Definition 5.2.1.
A matrix \(A\) is said to be diagonalizable if there exists a non singular matrix \(P\) such that \(P^{-1}AP\) is a diagonal matrix. That is, \(A\) is similar to a diagonal matrix.
Section 5.2 Diagonalization
Example 5.2.2.
Let \(A=\begin{pmatrix}1\amp 2\amp -2\\1\amp 1\amp 1\\1\amp 3\amp -1 \end{pmatrix}\) as in Example 5.1.9. Define \(P=\begin{pmatrix}1\amp 0\amp 8/7\\-1\amp 1\amp -5/7\\-1\amp 1\amp 1 \end{pmatrix}\text{,}\) whose columns are eigenvectors of \(A\text{.}\)
Check that \(\det(P)=\dfrac{12}{7}\) and \(\text{ adj } (P)=\begin{pmatrix}12/7 \amp 8/7 \amp -8/7\\12/7 \amp 15/7\amp -3/7\\0 \amp -1 \amp 1 \end{pmatrix}\text{.}\) Hence \(P^{-1}=\begin{pmatrix}1\amp 2/3 \amp -2/3\\1\amp 5/4\amp -1/4\\0\amp -7/4\amp 7/4 \end{pmatrix}\)
Then it is easy to check that \(P^{-1}AP=\begin{pmatrix}1\amp 0\amp 0\\0\amp 2\amp 0\\0\amp 0\amp -2 \end{pmatrix}\text{.}\)
In this case we can find any power of \(A\) quite easily. For example \(A^n=P^{-1}\begin{pmatrix}1\amp 0\amp 0\\0\amp 2^n\amp 0\\0\amp 0\amp (-2)^n \end{pmatrix} P\text{.}\)
Example 5.2.3.
Let \(A=\begin{pmatrix}1 \amp 1\\0 \amp 1 \end{pmatrix}\text{.}\) Then 1 is a repeated eigenvalue of \(A\) with eigenvector \(\begin{pmatrix}1\\ 0 \end{pmatrix}\text{.}\) It is easuy to see that \(A\) is non diagonalizable.
Checkpoint 5.2.4.
If \(A=[a_{ij}]\) has \(n\) distinct eigenvalues then \(A\) is diagonalizable. In this case, one can define \(P\) where columns of \(P\) are \(n\) eigenvectors of \(A\text{.}\)
Theorem 5.2.5.
A square matrix \(A\) of order \(n\) is diagonalizable if and only if \(A\) has \(n\) linearly independent eigenvectors.
Proof.
Let \(A\) be diagonalizable, and that there exists a non singular matrix \(P\) such that
\begin{equation}
P^{-1}AP=D=\diag(\lambda_1,\ldots,\lambda_n)\tag{5.2.1}
\end{equation}
Let us write \(P=[P_1,\ldots,P_n]\) where \(P_j\) is the \(j\)-th column of \(P\text{.}\) Then Eq. (5.2.1) implies
\begin{equation*}
AP=P\diag(\lambda_1,\ldots,\lambda_n)
\end{equation*}
This implies
\begin{equation*}
[AP_1,AP_2,\ldots,AP_n]=[\lambda_1P_1,\lambda_2P_2,\ldots,\lambda_nP_n]
\end{equation*}
Equivalently \(AP_i=\lambda_iP_i\) for \(1\leq i\leq n\text{.}\) That is same as saying columns of \(P\) are eigenvectors of \(A\) with respect to eigenvalue \(\lambda_i\text{.}\) This implies \(A\) has \(n\) linearly eigenvectors, namely columns of \(P\text{.}\)
Conversely, let \(A\) have \(n\) linearly independent eigenvectors \(v_1, \ldots,
v_n\) and that \(Av_j=\lambda_jv_j\text{.}\) Define \(P:=[v_1,v_2\ldots,v_n]\) and \(D:=\diag(\lambda_1,\ldots,\lambda_n)\text{.}\) Then
\begin{equation*}
AP=[Av_1,Av_2,\ldots,Av_n]=[\lambda_1v_1,\lambda_2v_2,\ldots,\lambda_nv_n]=PD\text{.}
\end{equation*}
Hence \(P^{-1}AP=D\text{.}\) Note that \(P\) has rank \(n\text{,}\) which implies \(P\) is invertible.
Corollary 5.2.6.
If \(A\) is a square matrix of order \(n\) has \(n\) distinct eigenvalues then \(A\) is diagonalizable.
Example 5.2.7.
Let \(A\) be a \(3\times 3\) real matrix with eigenvalues \(\lambda_1=2,\lambda_2=1,\lambda_3=-1\) and corresponding eigenvectors \(v_1=\left(1,\,\frac{1}{3},\,-\frac{1}{3}\right),
v_2=\left(1,\,0,\,-\frac{1}{3}\right),
v_3=\left(1,\,\frac{1}{2},\,-\frac{1}{2}\right)\) respectively. Then we have \(P=[v_1~v_2~v_3]=\left(\begin{array}{rrr} 1 \amp 1 \amp 1 \\ \frac{1}{3} \amp 0 \amp \frac{1}{2} \\ -\frac{1}{3} \amp -\frac{1}{3} \amp -\frac{1}{2} \end{array} \right)\text{.}\) \(P^{-1}AP=\left(\begin{array}{rrr} 2 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp -1 \end{array} \right)\text{.}\) Hence \(A = P\left(\begin{array}{rrr} 2 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp -1 \end{array} \right)P^{-1}\text{.}\) It is easy to see that \(P^{-1}=\left(\begin{array}{rrr} 3 \amp 3 \amp 9 \\ 0 \amp -3 \amp -3 \\ -2 \amp 0 \amp -6 \end{array} \right)\text{.}\)
Hence
\begin{align*}
A =\amp \left(\begin{array}{rrr} 1 \amp 1 \amp 1 \\
\frac{1}{3} \amp 0 \amp \frac{1}{2} \\
-\frac{1}{3} \amp -\frac{1}{3} \amp -\frac{1}{2} \end{array} \right)
\begin{pmatrix}2\amp 0 \amp 0\\0 \amp 1 \amp 0\\
0 \amp 0 \amp 1 \end{pmatrix}
\left(\begin{array}{rrr} 3 \amp 3 \amp 9 \\
0 \amp -3 \amp -3 \\ -2 \amp 0 \amp -6 \end{array} \right)\\
=\amp \left(\begin{array}{rrr} 8 \amp 3 \amp 21 \\
3 \amp 2 \amp 9 \\ -3 \amp -1 \amp -8 \end{array} \right)\text{.}
\end{align*}
Example 5.2.8.
Let \(A=\begin{pmatrix}-1\amp -1\amp -2\\8\amp -11\amp -8\\-10\amp 11\amp 7 \end{pmatrix}\) and \(B=\begin{pmatrix}1\amp -4\amp -4\\8\amp -11\amp -8\\-8\amp 8\amp 5 \end{pmatrix}\text{.}\)
It is easy to check that \(A\) and \(B\) have same characteristic polynomial \(\lambda^3+5\lambda^2+3\lambda-9=(\lambda-1)(\lambda+3)^2\text{.}\) Also We can show that \(A\) has only one linearly independent eigenvectors corresponding to eigenvalue \(-3\text{.}\) This implies \(A\) has only two eigenvectors and hence \(A\) is not diagonalizable.
Similarly, We can show that \(B\) has two linearly independent eigenvectors corresponding to eigenvalue \(-3\text{.}\) This implies \(B\) has three eigenvectors and hence \(B\) is diagonalizable.
We mention another criteria of diagonalizability without proof.
Theorem 5.2.9.
Let \(A\) be an \(n\times n\) real matrix with distinct eigenvalues \(\lambda_1,\ldots, \lambda_k\) and algebraic multiplicity \(m_1,\ldots,
m_k\) respectively. Then \(A\) is diagonalizable if and only if algebraic multiplicity is same as geometric multiplicity for each eigenvalue. That is \(m_i=\dim{(E_{\lambda_i})}\) for \(i=1,2,\ldots,k\text{.}\)
Example 5.2.10.
In view of theorem 5.2.9, the matrix in the example 5.1.14 is not diagonalizable.
