# 6.2 Diagonalization of Matrices

Given an $n\times n$ matrix **A**, we may ask *how many* linearly independent eigenvectors the matrix **A** has. In Section 6.1, we saw several examples (with $n=2$ and $n=3$) in which the $n\times n$ matrix **A** has `n` linearly independent eigenvectors—the largest possible number. By contrast, in Example 5 of Section 6.1, we saw that the $2\times 2$ matrix

has the single eigenvalue $\lambda =2$ corresponding to the single eigenvector $\mathbf{\text{v}}={\left[\begin{array}{rr}1& 0\end{array}\right]}^{T}.$

Something very nice happens when the $n\times n$ matrix **A** does have `n` linearly independent eigenvectors. Suppose that the eigenvalues ${\lambda}_{1},{\lambda}_{2},\dots ,{\lambda}_{n}$ (not necessarily distinct) of **A** correspond to the `n` linearly independent eigenvectors ${\mathbf{\text{v}}}_{1},{\mathbf{\text{v}}}_{2},\dots ,{\mathbf{\text{v}}}_{n}$, respectively. Let

be the $n\times n$ matrix having these eigenvectors ...

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