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Let $S$ be the subspace of any 3 by 3 skew symmetric matrix. Let $T$ be the subspace of all 3 by 3 matrices that are symmetric. Determine the dimension of the intersection of $T$ and $S$ and the dimension of the union of $T$ and $S$ .

What i tried

A symmetric matrix means that $A^T=A$ while skew matrix means $A^T=-A$

For $T$ intersect $S$ the intersection is $0$ since it is not possible for a symmetric and a skew symmetric matrix to intersect.

It is the union part which im unsure I know the ansewer $9$ dimension form my answer key but i couldnt figure out how did they get $9$ I only could think that they multiplied 3 by 3 to get 9. Could anyone explain thanks

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  • $\begingroup$ Do you know how to compute the dimensions of the subspaces $S$ and $T$? Do you know why the dimension of the space of $3\times 3$ matrices equals $9$? $\endgroup$ – B. Pasternak Nov 1 '15 at 16:19
  • $\begingroup$ Im unsure why. Could u explain. Thanks $\endgroup$ – ys wong Nov 1 '15 at 16:24
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Any (I'm assuming real entries) $3\times 3$ matrix can be written as a linear combination of the matrices $E_{ij}$, the $3\times 3$ matrix with a $1$ in the $(i,j)$th place and zeros everywhere else, i.e. if $A$ is a $3\times 3$ matrix we can write $A=\sum_{i=1}^3\sum_{j=1}^3a_{ij}E_{ij}$ for some coefficients $a_{ij}\in\mathbb{R}$. This is just a formal way of saying that for example we have \begin{align} \begin{pmatrix} 2&0&0\\ 0&0&6\\ 0&0&5 \end{pmatrix}=2E_{11}+6E_{23}+5E_{33}. \end{align} So the set $\{E_{ij}\}_{i,j=1}^3$ is a basis for the $3\times 3$ matrices, making this space $9$-dimensional. Now in order to span the skew-symmetric matrices, which are of the form \begin{align} \begin{pmatrix} 0&a&b\\ -a&0&c\\ -b&-c&0 \end{pmatrix} \end{align} for some $a,b,c\in\mathbb{R}$, we would only need the span of the elements $E_{12}-E_{21}$, $E_{13}-E_{31}$ and $E_{23}-E_{32}$, so this space is $3$-dimensional. I think you can figure out yourself which elements you need to span the symmetric matrices (hint: the only difference with the skew-symmetric matrices is that you also have elements on the diagonal), and then you'll see that that space is $6$-dimensional, so that the union is $9$-dimensional (and is in fact the whole space).

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  • $\begingroup$ Also, consider your way of phrasing. It is weird to say "it's impossible for a skew-symmetric matrix and a symmetric matrix" to intersect, they are just elements of sets and won't do anything like being intersected with each other. The correct way of saying what you mean is "The sets $S$ and $T$ don't intersect/have empty intersection, since no symmetric matrix is a skew-symmetric matrix and vice versa". $\endgroup$ – B. Pasternak Nov 1 '15 at 16:52

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