\ & = \sum_{n=0}^{\infty} \left ( c_n \hat{A}^n \ket{\psi} \right ) \qquad\text{by linearity} \\ & = \sum_{n=0}^{\infty} \left ( c_n a^n \ket{\psi} \right )\qquad \text{by the previous result} \\ & = \left ( \sum_{n=0}^{\infty} c_n a^n \right ) \ket{\psi}\qquad \text{by linearity again} \\ & = f(a) \ket{\psi}. \end{align*}

In particular, a special case of this that we use often in quantum mechanics is if $\hat{A}\ket{\psi} = a\ket{\psi}$ then $e^{i\hat{A}} \ket{\psi} = e^{ia}\ket{\psi}$.

The set of eigenvectors of $\hat{A}$ sharing the same eigenvalue $a$ forms a subspace of the Hilbert space, i.e. if $\hat{A}\ket{\psi} = a\ket{\psi}$ and $\hat{A}\ket{\phi} = a\ket{\phi}$ then \[\hat{A}\left ( c \ket{\psi} + d\ket{\phi} \right ) = a\left ( c \ket{\psi} + d\ket{\phi} \right ),\] for any scalars $c$ and $d$.

This property is a straightforward consequence of linearity. A subspace formed by eigenvectors sharing the same eigenvalue of $\hat{A}$ is called an eigenspace of $\hat{A}$.

Note that, since eigenvectors with a given eigenvalue form a subspace, if $\ket{\psi}$ is an eigenvector of $\hat{A}$ with eigenvalue $a$ then so is $c\ket{\psi}$ for any scalar $c$. For this reason, we can always choose to work with normalized eigenvectors such that $\|\psi \| = 1$. If you have an unnormalized eigenvector you can just multiply it by whatever scalar is needed to make it normalized. From now on, we will do this and by eigenvector I will mean normalized eigenvector unless otherwise stated. Note that this does not completely eliminate the ambiguity because if $\ket{\psi}$ is a normalized eigenvector of $\hat{A}$ with eigenvalue $a$ then so is $e^{i\theta}\ket{\psi}$ for any phase angle $\theta$.

An eigenvalue $a$ of $\hat{A}$ is called nondegenerate if the corresponding eigenspace is one-dimensional, i.e. up to multiplication by a scalar there is a unique vector such that $\hat{A}\ket{\psi} = a\ket{\psi}$. If the dimension of the eigenspace is $\geq 2$ then the eigenvalue is called degenerate. The operator $\hat{A}$ itself is called nondegenerate if all of its eigenspaces are nondegenerate and is otherwise called degenerate. Nondegenerate operators are much easier to deal with, but unfortunately we do often have to deal with degenerate operators in quantum mechanics.

Since the eigenspace corresponding to an eigenvalue $a$ is a subspace, we can form the projection operator $\hat{P}_a$ onto that subspace. This is defined as follows. For all eigenvectors $\ket{\psi}$ with eigenvalue $a$, $\hat{P}_a\ket{\psi} = \ket{\psi}$ and for any vector $\ket{\phi}$ that is orthogonal to all of these vectors $\hat{P}_a\ket{\psi} = 0$.

More explicitly, if $a$ is nondegenerate and $\ket{\psi}$ a (normalized) eigenvector then $P_a = \proj{\psi}$.