2

Prove that the set of real commuting matrices with the matrix $A= \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$ is a vector space relative to standard operations on matrices. Find the dimension and a basis for that space.

Question: Is it necessary to check the subtraction for commuting matrices.

What are the steps for proving the given statement?

user300048
  • 1,147
  • I suppose it means the set of real matrices $B$ such that $AB=BA$. For proving it is a vector space, it suffices to prove it is a subsapce, i.e. non-emptyness, closed under addition and scalar multiplication. – Empiricist Jan 16 '16 at 01:29

2 Answers2

3

You just have to show it is a subspace of the vector space of $n\times n$ matrices.

Clearly the zero matrix commutes with $M$.

Supose $A$ and $B$ commute with $M$. then $(A+B)M=AM+BM=MA+MB=M(A+B)$.

We also have $(cA)M=c(AM)=c(MA)=M(cA)$, so the matrices that commute with $M$ contain the zero matrix and are closed under addition and scalar multiplication, therefore they form a subspace of the vector space of matrices as desired.

Asinomás
  • 105,651
  • 2
    Indeed, it is important to note that definition of $A$ is irrelevant. The only condition is that $A$ is an $n\times n$ matrix. – sinbadh Jan 16 '16 at 01:40
  • 1
    The OP seems to have forgotten, but the question also asked for dimension and a basis. This can be done in an entirely elementary way (and I wish the OP had followed my suggestion), write another matrix $B$ with all entries distinct letters, compare $AB$ and $BA,$ equality gives many simple (linear) relations among the entries of $B.$ Conclusion is a case of http://math.stackexchange.com/questions/92480/given-a-matrix-is-there-always-another-matrix-which-commutes-with-it/92832#92832 – Will Jagy Jan 16 '16 at 18:43
1

For even $n$, the matrices that commute with each other has the following block form:

\begin{bmatrix} aI & M \\ 0 & aI \\ \end{bmatrix}

Each block $(0, aI, M)$ has $\frac{n}{2}\times\frac{n}{2}$ rows and columns. The matrix of the question is:

$A$= \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}

$A$ doesn't have the block form I mentioned above, so the only matrix that can commute with it is the zero matrix.

Mojee KD
  • 420