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Let $V=M_{2 \times 2}(\mathbb{R})$ be the set of all $2 \times 2$ real-valued matrices, and let the field $K = \mathbb{R}$. Then $V$ is a vector space under matrix addition and Scaler multiplication.

Let $$A = \begin{bmatrix}1&0\\1&2\end{bmatrix}$$

If $C(A) = \{B \mid B \in V$, such that $AB=BA$} find the dimension of the subspace $C(A)$ and determine a basis.

First of all I believe I would have to prove this is a subspace. So I would have to show that the subspace is closed under vector addition and scaler multiplication. Then I would have to construct a basis. The number of vectors in the basis is the dimension of the subspace.

It is the condition that is tripping me up. How do show all this with the condition that these $2 \times 2$ matrices are commutative?

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To show that $C(A)$ is a subspace, we need to show it closed under both addition and scalar multiplication; so let

$B_1, B_2 \in C(A); \tag 1$

then

$B_1A = AB_1, \; B_2A = AB_2; \tag 2$

thus,

$(B_1 + B_2)A = B_1A + B_2A = AB_1 + AB_2 = A(B_1 + B_2), \tag 3$

which of course implies

$B_1 + B_2 \in C(A); \tag 4$

likewise if $\alpha$ is any scalar, we have

$(\alpha B)A = \alpha (BA) = \alpha (AB) = A(\alpha B); \tag 5$

thus

$\alpha B \in C(A) \tag 6$

as well. Also, it is pretty easy to see that (6) implies

$0 \in C(A), \tag 7$

and

$B \in C(A) \Longleftrightarrow -B \in C(A); \tag 8$

since the rest of the vector space axioms are inherited by $C(A)$ from $M_{2 \times 2}(\Bbb R)$, it follows that $C(A)$ is indeed a subspace.

So, what do the elements of $C(A)$ look like? If

$B = \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix}, \tag 9$

then the condition

$AB = BA \tag{10}$

reads, with

$A = \begin{bmatrix}1&0\\1&2\end{bmatrix}, \tag{10}$

$\begin{bmatrix}1&0\\1&2\end{bmatrix}\begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix} = \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix}\begin{bmatrix}1&0\\1&2\end{bmatrix}; \tag{11}$

at this point, before proceeding further, we observe that the computations specified in (11) may be considerably simplified if we a priori write $A$ in the form

$A = \begin{bmatrix}1&0\\1&2\end{bmatrix} = I + \begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix}, \tag{12}$

since

$IB = BI; \tag{13}$

we are left with finding those $B$ such that

$ \begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix} = \begin{bmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{bmatrix}\begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix}; \tag{14}$

that is,

$\begin{bmatrix} 0 & 0 \\ b_{11} + b_{21} & b_{12} + b_{22} \end{bmatrix} = \begin{bmatrix} b_{12} & b_{12} \\ b_{22} & b_{22} \end{bmatrix}, \tag{15}$

we thus find that

$b_{12} = 0, \; b_{11} + b_{21} = b_{22} = b_{12} + b_{22}; \tag{16}$

by virtue of these equations, we see we may write $B$ in the form

$B = \begin{bmatrix} b_{11} & 0 \\ b_{22} - b_{11} & b_{22} \end{bmatrix} = b_{11} \begin{bmatrix} 1 & 0 \\ -1 & 0 \end{bmatrix} + b_{22} \begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix}; \tag{17}$

it is now clear that we may take $b_{11}$ and $b_{22}$ as free parameters, and that $C(A)$ is two dimensional, being spanned by matrices

$\begin{bmatrix} 1 & 0 \\ -1 & 0 \end{bmatrix}, \; \begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix}, \tag{18}$

which form a basis for $C(A)$.

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Let’s take $B\in C(A)$ to be

$$\begin{bmatrix} a & b \\ c & d\end{bmatrix},$$

and then compute $AB$ and $BA$. What constraints are there on $a,b,c,$ and $d$?

To show closure, let’s say that we have $B_1, B_2\in C(A)$. Then for all $x\in\mathbb{R}$ we want:

$$ (xB_1)A = A(xB_1); \quad (B_1 + B_2)A = A(B_1 + B_2).$$

Can you take it from here?

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