# Basis for subspace S

I would like some hints on how to do thees types of questions

a) Show that if $A$ is a fixed $n$ x $n$ matrix, then $S = \{B \in M_{n}(\mathbb{C}):AB=BA\}$ is a subspace.

I've done this part, just show that the zero $n$ x $n$ matrix is an element of $S$, by showing $A0 = 0A$ and therefore it is in the set. Then we let $C,D$ be some matrices in $S$ and $\alpha$ in our field and show that ($C+\alpha D) \in S$ by using properties of matrices upon multiplying by $A$

b) Find a basis for $S$ when $n = 3$ and $A = \begin{pmatrix} 1 &0 &0 \\ 0 &-1 &0\\ 0 &0 & i \end{pmatrix}$

I understand the definition of a basis and how to check whether a set is a basis of some subspace, but I'm not sure how to find a basis for $S$, but I think we have to use $AB=BA$ somehow.

c) Find a $3$ x $3$ matrix A such that $S$ has dimension $5$.

No clue how to do this one either, but I'm assuming we use part b...

Any hints would be nice. Thank you.

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For b), compute $AB$, compute $BA$, and see what conditions on $B$ imply $AB=BA$. –  Gerry Myerson Mar 29 '14 at 7:58

(b)We can write $B$ as $\left( {\begin{array}{*{20}{c}} {{b_{11}}}&{{b_{12}}}&{{b_{13}}}\\ {{b_{21}}}&{{b_{22}}}&{{b_{23}}}\\ {{b_{31}}}&{{b_{32}}}&{{b_{33}}} \end{array}} \right)$. By $AB=BA$, we have ${b_{ij}} = 0$ if $i \ne j$ (Check it!).

So $B$ is of the form $\left( {\begin{array}{*{20}{c}} {{b_1}}&{}&{}\\ {}&{{b_2}}&{}\\ {}&{}&{{b_3}} \end{array}} \right)$. Now we can easily find a basis of $S$.

(c)Obviously, this matrix A can be $\left( {\begin{array}{*{20}{c}} 1&0&0\\ 0&0&0\\ 0&0&0 \end{array}} \right)$. (Check it!)

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$(a)$ It seems that you succeed to do it, the basic idea is to use the definition of subspace and apply it directly.

$(b)$ You are looking for matrices which commute with $A$. You should use the fact that $A$ has three distinct eigenvalues, and observe that if $AB=BA$ and $v$ is a eigenvector of $A$ of eigenvalue $\lambda$, then $Bv$ is again an eingenvalue of $A$ with eigenvalue $\lambda$. This should help to describe $B$.

$(c)$ If $A$ is diagonal with three distinct eigenvalues, the dimension obtained for $S$ is three. Try with two distinct eigenvalues.

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