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.
 A: I will give more details to other answers.
For a specific $\lambda_i$, the idea is to transform the matrix $A$ (n by n) to matrix $B$ which shares the same eigenvalues as $A$. If $P_1=[v_1, \cdots, v_m]$ are eigenvectors of $\lambda_i$, we expand it to the basis $P=[P_1, P_2]=[v_1, \cdots, v_m, \cdots, v_n]$. Therefore, $AP=[\lambda_i P_1, AP_2]$. In order to make $P^{-1}AP=B$, we must have $AP=PB$. Let $B= \begin{bmatrix} B_{11} & B_{12} \\
 B_{21} & B_{22} \end{bmatrix}$, then $P_1B_{11}+P_2B_{21}=\lambda_i P_1$ and $P_1B_{12}+P_2B_{22}=AP_2$. Because $P$ is a basis, $B_{11}=\lambda_iI$(m by m), $B_{21}=\mathbf{0}$ ($(n-m)\times m$). So, we have $B=\begin{bmatrix} \lambda_iI & B_{12}\\ \mathbf{0} & B_{22} \end{bmatrix}$.
$\det(A-\lambda I)=\det(P^{-1}(A-\lambda I)P)=\det(P^{-1}AP-\lambda I)=det(B-\lambda I)$ $ = \det \Bigg(\begin{bmatrix} (\lambda_i-\lambda)I & B_{12}\\ \mathbf{0} & B_{22}-\lambda I \end{bmatrix} \Bigg)=(\lambda_i-\lambda)^{m}\det(B_{22}-\lambda I)$.
Obviously, $m$ is no bigger than the algebraic multiplicity of $\lambda_i$.
A: 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 $k$.
A: Another way to think about this, if it is your bag and you want to stick to algebraically closed, is that the algebraic multiplicity of an eigenvalue $\lambda$ for a matrix $A$ is the total number of times $\lambda$ shows up in the Jordan matrix $J$ associate to $A$, and the geometric multiplicity is the total number of Jordan blocks in $J$ associated to $\lambda$. More visually, suppose that 
$$A\sim J=\begin{pmatrix}J_{n_1}(\lambda) & & & & &\\ & J_{n_{2}}(\lambda) & & & & \\ & & \ddots & &\\\ & & & J_{n_{m}}(\lambda) & & & \\ & & & & \text{other Jordan blocks with different eigenvalue}\\ & & & & & &  \end{pmatrix}$$
then the geometric multiplicity of $\lambda$ is $m$ and the algebraic multiplicity is $n_1+\cdots+n_m$.
While this answer is definitively more sophisticated (maybe overly so) than the other answers, it is helpful (in the case of algebraically closed fields) to always think in terms of the Jordan matrix--it is the answer to all of your problems. For example, the multiplicity of $\lambda$ in the minimal polynomial for $A$ is just $\max\{n_i\}$.  
A: Suppose $\lambda$ has algebraic multiplicity $k$. You can find a basis $(a_1,\dots,a_k)$ of $\operatorname{Ker}\left(A-\lambda I\right)$. Complete it to get a basis of $\Bbb R^n$. Now with a change of basis (which conserves eigenvalues and both of their multiplicities), you can get a matrix of the form $B=\begin{pmatrix}\lambda I_k&C\\0&D\end{pmatrix}$. And you get your result since $\det (B-t I_n)=\det (\lambda I_k-tI_k)\det(D-tI_{n-k})=(\lambda - t)^k\det(D-tI_{n-k})$
