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Let $M$ be the vector space of all $n\times n$ matrices and $T:M\to M$ be a linear transformation such that $T(A) = 0$, where $A$ denotes all symmetric and skew symmetric matrices. Then what is the rank of $T$? $\mathrm{rank}(T) = \dim(M)-\mathrm{nullity}(T)$. Dimension of $M$ is $n^2$ but what is its nullity?

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    Can you think of any example of what transformation $T$ could be?2012-12-24
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    My previous question may be obsolete, but I will note: The idea that $T$ could be simply the $0$ map could come without knowing much. Even if there could be other maps that satisfy the given condition, the $0$ map is certainly an option. Hence if the question has a unique answer, you have it there. However, that in itself doesn't tell you why there is a unique answer, and for that Zev's approach is good.2012-12-24

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Hint: Considering the case of $n=2$, observe that any $2\times 2$ matrix can be written as $$\begin{pmatrix}a & b \\ c & d\end{pmatrix}=\begin{pmatrix}a & \tfrac{b+c}{2} \\ \tfrac{b+c}{2} & d\end{pmatrix}+\begin{pmatrix}0 & \tfrac{b-c}{2} \\ \tfrac{c-b}{2} & 0\end{pmatrix}$$ Can you generalize this observation? What does this imply about how $T$ acts on any matrix?

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    is rank 0? since dimension is 4 and nullity is also 42012-12-24
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    To make my point clearer: can you see what kinds of matrices the two on the right side of the equation are?2012-12-24
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    yes. 1st is symmetric and 2nd is skew symmetric.2012-12-24
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    every matrix can be written as a sum of symmetric and a skew symmetric matrices, so T maps every matrix to 0. is it correct?2012-12-24
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    Yup, that's right!2012-12-24