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Mathematics LibreTexts

13.3: Changing to a Basis of Eigenvectors

  • Page ID
    2083
  • [ "article:topic", "authortag:waldron", "authorname:waldron" ]

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    If we are changing to a basis of eigenvectors, then there are various simplifications:

    1. Since \(L:V\to V\), most likely you already know the matrix \(M\) of \(L\) using the same input basis as output basis \(S=(u_{1},\ldots ,u_{n})\) (say).

    2. In the new basis of eigenvectors \(S'(v_{1},\ldots,v_{n})\), the matrix \(D\) of \(L\) is diagonal because \(Lv_{i}=\lambda_{i} v_{i}\) and so

    $$
    \big(L(v_{1}),L(v_{2}),\ldots,L(v_{n})\big)=(v_{1},v_{2},\ldots, v_{n})
    \begin{pmatrix}
    \lambda_{1}&0&\cdots&0\\
    0&\lambda_{2}&&0\\
    \vdots&&\ddots&\vdots \\
    0&0&\cdots&\lambda_{n}\end{pmatrix}\, .
    $$

    3. If \(P\) is the change of basis matrix from \(S\) to \(S'\), the diagonal matrix of eigenvalues \(D\) and the original matrix are related by \(D=P^{-1}MP\).

    This motivates the following definition:

    Definition

    A matrix \(M\) is diagonalizable if there exists an invertible matrix \(P\) and a diagonal matrix \(D\) such that

    \[ D=P^{-1}MP. \]

    We can summarize as follows:

    1. Change of basis rearranges the components of a vector by the change of basis matrix \(P\), to give components in the new basis.
    2. To get the matrix of a linear transformation in the new basis, we \(\textit{conjugate}\) the matrix of \(L\) by the change of basis matrix: \(M\mapsto P^{-1}MP\).

    If for two matrices \(N\) and \(M\) there exists a matrix \(P\) such that \(M=P^{-1}NP\), then we say that \(M\) and \(N\) are \(\textit{similar}\). Then the above discussion shows that diagonalizable matrices are similar to diagonal matrices.

    Corollary

    A square matrix \(M\) is diagonalizable if and only if there exists a basis of eigenvectors for \(M\). Moreover, these eigenvectors are the columns of the change of basis matrix \(P\) which diagonalizes \(M\).

    Example 122

    Let's try to diagonalize the matrix

    \[M=\begin{pmatrix}
    -14 & -28 & -44 \\
    -7 & -14 & -23 \\
    9 & 18 & 29 \\
    \end{pmatrix}.\]

    The eigenvalues of \(M\) are determined by \[\det(M-\lambda I)=-\lambda^{3}+\lambda^{2}+2\lambda=0.\]

    So the eigenvalues of \(M\) are \(-1,0,\) and \(2\), and associated eigenvectors turn out to be 

    $$v_{1}=\begin{pmatrix}-8 \\ -1 \\ 3\end{pmatrix},~~ v_{2}=\begin{pmatrix}-2 \\ 1 \\ 0\end{pmatrix}, {\rm ~and~~} v_{3}=\begin{pmatrix}-1 \\ -1 \\ 1\end{pmatrix}.$$ 

    In order for \(M\) to be diagonalizable, we need the vectors \(v_{1}, v_{2}, v_{3}\) to be linearly independent. Notice that the matrix

    \[P=\begin{pmatrix}v_{1} & v_{2} & v_{3}\end{pmatrix}=\begin{pmatrix}
    -8 & -2 & -1 \\
    -1 & 1 & -1 \\
    3 & 0 & 1 \\
    \end{pmatrix}\]

    is invertible because its determinant is \(-1\). Therefore, the eigenvectors of \(M\) form a basis of \(\Re\), and so \(M\) is diagonalizable. 
    Moreover, because the columns of \(P\) are the components of eigenvectors, 

    $$
    MP=\begin{pmatrix}Mv_{1} &Mv_{2}& Mv_{3}\end{pmatrix}=\begin{pmatrix}-1.v_{1}&0.v_{2}&2.v_{3}\end{pmatrix}=\begin{pmatrix}v_{1}& v_{2} & v_{3}\end{pmatrix}\begin{pmatrix}
    -1 & 0 & 0 \\
    0 & 0 & 0 \\
    0 & 0 & 2 \\
    \end{pmatrix}\, .
    $$

    Hence, the matrix \(P\) of eigenvectors is a change of basis matrix that diagonalizes \(M\):
    \[P^{-1}MP=\begin{pmatrix}
    -1 & 0 & 0 \\
    0 & 0 & 0 \\
    0 & 0 & 2 \\
    \end{pmatrix}.\]

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