Exercise 4.1.9

Question

Use the two previous exercises to prove that geometric multiplicity of an eigenvalue cannot exceed its algebraic multiplicity.

Jing Huang · Accepted Answer

\begin{proof}
We consider the problem in the basis $\mathbf{v}_1, \mathbf{v}_2, ..., \mathbf{v}_n$ and $A$ has the block triangular form shown in Problem 1.8. Note $k$ is the number of linearly independent eigenvectors corresponding to $\lambda_k$, which is also the dimension of Ker$(A - \lambda_k I)$ (consider the queation $(A - \lambda_k I) \mathbf{x} = \mathbf{0}$). Namely, $k$ is the geometric multiplicity of $\lambda_k$.  \par
For the algebraic multiplicity, consider the determinant
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\begin{equation*}
    \begin{split}
    \det (A - \lambda I) &= 
    \begin{vmatrix}
    (\lambda_k - \lambda) I_k  &*  \
    \mathbf{0}  & B - \lambda I_{n-k}
    \end{vmatrix}  \
    &=
    (\lambda_k - \lambda)^k \det(B - \lambda I_{n-k}).
    \end{split}
\end{equation*}
%%%%%%%%%%%%%%%%
So the algebraic multiplicity of $\lambda_k$ is at least $k$. It is further possible that $\lambda_k$ is a root of the polynomial $\det(B - \lambda I_{n-k})$, then in this case the algebraic multiplicity will just exceed $k$. Thus geometric multiplicity of an eigenvalue cannot exceed its algebraic multiplicity.
\end{proof}

Exercise 4.1.9

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Exercise 4.1.9

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