[proofplan] We compare two decompositions of the finite-dimensional space $V$: a decomposition coming from an eigenbasis when $T$ is diagonalizable, and the internal direct sum of eigenspaces for distinct eigenvalues. In the diagonalizable direction, we write the matrix of $T$ in an eigenbasis and read the [characteristic polynomial](/page/Characteristic%20Polynomial) from its diagonal entries, which identifies the number of basis vectors of each eigenvalue with the exponent $m_j$. Conversely, if every eigenspace has dimension equal to its algebraic multiplicity, then the sum of all eigenspaces has dimension $\sum_{j=1}^r m_j=\dim_F V$; since distinct eigenspaces are linearly independent, this sum is direct and equals all of $V$. A basis obtained by concatenating bases of the eigenspaces is therefore an eigenbasis, so $T$ is diagonalizable. [/proofplan]

[step:Read the algebraic multiplicities from an eigenbasis when $T$ is diagonalizable] Since $V$ is nonzero and finite-dimensional, define $n:=\dim_F V \in \mathbb{N}$. Assume that $T$ is diagonalizable over $F$. By definition, there exists an ordered basis \begin{align*} \mathcal B=(v_1,\ldots,v_n) \end{align*} of $V$ consisting of eigenvectors of $T$. For each $i \in \{1,\ldots,n\}$, let $\mu_i \in F$ denote the eigenvalue satisfying \begin{align*} T v_i=\mu_i v_i. \end{align*} With respect to the ordered basis $\mathcal B$, the matrix of $T$ is the diagonal matrix \begin{align*} A=\operatorname{diag}(\mu_1,\ldots,\mu_n)\in F^{n\times n}. \end{align*} Therefore its characteristic polynomial is \begin{align*} \chi_T(X)=\det(XI_n-A)=\prod_{i=1}^{n}(X-\mu_i). \end{align*} Since the given factorisation \begin{align*} \chi_T(X)=\prod_{j=1}^{r}(X-\lambda_j)^{m_j} \end{align*} has distinct linear factors, uniqueness of factorisation in $F[X]$ implies that each $\mu_i$ is one of the $\lambda_j$, and that for each fixed $j \in \{1,\ldots,r\}$ exactly $m_j$ of the basis vectors $v_i$ have eigenvalue $\lambda_j$. For each $j \in \{1,\ldots,r\}$, define \begin{align*} \mathcal B_j:=\{v_i \in \mathcal B:T v_i=\lambda_j v_i\}. \end{align*} Then $\mathcal B_j \subset E_{\lambda_j}(T)$ and $|\mathcal B_j|=m_j$. Because $\mathcal B$ is the disjoint union of the sets $\mathcal B_1,\ldots,\mathcal B_r$, every vector $x \in V$ has a unique expression \begin{align*} x=\sum_{i=1}^{n} a_i v_i \end{align*} with $a_i \in F$. If $x \in E_{\lambda_j}(T)$, then comparing the coefficients in \begin{align*} T x=\lambda_j x \end{align*} gives $a_i=0$ whenever $v_i \notin \mathcal B_j$, because then $\mu_i-\lambda_j\neq 0$. Hence $\mathcal B_j$ spans $E_{\lambda_j}(T)$. Since $\mathcal B_j$ is linearly independent as a subset of the basis $\mathcal B$, it is a basis of $E_{\lambda_j}(T)$, and therefore \begin{align*} \dim_F E_{\lambda_j}(T)=|\mathcal B_j|=m_j. \end{align*} [/step]

[step:Use the dimension hypothesis to force the eigenspaces to fill $V$]Conversely, assume that \begin{align*} \dim_F E_{\lambda_j}(T)=m_j \end{align*} for every $j \in \{1,\ldots,r\}$. For each $j$, write \begin{align*} E_j:=E_{\lambda_j}(T). \end{align*} Because $\lambda_1,\ldots,\lambda_r$ are distinct, [citetheorem:8488] implies that the eigenspaces $E_1,\ldots,E_r$ form an internal direct sum: if $w_j \in E_j$ and \begin{align*} \sum_{j=1}^{r} w_j=0, \end{align*} then $w_j=0$ for every $j$. Indeed, after discarding the zero vectors among $w_1,\ldots,w_r$, the remaining nonzero vectors are eigenvectors belonging to distinct eigenvalues, so the cited theorem forces all their coefficients in the displayed relation to vanish. Hence \begin{align*} \dim_F(E_1+\cdots+E_r)=\sum_{j=1}^{r}\dim_F E_j. \end{align*} Using the assumed dimension equalities, we obtain \begin{align*} \dim_F(E_1+\cdots+E_r)=\sum_{j=1}^{r}m_j. \end{align*} Since $\chi_T$ has degree $\dim_F V$ and \begin{align*} \chi_T(X)=\prod_{j=1}^{r}(X-\lambda_j)^{m_j}, \end{align*} we have \begin{align*} \sum_{j=1}^{r}m_j=\dim_F V. \end{align*} Therefore \begin{align*} \dim_F(E_1+\cdots+E_r)=\dim_F V. \end{align*} Since $E_1+\cdots+E_r$ is a subspace of $V$ with the same finite dimension as $V$, it follows that \begin{align*} V=E_1\oplus\cdots\oplus E_r. \end{align*}[/step]

[guided]We now prove the converse in a way that explains why the dimension condition is exactly the right one. Assume that \begin{align*} \dim_F E_{\lambda_j}(T)=m_j \end{align*} for every $j \in \{1,\ldots,r\}$. To simplify notation, define the subspace \begin{align*} E_j:=E_{\lambda_j}(T) \end{align*} for each $j \in \{1,\ldots,r\}$. The first point is that eigenspaces corresponding to distinct eigenvalues cannot overlap in a way that creates a nontrivial linear relation. More precisely, let $w_j \in E_j$ for each $j \in \{1,\ldots,r\}$, and suppose that \begin{align*} \sum_{j=1}^{r}w_j=0. \end{align*} Discard the indices for which $w_j=0$. If any nonzero vectors remain, each remaining $w_j$ is an eigenvector of $T$ with eigenvalue $\lambda_j$, and the corresponding eigenvalues are distinct. Therefore [citetheorem:8488] applies to the remaining nonzero vectors and says that they are linearly independent. The displayed relation is a linear relation among those remaining vectors with all coefficients equal to $1$, so [linear independence](/page/Linear%20Independence) forces no nonzero vector to remain. Hence every $w_j$ is zero, and the sum $E_1+\cdots+E_r$ is an internal direct sum. For an internal direct sum of finite-dimensional subspaces, dimensions add. Hence \begin{align*} \dim_F(E_1+\cdots+E_r)=\sum_{j=1}^{r}\dim_F E_j. \end{align*} Now we use the hypothesis of the theorem: each geometric multiplicity equals the corresponding exponent $m_j$ in the characteristic polynomial. Therefore \begin{align*} \dim_F(E_1+\cdots+E_r)=\sum_{j=1}^{r}m_j. \end{align*} The remaining question is: why is this sum equal to the whole dimension of $V$? The characteristic polynomial of a linear operator on an $n$-dimensional [vector space](/page/Vector%20Space) has degree $n$, where $n=\dim_F V$. The displayed factorisation has degree \begin{align*} \sum_{j=1}^{r}m_j, \end{align*} because each factor $(X-\lambda_j)^{m_j}$ has degree $m_j$. Since these are two expressions for the same polynomial $\chi_T$, their degrees are equal, and so \begin{align*} \sum_{j=1}^{r}m_j=\dim_F V. \end{align*} Combining this with the preceding dimension computation gives \begin{align*} \dim_F(E_1+\cdots+E_r)=\dim_F V. \end{align*} The subspace $E_1+\cdots+E_r$ is contained in $V$, and both spaces have the same finite dimension. Therefore the subspace must be all of $V$: \begin{align*} V=E_1\oplus\cdots\oplus E_r. \end{align*}[/guided]

[step:Construct an eigenbasis from the direct sum decomposition] For each $j \in \{1,\ldots,r\}$, choose an ordered basis $\mathcal C_j$ of the finite-dimensional subspace $E_j$. Every vector in $\mathcal C_j$ is an eigenvector of $T$ with eigenvalue $\lambda_j$, because $E_j=E_{\lambda_j}(T)$. Since \begin{align*} V=E_1\oplus\cdots\oplus E_r, \end{align*} the ordered list obtained by concatenating the bases \begin{align*} \mathcal C_1,\ldots,\mathcal C_r \end{align*} is a basis of $V$. Every vector in this concatenated basis is an eigenvector of $T$. Hence $V$ has a basis consisting of eigenvectors of $T$, so $T$ is diagonalizable over $F$. This proves the converse implication, and together with the first step completes the proof. [/step]