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I've tried to solve the next problem, but I'm not quite sure about the 'a' part and don't know how to start with 'b'

Any improvements/clarifications/help is appreciated

The question:

Let $J$ be a $3*3$ matrix given by: \begin{bmatrix}0&1&0\\0&0&1\\0 &0&0\end{bmatrix}

(a) Is the following set a subspace?

$S=[A∈ℝ^{3*3}|AJ=JA]$

My attempt

  • S is nonempty since it contains the $0$ matrix, $0*J=0=J*0$

  • For $A∈S$ and $\alpha$ a real number scalar: $(\alpha A)J=\alpha (AJ)=\alpha(JA)=J(\alpha A)$

  • For $A∈S$, $B∈S$: $(A+B)J=AJ+BJ=JA+BA=J(A+B)$

So both closure properties are satisfied meaning S is a subspace of $ℝ^{3*3}$

(b) Find a basis for S and determine its dimension

Well I don't really get the question, should I find the basis of the matrix J? or something different?

I know that if the basis consists of $n$ vectors than S has dimension $n$ But how to get to this basis/these vectors is my main struggle

Thanks in advance :)

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    It means find a basis of $S$. It will consist of $ 3 \times 3$ matrices. – 2017-01-03
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    This comment might make more sense once you've had some group theory, but I'll leave it anyways. Note that $S$ is known as the centralizer of $A$. As $\mathbb{R}^{3 \times 3}$ is a group, $S$ is a subgroup of $\mathbb{R}^{3 \times 3}$, which means it has the $0$ matrix and is closed under addition. The real check is whether it is closed under the scalar multiplication (which you showed it is). – 2017-01-03

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Hint for part (b)

You should first find a typical matrix that commutes with the given $J$. For that you need to see when will $$ \begin{bmatrix} a&b&c\\ d&e&f\\ g&h&i \end{bmatrix} \begin{bmatrix} 0&1&0\\ 0&0&1\\ 0&0&0 \end{bmatrix} = \begin{bmatrix} 0&1&0\\ 0&0&1\\ 0&0&0 \end{bmatrix} \begin{bmatrix} a&b&c\\ d&e&f\\ g&h&i \end{bmatrix} $$ This gives $$ \begin{bmatrix} 0&a&b\\ 0&d&e\\ 0&g&h \end{bmatrix} = \begin{bmatrix} d&e&f\\ g&h&i\\ 0&0&0 \end{bmatrix} $$ This results in $$d=g=h=0 \qquad \text{and} \quad b=f \quad \text{and} \quad a=e=i$$ So a typical matrix in $S$ looks like $$ \begin{bmatrix} a&b&c\\ 0&a&b\\ 0&0&a \end{bmatrix} $$

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    You must have a mistake here as $I$ commutes with all square matrices... – 2017-01-03
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    @copper.hat You are absolutely right. Thanks for pointing it. I have edited the answer. – 2017-01-03
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    Heey, (I'm on my phone instead of computer so therefor I'm not able to write the matrices, sorry for the inconvenience) so this means the basis is 3 matrices 1. Looks like I 2. All zeros entries except the b's 3. All zero entries except the c's right? Do I then still have to prove these matrices a. Span S and b. Are linearly independent? I guess not since the method to get the basis already 'includes' this data – 2017-01-05
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    @Amaluena you are right the three matrices that can span are $I=[e_1 \, e_2 \, e_3]$ and $ [0 \, e_1 \, e_2]$ and $[0 \, 0 \, e_1]$. Linear independence is easy to check, just look at the zero columns and the corresponding non-zero ones. – 2017-01-05
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    Now I looked at it a little better and for the typical matrix in S you say $\begin{bmatrix}a&b&c\\0&a&b\\0&0&a\end{bmatrix}$ but isnt the diagonal consisting of zeroes? meaning it would be \begin{bmatrix}0&a&b\\ 0&0&a\\ 0&0&0 \end{bmatrix}? – 2017-01-09
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    @Amaluena you are looking at $J$, instead of looking at $A$. – 2017-01-10