Linear Algebra – Eigenvalues and eigenvectors – Diagonalization

Definition: A square matrix \(A\) is called diagonalizable if it is similar to a diagonal matrix \(D\), that is: if there exists a diagonal matrix \(D\) and an invertible matrix \(P\) such that \(A=PDP^{-1}\).

Theorem: An \(n\times n\) matrix \(A\) is diagonalizable if and only if \(A\) has \(n\) linearly independent eigenvectors.

In fact, \(A=PDP^{-1}\) with \(D\) a diagonal matrix if and only if the columns of \(P\) are \(n\) linearly independent eigenvectors of \(A\). In this case, the diagonal entries of \(D\) are eigenvalues of \(A\) that correspond, respectively, to the eigenvectors in \(P\).

Proof: Let \(P\) be any \(n\times n\) matrix with columns \(\mathbf{v}_1,\ldots,\mathbf{v}_n\) and let \(D\) be any diagonal matrix with diagonal entries \(\lambda_1,\ldots,\lambda_n\), then:

\[AP=A\Bigg(\mathbf{v}_1\;\mathbf{v}_2\;\ldots\;\mathbf{v}_n\Bigg)=\Bigg(A\mathbf{v}_1\;A\mathbf{v}_2\;\ldots\;A\mathbf{v}_n\Bigg) \quad\text{and}\quad PD=P\begin{pmatrix}\lambda_1&0&\ldots&0\\0&\lambda_2&\ldots&0\\\vdots&&\ddots&\vdots\\0&0&\ldots&\lambda_n\end{pmatrix} =\Bigg(\lambda_1\mathbf{v}_1\;\lambda_2\mathbf{v}_2\;\ldots\;\lambda_n\mathbf{v}_n\Bigg).\]

So, if \(A\) is diagonalizable and \(A=PDP^{-1}\), then we have \(AP=PD\) and therefore \(A\mathbf{v}_k=\lambda_k\mathbf{v}_k\) with \(k=1,2,\ldots,n\). Since \(P\) is invertible, its columns \(\mathbf{v}_1,\ldots,\mathbf{v}_n\) are linearly independent.

The other way around, if \(\mathbf{v}_1,\ldots,\mathbf{v}_n\) are \(n\) linearly independent eigenvectors of \(A\) corresponding to the eigenvalues \(\lambda_1,\ldots,\lambda_n\) respectively, then we can construct the matrices \(P\) and \(D\) as above. Then we have: \(AP=PD\). Since the columns of \(P\) are linearly independent, \(P\) is invertible which implies that \(A=PDP^{-1}\).

Theorem: An \(n\times n\ matrix with \(n\) different (real) eigenvalues is diagonalizable.

Theorem: An \(n\times n\) matrix with less than \(n\) (real) eigenvalues, counted with multiplicity, is not diagonalizable.

Theorem: An \(n\times n\) matrix with \(n\) real eigenvalues, counted with multiplicity, is only diagonalizable if for each eigenvalue its algebraic multiplicity is equal to its geometric multiplicity.

Examples

1) Consider \(A=\begin{pmatrix}2&3\\3&-6\end{pmatrix}\). Then we have: \(\det(A-\lambda I)=\begin{vmatrix}2-\lambda&3\\3&-6-\lambda\end{vmatrix} =\lambda^2+4\lambda-21=(\lambda+7)(\lambda-3)\). This implies that \(A\) has two different eigenvalues \(\lambda_1=-7\) and \(\lambda_2=3\). Hence: \(A\) is diagonizable. In order to find a matrix \(P\) such that \(A=PDP^{-1}\) we obtain:

\[\lambda_1=-7:\quad\begin{pmatrix}9&3\\3&1\end{pmatrix}\sim\begin{pmatrix}3&1\\0&0\end{pmatrix}\quad\Longrightarrow\quad \mathbf{v}_1=\begin{pmatrix}-1\\3\end{pmatrix}\]

and

\[\lambda_2=3:\quad\begin{pmatrix}-1&3\\3&-9\end{pmatrix}\sim\begin{pmatrix}-1&3\\0&0\end{pmatrix}\quad\Longrightarrow\quad \mathbf{v}_2=\begin{pmatrix}3\\1\end{pmatrix}.\]

This implies that for instance \(A=PDP^{-1}\) with \(P=\begin{pmatrix}-1&3\\3&1\end{pmatrix}\) and \(D=\begin{pmatrix}-7&0\\0&3\end{pmatrix}\). However, for instance we also have \(A=PDP^{-1}\) with \(P=\begin{pmatrix}3&1\\1&-3\end{pmatrix}\) and \(D=\begin{pmatrix}3&0\\0&-7\end{pmatrix}\) or for instance \(A=PDP^{-1}\) with \(P=\begin{pmatrix}9&-1\\3&3\end{pmatrix}\) and \(D=\begin{pmatrix}3&0\\0&-7\end{pmatrix}\).

2) Consider \(A=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\). Then we have: \(\det(A-\lambda I)=\begin{vmatrix}-\lambda&-1\\1&-\lambda\end{vmatrix} =\lambda^2+1\). This implies that \(A\) does not have real eigenvalues, which implies that \(A\) is not diagonizable.

3) Consider \(A=\begin{pmatrix}1&-1\\1&3\end{pmatrix}\). Then we have: \(\det(A-\lambda I)=\begin{vmatrix}1-\lambda&-1\\1&3-\lambda\end{vmatrix} =\lambda^2-4\lambda+4=(\lambda-2)^2\). This implies that \(\lambda=2\) is the only eigenvalue with algebraic multiplicity \(2\). Now we obtain:

\[\lambda=2:\quad\begin{pmatrix}-1&-1\\1&1\end{pmatrix}\sim\begin{pmatrix}1&1\\0&0\end{pmatrix}\quad\Longrightarrow\quad E_2=\text{Span}\left\{\begin{pmatrix}1\\-1\end{pmatrix}\right\}.\]

So, the geometric multiplicity of the eigenvalue \(\lambda=2\) equals \(1\). Hence, the matrix \(A\) is not diagonizable.

If a square matrix \(A\) is diagonizable, then it is quite easy to compute powers of \(A\). Suppose that \(A=PDP^{-1}\) with \(P\) an invertible matrix and \(D=\text{diag}(\lambda_1,\ldots,\lambda_n)\), then we have:

\[A^k=(PDP^{-1})^k=\underbrace{PDP^{-1}PDP^{-1}\cdots PDP^{-1}}_{k\;\text{factors}}=PD^kP^{-1}\]

with \(D^k=\text{diag}(\lambda_1^k,\ldots,\lambda_n^k)\).

Examples

1) If \(A=\begin{pmatrix}1&0&0\\-2&1&2\\2&0&-1\end{pmatrix}\), then we have

\[\det(A_\lambda I)=\begin{vmatrix}1-\lambda&0&0\\-2&1-\lambda&2\\2&0&-1-\lambda\end{vmatrix}=(1-\lambda)\begin{vmatrix}1-\lambda&2\\0&-1-\lambda\end{vmatrix} =(1-\lambda)^2(-1-\lambda).\]

So, the eigenvalues of \(A\) are \(\lambda_1=1\) with algebraic multiplicity \(2\) and \(\lambda_2=-1\) with algebraic multiplicity \(1\). Then we have:

\[\lambda_1=1:\quad\begin{pmatrix}0&0&0\\-2&0&2\\2&0&-2\end{pmatrix}\sim\begin{pmatrix}1&0&-1\\0&0&0\\0&0&0\end{pmatrix} \quad\Longrightarrow\quad\text{E}_1=\text{Span}\left\{\begin{pmatrix}1\\0\\1\end{pmatrix},\begin{pmatrix}0\\1\\0\end{pmatrix}\right\}\]

and

\[\lambda_2=-1:\quad\begin{pmatrix}2&0&0\\-2&2&2\\2&0&0\end{pmatrix}\sim\begin{pmatrix}1&0&0\\0&1&1\\0&0&0\end{pmatrix} \quad\Longrightarrow\quad\text{E}_{-1}=\text{Span}\left\{\begin{pmatrix}0\\1\\-1\end{pmatrix}\right\}.\]

Hence we have \(A=PDP^{-1}\) with \(P=\begin{pmatrix}1&0&0\\0&1&1\\1&0&-1\end{pmatrix}\) and \(D=\begin{pmatrix}1&0&0\\0&1&0\\0&0&-1\end{pmatrix}\). For instance, this implies that \(A^{100}=PD^{100}P^{-1}=PIP^{-1}=PP^{-1}=I\) and \(A^{151}=PD^{151}P^{-1}=PDP^{-1}=A\).

2) If \(A=\begin{pmatrix}5&-3\\6&-4\end{pmatrix}\), then we have

\[\det(A-\lambda I)=\begin{vmatrix}5-\lambda&-3\\6&-4-\lambda\end{vmatrix}=\lambda^2-\lambda-2=(\lambda-2)(\lambda+1).\]

So, the eigenvalues of \(A\) are \(\lambda_1=2\) and \(\lambda_2=-1\), both with algebraic multiplicity \(1\). Then we have:

\[\lambda_1=2:\quad\begin{pmatrix}3&-3\\6&-6\end{pmatrix}\sim\begin{pmatrix}1&-1\\0&0\end{pmatrix}\quad\Longrightarrow\quad \text{E}_2=\text{Span}\left\{\begin{pmatrix}1\\1\end{pmatrix}\right\}\]

and

\[\lambda_2=-1:\quad\begin{pmatrix}6&-3\\6&-3\end{pmatrix}\sim\begin{pmatrix}2&-1\\0&0\end{pmatrix}\quad\Longrightarrow\quad \text{E}_{-1}=\text{Span}\left\{\begin{pmatrix}1\\2\end{pmatrix}\right\}.\]

Hence we have \(A=PDP^{-1}\) with \(P=\begin{pmatrix}1&1\\1&2\end{pmatrix}\) and \(D=\begin{pmatrix}2&0\\0&-1\end{pmatrix}\) for instance. As an example we now have that

\[A^{10}=PD^{10}P^{-1}=P\begin{pmatrix}2^{10}&0\\0&(-1)^{10}\end{pmatrix}P^{-1} =\begin{pmatrix}1&1\\1&2\end{pmatrix}\begin{pmatrix}1024&0\\0&1\end{pmatrix}\begin{pmatrix}2&-1\\-1&1\end{pmatrix} =\begin{pmatrix}1&1\\1&2\end{pmatrix}\begin{pmatrix}2048&-1024\\-1&1\end{pmatrix}=\begin{pmatrix}2047&-1023\\2046&-1022\end{pmatrix}.\]

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Last modified on April 5, 2021