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Let $A = -A^T$ be a $3 \times 3$ real skew-symmetric matrix.

  1. Prove that $A$ has an eigenvalue $\lambda = 0$ and two other pure imaginary eigenvalues, and that the eigenvectors of $A$ are orthogonal in $\mathbb{C}^3$.

  2. Find $e^A$, with: $$ A = \left[ \begin{matrix} 0 & 1 & 0 \\ -1 &0 & 2 \\ 0 & -2 & 0 \end {matrix} \right ]$$

I've found a similar question here: Real Skew Symmetric $3\times 3$ matrix has one eigen value $2i$

In one of the answers it is said: "In general, a real skew-symmetric 3×3 matrix K looks like: $$\left[ \begin{matrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end {matrix} \right ]$$ "

I don't understand why the general form of the matrix has that diagonal of zeros. I would have thought about something like: $$\left[ \begin{matrix} d & a & b \\ -a & e & c \\ -b & -c & f \end {matrix} \right ]$$

Wich of course has much worse determinant & characteristic polynomial...

Community
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agneau
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3 Answers3

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Note that$$-\begin{pmatrix}d & a & b \\ -a & e & c \\ -b & -c & f\end{pmatrix}^T=\begin{pmatrix}-d & a & b \\ -a & -e & c \\ -b & -c & -f\end{pmatrix},$$which will be equal to$$\begin{pmatrix}d & a & b \\ -a & e & c \\ -b & -c & f\end{pmatrix}$$if and only if $d=e=f=0$. Can you take it from here?

José Carlos Santos
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  • That's right! Thank you! I think I can manage to do the first part, I'll have some problems with the exponential of the matrix – agneau Feb 12 '18 at 20:19
  • Sorry I have some problems also with the eigenvectors part. I can't find an easy way to show that they are orthogonal to each other! – agneau Feb 12 '18 at 21:04
  • maybe I've got something: $( A \vec{v} | \vec{w}) = (\vec{v}| A^T \vec{w})=(\vec{v}|-A^T \vec{w})$ then if $\vec{v}, \vec{w}$ are eigenvectors for, respectively, $\lambda_0 = 0$ and $\lambda_1 = 1$, $( \lambda_0 \vec{v} | \vec{w}) = (\vec{v}| - \lambda_1 \vec{w})$, wich means that $(\lambda_0+lambda_1)(\vec{v}|\vec{w})=0$ and that's never true unless $v \perp w$ (the same for the others eigenvalues) – agneau Feb 12 '18 at 21:10
  • @Gitana You'll find a proof of a more general statement here. – José Carlos Santos Feb 12 '18 at 21:16
  • @ José Thank you, very useful proof, and to solve the exponential? I was thinking about something like $e^A=Ve^\lambda V^{-1}$ but I have some difficulties in the determination of the eigenvectors. Is there a simpler way? $e^\lambda$ where $ \lambda$ is the matrix of the eigenvaules – agneau Feb 12 '18 at 21:39
  • @Gitana Use the fact that yor matrix is similar to$$\begin{pmatrix}i\sqrt5&0&0\0&-i\sqrt5&0\0&0&0\end{pmatrix}.$$ – José Carlos Santos Feb 12 '18 at 21:55
  • still, I have to determine the change of basis matrix V, which is formed by the eigenvectors but to find the eigenvectors is not so simple (or maybe I'm too tired and i'm missing something) – agneau Feb 12 '18 at 22:02
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$iA$ is a Hermitian matrix.

Hermitian matrices are always diagonalizable and always have real eigenvalues, and have orthogonal eigen-vectors.

The eigenvalues of $A$ are either $0$ or pure imaginary.

The characteristic equation of $A$ will be a degree 3 polynomial with real coefficients. It must have one real root. And it must be $0.$

Doug M
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If you negate every value on the diagonal, you must have the same value. That is only true of zero. Therefore you get 0 on the diagonal.

The eigenvalues of a real matrix must be either real or in complex conjugate pairs. If you have three eigenvalues, one of them must be a real one. Since you can prove, that skew syymetric matrices have all imaginary eigenvalues, you can directly see, that there is only one number, that is both on the real and the imaginary axis. That is 0.

Laray
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