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If $M$ is the vector space of $2\times 2$ real matrices, then I can show that

$$ \{A \in M \mid A^\mathrm{T}=-A \} $$

is a subspace of $M$, since

$$ \left[ \begin{array}{cc} x & z \\ -z & y \end{array} \right]+\left[ \begin{array}{cc} x' & z' \\ -z' & y' \end{array} \right] = \left[ \begin{array}{cc} x+x' & z+z' \\ -(z+z') & y+y' \end{array} \right] $$

and for some $\lambda\in\mathbb{R}$

$$ \lambda\left[ \begin{array}{cc} x & z \\ -z & y \end{array} \right] = \left[ \begin{array}{cc} \lambda x & \lambda z \\ -\lambda z & \lambda y \end{array} \right] $$

But I'm not sure if I'm correct on finding the dimension and a basis of the subspace:

$$ \left[ \begin{array}{cc} x & z \\ -z & y \end{array} \right] = x\left[ \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right]+y\left[ \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} \right]+z\left[ \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right] $$

This makes me think that a basis is made up of

$$ \left(\left[ \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right],\left[ \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} \right],\left[ \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right] \right) $$

and so the dimension is three. Is that right?

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    $\begingroup$ I like how you put the "T" for transpose in roman type, that's good latex! $\endgroup$ – Gregory Grant May 16 '15 at 18:02
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    $\begingroup$ Note that if $A^{\mathrm T}=-A$, then the elements on the diagonal are $0$. So the basis is just $$\left(\left[ \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right]\right)$$ $\endgroup$ – Kitegi May 16 '15 at 18:09
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$$ A^\mathrm{T}=-A $$ implies that $A$ has zeroes on the main diagonal, since they change sign on the RHS, but remain unchanged on the LHS. if you incorporate that into your representation, it should be easy to answer the question about dimension and basis.

as a related exercise you might like to consider the $2 \times 2$ matrices with complex entries. these are isomorphic to $\Bbb{R}^8$ as a real vector space, and to $\Bbb C^4$ as a complex vector space. the skew-hermitian matrices satisfy: $$ A^{\mathrm{T}}+A^* =0 $$ where $A^*$ is the complex conjugate of $A$. the skew-hermitian matrices are isomorphic to a $4$-dimensional subspace of $\Bbb R^8$. however in $\Bbb C^4$ they are only a subgroup, but do not form a subspace - the diagonal elements must be purely imaginary and this property is not preserved under multiplication by an arbitrary complex scalar.
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  • $\begingroup$ Ah, forgot about the minus and the diagonals. Thanks. $\endgroup$ – jonbaldie May 16 '15 at 18:31
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Note that $$ A=\begin{bmatrix}a&b\\c&d\end{bmatrix} $$ satisfies $A^\top=-A$ if and only if $$ \begin{bmatrix} a&c\\b&d \end{bmatrix} = \begin{bmatrix} -a&-b\\-c&-d \end{bmatrix} $$ That is, $A^\top=-A$ if and only if $$ A= \begin{bmatrix} a&b\\c&d \end{bmatrix} = \begin{bmatrix} 0&b\\-b&0 \end{bmatrix} = b \begin{bmatrix} 0&1\\-1&0 \end{bmatrix} $$ Hence $$ \DeclareMathOperator{Skew}{Skew}\Skew_2=\DeclareMathOperator{Span}{Span}\Span\left\{\begin{bmatrix}0&1\\-1&0\end{bmatrix}\right\} $$ In particular, $\Skew_2$ is a subspace of $M_2$ with $\dim\Skew_2=1$.

One of the advantages of the above argument is by showing that $\Skew_2$ is spanned by a subset of $M_2$ we don't have to manually check that $\Skew_2$ is a subspace. The disadvantage, of course, is that the formulas are harder to work out in the general case $\Skew_n$. @math.noob's answer gives the most elegant proof that $\Skew_n$ is a subspace of $M_n$.

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  • $\begingroup$ Yes, but also does the dimension not depend on the considered characteristic? $\endgroup$ – Quality May 16 '15 at 18:33
  • $\begingroup$ @Quality OP is working over the reals so considering a characteristic other than zero seems inappropriate. $\endgroup$ – Brian Fitzpatrick May 16 '15 at 18:41
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More generally, even if $A$ and $B$ are $n \times n$ matrices, they still form a subspace because: $(A+\lambda B)^T = A^T + \lambda B^T$

So, if $A^T = -A$ and $B^T = -B$ we get $(A+\lambda B)^T = -(A + \lambda B)$ which proves that $A + \lambda B$ belongs to the set of skew symmetric $n \times n$ matrices.

About the dimension, well, you just need to determine the upper triangular part of the matrix. The diagonal entries are going to be zero because for any $i$ we must have: $a_{ii} = -a_{ii}$ which implies $a_{ii} = 0$ So, how many entries you should fill in order to form an upper triangular matrix with zero diagonal entries?

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