[proofplan] We compare the columns of the matrix $[T]_{\mathcal B \leftarrow \mathcal B}$ with the coordinates of the vectors $T(v_j)$ in the basis $\mathcal B$. A diagonal matrix has each column supported only in its diagonal entry, which says exactly that $T(v_j)$ is a scalar multiple of $v_j$. Conversely, if every basis vector is an eigenvector, then each coordinate column has only one nonzero possible entry, so the matrix is diagonal. [/proofplan] [step:Express each matrix column as the coordinate vector of $T(v_j)$] Let $A=[T]_{\mathcal B \leftarrow \mathcal B}=(a_{ij})_{1\leq i,j\leq n}\in k^{n\times n}$. By the definition of the matrix of a [linear map](/page/Linear%20Map) with respect to the ordered basis $\mathcal B$, the $j$th column of $A$ is the coordinate column of $T(v_j)$ in $\mathcal B$. Therefore, for each $j\in\{1,\ldots,n\}$, $T(v_j)=\sum_{i=1}^{n} a_{ij}v_i$. [guided] Let $A=[T]_{\mathcal B \leftarrow \mathcal B}=(a_{ij})_{1\leq i,j\leq n}\in k^{n\times n}$. The role of $A$ is to record how $T$ acts on basis vectors, then extend that information linearly to all vectors in $V$. More precisely, the $j$th column of $A$ is, by definition, the coordinate vector of $T(v_j)$ in the same ordered basis $\mathcal B=(v_1,\ldots,v_n)$. Thus the scalars $a_{1j},\ldots,a_{nj}\in k$ are exactly the coefficients in the basis expansion of $T(v_j)$: $T(v_j)=\sum_{i=1}^{n} a_{ij}v_i$. This column-by-column interpretation is the bridge between the matrix statement and the eigenvector statement. [/guided] [/step] [step:Use diagonality to show each basis vector is an eigenvector] Assume that $A$ is diagonal. Then $a_{ij}=0$ whenever $i\neq j$. Hence, for each $j\in\{1,\ldots,n\}$, the coordinate formula above becomes $T(v_j)=a_{jj}v_j$. Since $v_j$ is a basis vector, $v_j\neq 0$. Therefore $v_j$ is an eigenvector of $T$ with eigenvalue $a_{jj}$. Since this holds for every $j$, the ordered basis $\mathcal B$ is an eigenbasis for $T$. [/step] [step:Use the eigenbasis property to force every off-diagonal entry to vanish] Conversely, assume that $\mathcal B$ is an eigenbasis for $T$. Then, for each $j\in\{1,\ldots,n\}$, there exists a scalar $\lambda_j\in k$ such that $T(v_j)=\lambda_j v_j$. Writing the right-hand side in the basis $\mathcal B$, its $i$th coordinate is $\lambda_j$ if $i=j$ and $0$ if $i\neq j$. Comparing this coordinate expansion with $T(v_j)=\sum_{i=1}^{n}a_{ij}v_i$ and using uniqueness of coordinates in a basis, we obtain $a_{jj}=\lambda_j$ and $a_{ij}=0\quad\text{for every }i\neq j$. Thus every off-diagonal entry of $A$ is zero, so $A=[T]_{\mathcal B \leftarrow \mathcal B}$ is diagonal. [/step]