$ \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\abs}[2][]{\left\lvert#2\right\rvert_{\text{#1}}} \newcommand{\ket}[1]{\left\lvert#1 \right.\rangle} \newcommand{\bra}[1]{\langle\left. #1\right\rvert} \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\dd}{\text{d}} \newcommand{\dv}[2]{\frac{\dd #1}{\dd #2}} \newcommand{\pdv}[2]{\frac{\partial}{\partial #1}} $
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Let $T:V\to V$. The matrix $T$ is diagoniable if and only if there is a basis for $V$ consisting of eigenvectors of $T$. In fact, for a finite dimensional vector space and $T:V\to V$ linear, the following are equivalent:
  1. $T$ is diagonalizable
  2. The characteristic polynomial of $T$ is $P_T(t)=(\mu_1-t)^{n_1}\dots (\mu_k-t)^{n_k}$, and $\text{dim}E_{\mu_i}=n_i$, where the $\mu_i$'s are distinct
  3. $\text{dim}E_{\mu_1}+\text{dim}E_{\mu_k}=\text{dim}V$

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Let $T:V\to V$. The matrix $T$ is diagoniable if and only if there is a basis for $V$ consisting of eigenvectors of $T$. In fact, for a finite dimensional vector space and $T:V\to V$ linear, the following are equivalent:
  1. $T$ is diagonalizable
  2. The characteristic polynomial of $T$ is $P_T(t)=(\mu_1-t)^{n_1}\dots (\mu_k-t)^{n_k}$, and $\text{dim}E_{\mu_i}=n_i$, where the $\mu_i$'s are distinct
  3. $\text{dim}E_{\mu_1}+\text{dim}E_{\mu_k}=\text{dim}V$

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result
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If
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proof
Let $T:V\to V$. The matrix $T$ is diagoniable if and only if there is a basis for $V$ consisting of eigenvectors of $T$. In fact, for a finite dimensional vector space and $T:V\to V$ linear, the following are equivalent:
  1. $T$ is diagonalizable
  2. The characteristic polynomial of $T$ is $P_T(t)=(\mu_1-t)^{n_1}\dots (\mu_k-t)^{n_k}$, and $\text{dim}E_{\mu_i}=n_i$, where the $\mu_i$'s are distinct
  3. $\text{dim}E_{\mu_1}+\text{dim}E_{\mu_k}=\text{dim}V$

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Proof

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Let $T:V\to V$. The matrix $T$ is diagoniable if and only if there is a basis for $V$ consisting of eigenvectors of $T$. In fact, for a finite dimensional vector space and $T:V\to V$ linear, the following are equivalent:
  1. $T$ is diagonalizable
  2. The characteristic polynomial of $T$ is $P_T(t)=(\mu_1-t)^{n_1}\dots (\mu_k-t)^{n_k}$, and $\text{dim}E_{\mu_i}=n_i$, where the $\mu_i$'s are distinct
  3. $\text{dim}E_{\mu_1}+\text{dim}E_{\mu_k}=\text{dim}V$

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