
# 7.3 Diagonal matrices

Note that if $$T$$ has $$n=\dim(V)$$ distinct eigenvalues, then there exists a basis $$(v_1,\ldots,v_n)$$ of $$V$$such that
\begin{equation*}  Tv_j = \lambda_j v_j, \quad \text{for all $$j=1,2,\ldots,n$$.}
\end{equation*}

Then any $$v\in V$$ can be written as a linear combination $$v=a_1v_1+\cdots+a_nv_n$$of $$v_1,\ldots,v_n$$. Applying $$T$$ to this, we obtain \begin{equation*}
Tv = \lambda_1 a_1 v_1 + \cdots + \lambda_n a_n v_n.
\end{equation*}

Hence the vector \begin{equation*}
M(v) = \begin{bmatrix} a_1 \\ \vdots \\ a_n \end{bmatrix}
\end{equation*}

is mapped to \begin{equation*}
M(Tv) = \begin{bmatrix} \lambda_1 a_1 \\ \vdots \\ \lambda_n a_n \end{bmatrix}.
\end{equation*}

This means that the matrix $$M(T)$$for $$T$$ with respect to the basis of eigenvectors $$(v_1,\ldots,v_n)$$ is diagonal, and so we call $$T$$ diagonalizable: \begin{equation*}
M(T) = \begin{bmatrix} \lambda_1 & & 0 \\  & \ddots & \\
0& & \lambda_n \end{bmatrix}.
\end{equation*}

We summarize the results of the above discussion in the following Proposition.

Proposition 7.3.1. If $$T\in \mathcal{L}(V,V)$$ has $$\dim(V)$$ distinct eigenvalues, then $$M(T)$$ is diagonal with respect to some basis of $$V$$. Moreover, $$V$$ has a basis consisting of eigenvectors of $$T$$. ​

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