0
$\begingroup$

a.) How many orthonormal eigenvector bases does a symmetric $n$ x $n$ matrix have? Now let $A=\pmatrix{a&b\\c&d}$, write down necessary and sufficient conditions on the entries a, b, c, d that ensures that A has only real eigenvalues.

b.) Let $A^T =-A$ be a real, skew-symmetric $n$ x $n$ matrix. Prove that the only possible real eigenvalue of A is $\lambda = 0$?

Answer for a:

If all eigenvalues are distinct there are $2^n$ different bases. If the eigen values are repeated there are infinitely many.

How did they get that? Lets say I have a $2$ x $2$ matrix and it has distinct eigenvalues (lets say 1 and 2 are the eigenvalues) wouldn't the eigenvectors be equal to the amount of eigenvalues, so in this case it will equal 2? But the answer says it equals 4?

  • 0
    @GerryMyerson thank you very much for clearing that up for me.2012-11-16

1 Answers 1

1

Let's say a symmetric matrix $A \in \mathbb R^{2 \times 2}$ has distinct eigenvalues $\lambda_1$ and $\lambda_2$, and assume $\{v_1,v_2\}$ is a corresponding orthonormal basis of eigenvectors for $\mathbb R^2$. Then the following are also orthonormal eigenbases of $\mathbb R^2$: $\{ v_1,-v_2 \},\{ -v_1,v_2\},\{-v_1,-v_2\}$.

For part b): suppose $A \in \mathbb R^{n \times n}$ is skew-symmetric and $\lambda \in \mathbb R$ is an eigenvalue of $A$ with corresponding (nonzero) eigenvector $x$. Then

\begin{align*} \langle Ax,x \rangle &= \langle \lambda x, x \rangle \\ &= \lambda \|x\|_2^2. \end{align*}

On the other hand, \begin{align*} \langle Ax,x \rangle &= \langle x, A^T x \rangle \\ &= \langle x, -Ax \rangle \\ &= \langle x, -\lambda x \rangle \\ &= -\lambda \|x\|_2^2. \end{align*} It follows that $\lambda = -\lambda$, which implies that $\lambda = 0$.

  • 1
    @diimension Yes, that's correct.2012-11-26