ask questions - get instant answers - receive/send p2p payments - get tips - get points - get fundraising support - meet community tutors
First time here? Checkout the FAQs!
x
1 like 0 dislike
472 views
Let \(A\) and \(B\) be \(n \times n\) complex matrices that commute: \(A B=B A\). If \(\lambda\) is an eigenvalue of \(A\), let \(\mathcal{V}_{\lambda}\) be the subspace of all eigenvectors having this eigenvalue.
a) Show there is an vector \(v \in \mathcal{V}_{\lambda}\) that is also an eigenvector of \(B\), possibly with a different eigenvalue.
b) Give an example showing that some vectors in \(\mathcal{V}_{\lambda}\) may not be an eigenvectors of B.
c) If all the eigenvalues of \(A\) are distinct (so each has algebraic multiplicity one), show that there is a basis in which both \(A\) and \(B\) are diagonal. Also, give an example showing this may be false if some eigenvalue of \(A\) has multiplicity greater than one.
in Mathematics by Platinum (108k points) | 472 views

1 Answer

0 like 0 dislike
Best answer
a) Let \(v\) be an eigenvector of \(A\) with eigenvalue $\lambda$, so \(A v=\lambda v\). Then,

\[B A v = B (\lambda v) = \lambda B v\].

Since \(A B=B A\), we have
\[\mathrm{ABv}=\mathrm{BAv}\],
which implies that \(B v\) is also an eigenvector of \( A \) with eigenvalue \(\lambda \).
b) Consider the \(2 \times 2\) matrices

\[A = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}\]

\[B = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\].

Then, \(A\) and \(B\) commute, and both have eigenvalue 1 with eigenvector \((1,0)\) in the subspace \( \mid\) mathcal \(\{V\}_{-} 1 \). However, this vector is not an eigenvector of \(B\).

c) The statement is true by the spectral theorem, which states that any normal matrix (a matrix that commutes with its conjugate transpose) can be diagonalized by a unitary matrix. If all the eigenvalues of \(A\) are distinct, then the eigenvectors form a linearly independent set, and can be used to form a basis for the vector space. In this basis, both \( A \) and \( B \) will be diagonal.

However, if some eigenvalue of \( A \) has multiplicity greater than one, then there are linearly dependent eigenvectors, and it may not be possible to find a basis in which both \( \mathrm{~A} \) and \(\mathrm{~B} \) are diagonal. For example, consider the \(2 \times 2\) matrices

\[A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\]

\[B = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\].

Both \(A\) and \(B\) commute, but there is no basis in which both are diagonal.
by Diamond (64.2k points)

Related questions

0 like 0 dislike
1 answer
asked Jan 27, 2023 in Mathematics by AstraNova Diamond (64.2k points) | 350 views
0 like 0 dislike
1 answer
asked May 6, 2022 in Mathematics by AstraNova Diamond (64.2k points) | 519 views
0 like 0 dislike
1 answer
asked Jan 27, 2023 in Mathematics by AstraNova Diamond (64.2k points) | 421 views
0 like 0 dislike
0 answers
asked Jan 21, 2022 in Mathematics by MathsGee Platinum (108k points) | 235 views

Join MathsGee for AI-powered Q&A, tutor insights, P2P payments, interactive education, live lessons, and a rewarding community experience.

On MathsGee, you can:

1. Ask Questions on Various Topics


2. Request a Tutor


3. Start a Fundraiser


4. Become a Tutor


5. Create Tutor Session - For Verified Tutors


6. Host Tutor Session - For Verified Tutors


7. Join Tutor Session


8. Enjoy our interactive learning resources


9. Get tutor-verified answers

10. Vote on questions and answers

11. Tip/Donate your favorite community members

12. Earn points by participating


Posting on the MathsGee

1. Remember the human

2. Act like you would in real life

3. Find original source of content

4. Check for duplicates before publishing

5. Read the community guidelines


MathsGee Rules

1. Answers to questions will be posted immediately

2. Questions will be queued for posting immediately after moderation

3. Depending on the number of messages we receive, you could wait up to 24 hours for your message to appear. But be patient as posts will appear after passing our moderation.


MyLinks On Acalytica | Social Proof Widgets | Web Analytics | SEO Reports | Learn | Uptime Monitoring