why geometric multiplicity is bounded by algebraic multiplicity? Define * The _algebraic multiplicity_ of $\lambda_{i}$ to be the degree of the root $\lambda_i$ in the polynomial $\det(A-\lambda I)$. * The _geometric multiplicity_ the be the dimension of the eigenspace associated with the eigenvalue $\lambda_i$. For example: $\begin{bmatrix}1&1\\\0&1\end{bmatrix}$ has root $1$ with algebraic multiplicity $2$, but the geometric multiplicity $1$. **My Question** : Why is the geometric multiplicity always bounded by algebraic multiplicity? Thanks.

Suppose the geometric multiplicity of the eigenvalue $\lambda$ of $A$ is $k$. Then we have $k$ linearly independent vectors $v_1,\ldots,v_k$ such that $Av_i=\lambda v_i$. If we change our basis so that the first $k$ elements of the basis are $v_1,\ldots,v_k$, then with respect to this basis we have $$A=\begin{pmatrix} \lambda I_k & B \\\ 0 & C \end{pmatrix}$$ where $I_k$ is the $k\times k$ identity matrix. Since the characteristic polynomial is independent of choice of basis, we have $$\mathrm{char}_A(x)=\mathrm{char}_{\lambda I_k}(x)\mathrm{char}_{C}(x)=(x-\lambda)^k\mathrm{char}_{C}(x)$$ so the algebraic multiplicity of $\lambda$ is at least $k$.