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Mathematics LibreTexts

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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