Geometric multiplicity of an eigenvalue

Geometric multiplicity of an eigenvalue of a matrix is the measurement of the equivalent eigenspace. The algebraic multiplicity is its multiplicity as an origin of the particular polynomial.

It is recognized that the geometric multiplicity of an eigenvalue can not be more than the algebraic multiplicity . This reality can be revealed conveniently making use of the Jordan normal form of a matrix.

I was asking yourself if there is an extra primary means to confirm this reality, perhaps much longer yet without making use of the Jordan regular kind? (This is a workout in Kreyszig's book on functional analysis, and also offered the writer is design, I believe that he did not plan the remedy to make use of Jordan kind, due to the fact that or else I presume he would certainly have offered a tip concerning that. Yet I could be incorrect.)

0
2019-05-19 00:05:46
Source Share
Answers: 1

You do not require the Jordan kind: intend the geometric multiplicity of $\lambda$ is $k$, and also allow $\gamma=\{\mathbf{v}_1,\ldots,\mathbf{v}_k\}$ be a basis for the equivalent eigenspace. Expand the basis $\gamma$ to a basis $\beta$ for $F^n$, and also allow $Q$ be the adjustment - of - basis matrix. After that the particular polynomials of $A$ and also $Q^{-1}AQ$ coincide. The upper left $k\times k$ block of $Q^{-1}AQ$ is merely $\lambda I_k$, and also, the $(n-k)\times k$ block under it is all absolutely nos. So the particular polynomial of $Q^{-1}AQ$ is a numerous of $(\lambda - t)^k$ therefore the algebraic multiplicity of $\lambda$ goes to the very least $k$.

0
2019-05-21 10:06:38
Source