0
$\begingroup$

I was reading about eigenvalues and eigenvectors.

Consider an eigenvalue $ \lambda $ of a geometric multiplicity of 2.

$$Ax=\lambda x$$

If the geometric multiplicity of $ \lambda $ is 2, therefore, we can write the eigenvector $ x $ as a sum of two independent vectors say $$ x=a+b $$

If I am thinking in the correct direction these vectors should also give the same eigenvalue of $\lambda$ for $A$. Therefore putting the value of $ x $ in the first equation we get

$$ A(a+b)= \lambda(a+b) $$ $$ Aa+Ab= \lambda a +\lambda b $$

I am stuck here. How can I prove that $ a $ and $ b $ have the same eigenvalue $ \lambda $?

A similar example from the book I am reading

enter image description here

Here also we get an eigenvector corresponding to $ F $ as a linear combination of two linearly independent vectors $ e1 $ and $ e2 $. I just wanna confirm whether these linear independent vectors by themself give the same eigenvalue $\lambda$.

$\endgroup$
3
  • $\begingroup$ You need a bit more than what you say. Use the definition of geometric multiplicity in its entirety. $\endgroup$
    – Exodd
    Apr 25 '20 at 21:41
  • $\begingroup$ I am really stuck with the concept of Geometric Multiplicity. After seeing this particular video youtube.com/watch?v=NffFdxiQFMM I did get some intuition about GM but it's not in a way its taught in my uni. I would be grateful if you could point me in the right direction. $\endgroup$
    – daniel
    Apr 25 '20 at 21:44
  • $\begingroup$ Does GM for a particular eigenvalue means that those are the number of independent vectors which form the basis for all the possible eigenvectors corresponding to that eigenvalue? Even if that is the case i should be able to solve my above equation and each of the independent vectors should give same eigenvalue? $\endgroup$
    – daniel
    Apr 25 '20 at 21:51
0
$\begingroup$

Take for example the diagonal matrix $$ A = \begin{pmatrix} 1 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2\\ \end{pmatrix} $$ It has the eigenvalue $2$ of geometric multiplicity 2, and one of its eigenvectors is $$ v = \begin{pmatrix} 0\\ 0\\ 1\\ \end{pmatrix} \implies Ax = 2x. $$ If you take any $x=a+b$ with $a,b$ independent, it isn't true that they are eigenvectors: $$ v = a+b = \begin{pmatrix} 1\\ 0\\ 1\\ \end{pmatrix}+ \begin{pmatrix} -1\\ 0\\ 0\\ \end{pmatrix} $$ Notice that $a,b$ are independent, but $$ Aa = \begin{pmatrix} 1\\ 0\\ 2\\ \end{pmatrix}\ne 2a \qquad Ab = \begin{pmatrix} -1\\ 0\\ 0\\ \end{pmatrix}\ne 2b. $$

The definition of Geometric multiplicity of $\lambda$ is the dimension of the space of $\lambda$ eigenvectors. If $\lambda$ has multiplicity $k$, then there exist $k$ fixed independent eigenvectors $v_1,\dots,v_k$ such that any $\lambda$-eigenvector is a linear combination of $v_1,\dots,v_k$.

$\endgroup$
3
  • $\begingroup$ If I am right, this matrix stretches all the vectors in y and z-direction. And you have mentioned one the vectors in the Z direction(v) which is an eigenvector. Is any vector which is a linear combination of these two eigenvectors also an eigenvector with the same eigenvalue? $\endgroup$
    – daniel
    Apr 25 '20 at 22:14
  • $\begingroup$ any vector that is a combination of $(0 0 1)$ and $(0 1 0)$ is an eigenvector $\endgroup$
    – Exodd
    Apr 25 '20 at 22:15
  • $\begingroup$ Thank you, professor. $\endgroup$
    – daniel
    Apr 25 '20 at 22:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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