# Thread: problem with invertible matrices

1. ## problem with invertible matrices

Question:
If A and B are similar matrices
the geometric multiplicity of an eigenvalue $\lambda$ of A is the same as its geometric multiplicity as an eigenvalue of B.

Proof:
${\lambda} I - B = Q({\lambda} I -A)Q^{-1}$
since Q is invertible, it follows that $nullity({\lambda}I - B) = nullity({\lambda}I - A)$

I dont get the bit about "since Q is invertible" I do not see how this is relevant

2. We can see that $x\in \ker (\lambda I-B)$ if and only if $Q^{-1}x\in\ker(\lambda I-A)$. We deduce that we have an isomorphism between these two space.

3. An invertible matrix is one-to-one. Consequently, the null spaces of A and QA, where Q is invertible, are the same. QAv is 0 if and only if Av is 0- v is in the null space of A. Similarly, AQv= 0 only if Qv is in the null space of A. Since Q is invertible, the dimension of "all v such that Qv is in the null space of A" is the same as the dimension of the null space of A.

Personally, I wouldn't prove this statement in that way. Instead (longer but more "fundamental"):
Suppose $\lambda$ is an eigenvalue of A with geometric multiplicity n- that is, there exist n independent vectors, $\{v_1, v_2, \cdot\cdot\cdot, v_n\}$ vectors such that $Av_i= \lambda v_i$ for i= 1, 2, ..., n. Suppose B is similar to A: $A= Q^{-1}BQ$ for some invertible matrix Q.

Then, for every i from 1 to n, $B(Qv_i)= Q(Q^{-1}BQ)v_i= QAv_i= Q(\lambda v_i)= \lambda (Qv_i)$
so that every $Qv_i$ is an eigenvector of B corresponding to eigenvalue $\lambda$. To show that they are independent, write $a_1(Qv_1)+ a_2(Qv_2)+ \cdot\cdot\cdot+ a_n(Qv_n)= Q(a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n)= 0$. Since Q is invertible, it is one-to-one: we have $a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n= 0$. Since the $v_i$ vectors are independent, we have $a_1= a_2= \cdot\cdot\cdot= a_n= 0$ so the $\{Qv_1, Qv_2, \cdot\cdot\cdot, Qv_n\}$ are n independent eigenvectors of B corresponding to eigenvalue $\lambda$.