13.3: Changing to a Basis of Eigenvectors

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Page ID: 2083

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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:

Change of basis rearranges the components of a vector by the change of basis matrix $P$, to give components in the new basis.
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 $\PageIndex{1}$:

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 $M$:
\[P^{-1}MP=\begin{pmatrix}
-1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 2 \\
\end{pmatrix}.\]

$eigenbasis.jpg$

Contributor

David Cherney, Tom Denton, and Andrew Waldron (UC Davis)

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