### CSIR JUNE 2011 PART C QUESTION 75 SOLUTION (Which are subspaces ? $\{X \in M: \text{trace}(AX)=0\}$, $\{X \in M: \text{det}(AX)=0\}$.)

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

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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: (TrueLet $X_1,X_2 \in W_3$ 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.

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