# Eigenvalues and eigenvectors of Deformation Tensor

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

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 $$\lambda$$.

• You need a bit more than what you say. Use the definition of geometric multiplicity in its entirety. Apr 25 '20 at 21:41
• 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. Apr 25 '20 at 21:44
• 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? Apr 25 '20 at 21:51

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 $$v_1,\dots,v_k$$.
• any vector that is a combination of $(0 0 1)$ and $(0 1 0)$ is an eigenvector Apr 25 '20 at 22:15