I have the following linear algebra proof:
Question: Let $V$ be an n-dimensional vector space, $\lambda \in \mathbb{R}$, and $T : V \rightarrow V$ a linear map with det($T − tI$) = ($\lambda − t)^n$. Show that $T$ is diagonalizable if and only if $T = λI$.
I'm not entirely sure where, to begin with, this, in particular how to use the fact that det($T − tI$) = ($\lambda − t)^n$. How does that relate to diagonalizability?
It's helpful to recall some definitions. A linear map is diagonalisable if the algebraic multiplicity of each eigenvalue $\lambda$ (its multiplicity as a solution of the characteristic equation $\det (T-\lambda I)=0$) equals its geometric multiplicity of $\lambda$ (how many linearly independent eigenvectors it has). In this case, $\det (T-\lambda I) = (\lambda -t )^n$, so $T$ only has one eigenvalue ($\lambda$) with algebraic multiplicity $n$.
Hence, proving $T$ is diagonal is equivalent to proving there is a basis $\{ v_j\}_{j=1}^n$ of $V$ with each $v_j$ an eigenvector with corresponding eigenvalue $\lambda$.
Sufficient direction: Suppose $T$ is diagonalisable. Then there is a basis $\{ v_j\}_{j=1}^n$ of $V$ with each $v_j$ an eigenvector with corresponding eigenvalue $\lambda$. Hence, for each $x \in V$ we may write $$x =\sum_{j=1}^n x_jv_j $$ for some $x_j \in \mathbb{R}$ (or $\mathbb{F}$ depending on the field $V$ is over). It follows \begin{align*} Tx &= \sum_{j=1}^n x_j T v_j \\ &= \sum_{j=1}^n x_j (\lambda v_j) \\ &= \lambda x . \end{align*} Thus, $T = \lambda I$.
Necessary direction: Suppose $T=\lambda I$. Let $\{v_j\}_{j=1}^n$ be any basis for $V$. Then for each $j$, $(T-\lambda I)v_j = 0$, so each $v_j$ is an eigenvector with corresponding eigenvalue $\lambda$. Thus, $T$ is diagonalisable.