0
$\begingroup$

$\textbf{Question:}$ Find a basis for the vector space of all 2x2 matrices that commute with $\begin{bmatrix}3&2\\4&1\end{bmatrix}$, which is the matrix $B$. You must find two ways of completing this problem for full credit.

$\textbf{My Attempt:}$ I found that $B$ is diagonalizable, and so any other diagonalizable 2x2 matrix $A$ will satisfy $AB=BA$. However, I cannot think of a way to form a basis for all 2x2 diagonalizable matrices. I tried to start with a diagonal matrix with distinct entries on its diagonal, but ended up running into a lot of dead ends.

Does anyone else have any ideas on how I might find this basis? Does anyone have any other potential methods of solving this problem?

$\endgroup$
  • $\begingroup$ $AB = BA$ only if $A$ and $B have the same eigenvectors. So you should find these. $\endgroup$ – Hans Engler Nov 20 '16 at 22:29
  • $\begingroup$ Another approach is to write out the condition $AB - BA = 0$ as a system of homogeneous linear equations for the entries of $A$. Then solve this system. $\endgroup$ – Hans Engler Nov 20 '16 at 22:31
  • $\begingroup$ Yeah, you had it the wrong way : $AB=BA$ and $B$ diagonalizable means the eigenspaces of $B$ are stable for $A$. As those eigenspaces are of dimension one, it means $A$ is also diagonalizable, in the same basis. $\endgroup$ – Nicolas FRANCOIS Nov 20 '16 at 22:32
  • 1
    $\begingroup$ No, not every diagonalizable $2 \times 2$ matrix will commute with $B$. Moreover, the diagonalizable matrices do not form a vector space, so there is no such thing as a basis for them. $\endgroup$ – Robert Israel Nov 20 '16 at 22:33
  • $\begingroup$ You need to change your title; it makes no sense (by the previous comment). $\endgroup$ – Marc van Leeuwen Nov 22 '16 at 7:44
1
$\begingroup$

Here is a way of finding one basis:

Let $L(A) = AB-BA$, then $A$ commutes with $B$ iff $A \in \ker L$. Using a standard basis, find the null space of $L$ and use this to determine a basis of $\ker L$.

This can be simplified a little since $B$ has a full set of eigenvectors.

Suppose $v_k,u_k$ are the left and right eigenvectors of $B$ corresponding to $\lambda_k$. Show that $u_i v_j^T$ is a basis and that $L(u_i v_i^T) = (\lambda_i - \lambda_j) u_i v_j^T$. In particular, this shows that $\ker L = \operatorname{sp} \{ u_1 v_1^T, u_2 v_2^T \} $.

By inspection, we can choose $v_1 = (2,1)^T, v_2 = (-1,1)^T$ and $u_1 =(1,1)^T, u_2 = (-1,2)^T$ to get a basis $\begin{bmatrix} 2 & 1 \\ 2 & 1 \end{bmatrix}$, $\begin{bmatrix} 1 & -1 \\ -2 & 2 \end{bmatrix}$.

Here is another way: Suppose $V^{-1} B V = \Lambda$, where $\Lambda$ is diagonal (with different entries). Then $AB=BA$ iff $ V^{-1} A V V^{-1} B V = V^{-1} B V V^{-1} A V$ iff $V^{-1} A V \Lambda = \Lambda V^{-1} A V$.

In particular, $C$ commutes with $\Lambda$ iff $V C V^{-1}$ commutes with $B$. Since $\Lambda$ is diagonal with distinct eigenvalues, we see that $C$ commutes with $\Lambda$ iff $C$ is diagonal.

Hence a basis for the set of commuting matrices is $V \operatorname{diag}(1,0) V^{-1}$, $V \operatorname{diag}(0,1) V^{-1}$.

$\endgroup$
  • $\begingroup$ Thank you for the thoroughness of your answer :) $\endgroup$ – riley lyman Nov 21 '16 at 0:41
1
$\begingroup$

As you noted, the matrix $B$ is diagonalizable, and we have: $$ B=\begin{bmatrix} 3 & 2\\ 4 & 1 \end{bmatrix}=SDS^{-1}= \begin{bmatrix} -1 & 1\\ 2 & 1 \end{bmatrix} \begin{bmatrix} -1 & 0\\ 0 & 5 \end{bmatrix} \begin{bmatrix} -1/3 & 1/3\\ 2/3 & 1/3 \end{bmatrix} $$

A matrix $A$ commutes with $B$ iff they are simultaneously diagonalizable, and this means that $A$ has the form:

$$ A=\begin{bmatrix} -1 & 1\\ 2 & 1 \end{bmatrix} \begin{bmatrix} a & 0\\ 0 & b \end{bmatrix} \begin{bmatrix} -1/3 & 1/3\\ 2/3 & 1/3 \end{bmatrix} =\frac{1}{3}\left\{ a\begin{bmatrix} 1 & -1\\ -2 & 2 \end{bmatrix} +b\begin{bmatrix} 2 & 1\\ 2 & 1 \end{bmatrix} \right\} $$ so the matrices $$ \begin{bmatrix} 1 & -1\\ -2 & 2 \end{bmatrix} \qquad\begin{bmatrix} 2 & 1\\ 2 & 1 \end{bmatrix} $$ are a basis for the space of the matrices that commute with $B$.

$\endgroup$
  • $\begingroup$ Thanks so much, I wish I could accept both answers. $\endgroup$ – riley lyman Nov 21 '16 at 0:41
1
$\begingroup$

Adapted from this answer to a very similar question.

That matrix $B$ is clearly not a multiple of the identity matrix, so its minimal polynomial is of degree${}>1$, hence equal to its characteristic polynomial (which you do not have to compute). Then by the result of this question, matrices that commute with $B$ are just the polynomials in$~B$. Given that the minimal polynomial has degree$~2$, the polynomials in $B$ are just the linear combinations of $B$ and the $2\times2$ identity matrix (filling a $2$-dimensional subspace of matrices).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.