Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 1 down vote accepted

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 $\lambda = 0$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.