## Do commuting operators share eigenvectors?

Commuting matrices do not necessarily share all eigenvector, but generally do share a common eigenvector. Let A,B∈Cn×n such that AB=BA. There is always a nonzero subspace of Cn which is both A-invariant and B-invariant (namely Cn itself).

## How do you know if two operators commute?

If two operators commute, then they can have the same set of eigenfunctions. By definition, two operators ˆA and ˆBcommute if the effect of applying ˆA then ˆB is the same as applying ˆB then ˆA, i.e….Consider the following operators:

- ˆA=ddx.
- ˆE=x2.
- ˆB=hx.
- ˆC{f(x)}=f(x)+3.
- ˆJ=3x.
- ˆO=x−1.

**What does it mean if two operators commute?**

If two operators commute then both quantities can be measured at the same time, if not then there is a tradeoff in the accuracy in the measurement for one quantity vs. the other.

**What are the appropriate eigenvector of Sz?**

The only possible result of a measurement of an observable is one of the eigenvalues an of the corresponding operator A. These equations tells us that + 2 is the eigenvalue of Sz corresponding to the eigenvector + and “ 2 is the eigenvalue of Sz corresponding to the eigenvector “ .

### Is V an eigenvector of AB BA?

Conversely, show that if AB = BA, B is invertible and Bv is an eigenvector of A, then v is an eigenvector of A. Conversely, suppose that B is invertible and Bv is an eigenvector of A, i.e. ABv = λBv for some λ. Then BAv = ABv = B(λv).

### Do AB and BA have the same eigenvectors?

In other words AB and B A have the same eigenvalues. nonzero eigenvalue of AB has the same geometric multiplicity as it has as an eigenvalue of BA. This may not be true for a zero eigenvalue.

**What does it mean when an operator commutes with the Hamiltonian?**

Since the operator that commutes with the Hamiltonian do not change in time, the corresponding observables (or their expectation values) are independent of time. The expectation value of observable A do not vary with time.

**What is an eigenstate quantum mechanics?**

An eigenstate is the measured state of some object possessing quantifiable characteristics such as position, momentum, etc.

#### What are simultaneous eigenfunctions?

Hence commuting operators have simultaneous eigen- states. That is these can be exactly measured simultaneously. In classical mechanics you can measure any two observables simultaneously. In quan- tum mechanics, only variables whose (Hermitian) operators commute can be observed simultaneously.

#### Do spin operators commute?

In general, every two operators that act on different parts of the system should commute.

**Does s 2 commute with Sz?**

Because S2 commutes with Sz, there must exist an orthonormal basis consisting entirely of simultaneous eigenstates of S2 and Sz.

**How do you prove V is an eigenvector?**

Since v and Av both lie in the one- dimensional eigenspace of B corresponding to the eigenvalue λ, v and Av must be linearly dependent. Since v = 0, this means that Av = µv for some scalar µ. Therefore, v is an eigenvector of A corresponding to the eigenvalue µ.

## Are there any commuting matrices that share the same eigenvector?

In other words x is an eigenvector of B as well as A. There’s another proof using diagonalization in the book. Commuting matrices do not necessarily share all eigenvector, but generally do share a common eigenvector. Let A, B ∈ C n × n such that A B = B A.

## Is there a common basis of eigenvectors?

Let be a set of commuting matrices over an algebraically closed field . Then there may not be a common basis of eigenvectors (since any of them may not be diagonalizable!) but there must be at least a common eigenvector:

**Are there eigenvectors for k 2 in linear algebra?**

Now for K we have no real eigenvectors at all, but for K 2 (as it is symmetric) there are eigenvectors which can be presented as only real. The matrices do not share any eigenvectors.

**How to find the roots of the eigenvalues?**

The roots are λ 1 = 1, λ 2 = 3 (for steps, see equation solver ). These are the eigenvalues. Next, find the eigenvectors. The null space of this matrix is { [ 1 0] } (for steps, see null space calculator ). This is the eigenvector.