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I'm trying to find a basis for the kernel for the following mapping: Considering the linear transformation T: $M_{33} \rightarrow M_{33} $ defined by $T(A) = .5(A + A^T)$. I know that this is basically asking for the basis under the condition that $T(A)=0$ which means that $A+A^T=0$ so $A^T = -A$. I found that matrices that fit this condition are Skew Symmetric Matrices. However, I'm not sure how to find the basis for the kernel of these matrices.

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    $\begingroup$ Any $3\times 3$ skew-symmetric matrix has the form $\pmatrix{0&a_{12}&a_{13}\\-a_{12}&0&a_{23}\\-a_{13}&-a_{23}&0}$, so the basis would consist of $\pmatrix{0&1&0\\-1&0&0\\0&0&0}$, $\pmatrix{0&0&1\\0&0&0\\-1&0&0}$, $\pmatrix{0&0&0\\0&0&1\\0&-1&0}$. $\endgroup$ Commented Dec 10, 2015 at 20:20
  • $\begingroup$ @AlgebraicPavel So then that would be the basis for the kernel as well, right? $\endgroup$
    – tibsar
    Commented Dec 10, 2015 at 20:22
  • $\begingroup$ Yes$\,\,\,\,\,$ $\endgroup$ Commented Dec 10, 2015 at 20:29

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Let $a_{ij}$ denote the entries of $A$. If $A \in \ker T$, then all of the entries of $T(A)$ are zero. In other words, $$ a_{ij} + a_{ji} = 0. $$ This forces diagonal entries to vanish: $$ a_{ii} = 0. $$ Define the matrix unit $E_{ij}$ to be the $3 \times 3$ matrix, all of whose entries are $0$ except for the $(i,j)$ entry, which is $1$. These nine matrices form a basis for $M_{3,3}$, the space of all $3 \times 3$ matrices.

Now, we can build a basis $\{ B_{12}, B_{13}, B_{23} \}$ for the space of skew symmetric matrices out of the matrix units: \begin{align} B_{12} = E_{12} - E_{21} &= \begin{pmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}\!, \\[2pt] B_{13} = E_{13} - E_{31} &= \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{pmatrix}\!, \\[2pt] B_{23} = E_{23} - E_{32} &= \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & -1 & 0 \end{pmatrix}\!. \end{align} An arbitrary skew symmetric matrix decomposes as $$ \begin{pmatrix} 0 & a_{12} & a_{13} \\ -a_{12} & 0 & a_{23} \\ -a_{13} & -a_{23} & 0 \end{pmatrix} = a_{12} B_{12} + a_{13} B_{13} + a_{23} B_{23}\!, $$ showing that the set $\{ B_{12}, B_{13}, B_{23} \}$ spans. It's pretty clear that these three are linearly independent as well: if we set the arbitrary linear combination to zero on the right, then each entry of the matrix is $0$, so $a_{12} = a_{13} = a_{23} = 0$ which is the trivial combination. In other words, the decomposition of any skew symmetric matrix is unique.

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