# Eigenvector Bases

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 $\lambda = 0$?

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?

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You can multiply an eigenvector by $-1$ to obtain a new eigenvector,while preserving orthonormality. (If our field of scalars is $\mathbb C$, you can multiply by any unit.) – littleO Nov 16 '12 at 3:55
@littleO so all distinct eigenvalues have the property of $^+_-$ ? And how is repeated eigenvalues infinite? – diimension Nov 16 '12 at 4:03
Well, think about the identity matrix, a symmetric matrix with repeated eigenvalue 1. Any pair of orthogonal unit vectors will be an orthonormal basis consisting of eigenvectors, right? – Gerry Myerson Nov 16 '12 at 4:35
@GerryMyerson Can you elaborate more on your second sentence. I am really trying to understand this but need more help if you don't mind. – diimension Nov 16 '12 at 4:44
Start with this: what are the eigenvectors of the identity matrix? – Gerry Myerson Nov 16 '12 at 4:49

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 $\lambda = 0$.

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Thank you very much! For part b. does a skew matrix always have $0$ as its diagonal? If not, why does the negative make it zero? I cant see that in your proof. – diimension Nov 16 '12 at 6:31
Yes, REAL skew symmetric matrices have zeros on the diagonal. That's because $a_{ii} = -a_{ii}$ – bartgol Nov 16 '12 at 6:34
@littleO Beautiful. Thank you very much! I understand it now with your help. – diimension Nov 16 '12 at 6:55
@littleO Sry to bother you on this question again but for part b I was looking at your proof again and I am wondering will all the eigenvalues of A be imaginary then? – diimension Nov 26 '12 at 6:58
@diimension Yes, that's correct. – littleO Nov 26 '12 at 11:39