Let $M$ be the vector space of all $3 \times 3$ real matrices and let $$A = \begin{bmatrix}2&1&0\\ 0&2&0\\ 0&0&3\end{bmatrix}.$$ Which of the following are subspaces of $M$?

1) $W_1 = \{X \in M: XA = AX\}$,

2) $W_2 = \{X \in M: X+A = A + X\}$,

3) $W_3 = \{X \in M: \text{trace}(AX)=0\}$,

4) $W_4 = \{X \in M: \text{det}(AX)=0\}$.

Also, since $\alpha I$ commutes with all the matrices, we have $$(\alpha X_1)A = (\alpha I) (X_1 A) = (\alpha I) (A X_1) = A (\alpha I)X_1 = A (\alpha X_1).$$ Therefore

Also, since $\alpha I$ commutes with all the matrices and $\text{trace}(\alpha A) = \text{trace }((\alpha I)(A)) = \alpha \cdot \text{trace }A$, we have $$\text{trace}(A (\alpha X_1)) = \text{trace}(A ((\alpha I) (X_1)) ) = \text{trace}((\alpha I) (A X_1))\\ = \alpha \cdot \text{trace}(A X_1) = 0.$$ Therefore

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Share to your groups: 1) $W_1 = \{X \in M: XA = AX\}$,

2) $W_2 = \{X \in M: X+A = A + X\}$,

3) $W_3 = \{X \in M: \text{trace}(AX)=0\}$,

4) $W_4 = \{X \in M: \text{det}(AX)=0\}$.

**I am spending my crucial Ph.D. time in writing this blog to help others to achieve in Mathematics. So please encourage me. Visit my blog every day and share the solutions with friends. Follow the blog by email. Thank you.****Solution**:**Observation**: Let $\alpha$ be a scalar then $\alpha A = (\alpha I) A$. That is multiplying every element of $A$ is the same as pre multiplying the matrix $A$ be the scalar matrix $\alpha I$ (diagonal matrix with all the diagonal entries are $\alpha$). Also, these scalar matrices commute with all the matrices in $M$ (Center of the ring $M$).**option 1**:(**True**) Let $X_1,X_2 \in W_1$ then $X_1 A = A X_1$ and $X_2 A = A X_2$. Now, $$(X_1+X_2)A = X_1 A + X_2 A = A X_1 + A X_2 = A (X_1 + X_2).$$**Therefore $X_1+X_2 \in W_1$.**Also, since $\alpha I$ commutes with all the matrices, we have $$(\alpha X_1)A = (\alpha I) (X_1 A) = (\alpha I) (A X_1) = A (\alpha I)X_1 = A (\alpha X_1).$$ Therefore

**$\alpha X_1 \in W_1$**and hence $W_1$ is a subspace of $M$.**option 2**: (**True**) Matrix addition is commutative. Therefore every matrix in $M$ commutes with $A$ and we have $W_2 = M$. Hence it is a subspace.**option 3**: (**True**) Let $X_1,X_2 \in W_3$ then $\text{trace}(A X_1) = 0$ and $\text{trace}(A X_2)=0$. Now, $$\text{trace}(A(X_1+X_2)) = \text{trace}(AX_1) + \text{trace}(AX_2) = 0.$$**Therefore $X_1+X_2 \in W_3$.**Also, since $\alpha I$ commutes with all the matrices and $\text{trace}(\alpha A) = \text{trace }((\alpha I)(A)) = \alpha \cdot \text{trace }A$, we have $$\text{trace}(A (\alpha X_1)) = \text{trace}(A ((\alpha I) (X_1)) ) = \text{trace}((\alpha I) (A X_1))\\ = \alpha \cdot \text{trace}(A X_1) = 0.$$ Therefore

**$\alpha X_1 \in W_3$**and hence $W_3$ is a subspace of $M$.**option 4**:(**False**) We have $\text{det}A = 12 \ne 0$. Therefore $\text{det}(AX) = \text{det }A \cdot \text{det}X = 0$ if and only if $\text{det} X =0$. This shows that $W_4 = \{X \in M: \text{det}X = 0\}$.**The set of all singular matrices is not a vector subspace of $M$**. For example, consider the matrices which are of determinant zero $\begin{bmatrix}1&0&0\\ 0&1&0\\ 0&0&0\end{bmatrix}$ and $\begin{bmatrix}0&0&0\\ 0&0&0\\ 0&0&1\end{bmatrix}$. Clearly their sum is the identity matrix whose determinant is non-zero. Therefore the set of all matrices with determinant zero is not even closed under addition.**FOLLOW BY EMAIL TO GET NOTIFICATION OF NEW PROBLEMS. SHARE YOUR DOUBTS IN THE COMMENTS BELOW. ALSO, YOU CAN SUGGEST PROBLEMS TO SOLVE WHICH WILL BE SOLVED IMMEDIATELY.**

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