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# 13.3: Changing to a Basis of Eigenvectors

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