Let \(A \in \mathbb{C}^{n \times n}\). \(A\) is said to be diagonalizable if there exist \(P\) and \(D\) in \(\mathbb{C}^{n \times n}\) such that \(D\) is a diagonal matrix and \(A = PDP^{-1}\).
\(A\) is diagonalizable if and only if for every eigenvalue \(\lambda\) of \(A\), the algebraic multiplicity of \(\lambda\) is equal to the geometric multiplicity of \(\lambda\).
An equivalent characterization is that the sum of the geometric multiplicities of the eigenvalues of \(A\) is \(n\).
Let \(A = \begin{bmatrix} 1 & 2\\ 0 & 1\end{bmatrix}.\) Note that \(p_A = (1-\lambda)^2\). Hence, the only eigenvalue of \(A\) is \(1\). Now \(A - I = \begin{bmatrix} 0 & 2\\ 0 & 0\end{bmatrix}.\) The nullity of this matrix is \(1\), implying that the geometric multiplicity is \(1\), not \(2\). So \(A\) is not diagonalizable.
Let \(A = \begin{bmatrix} 4 & 0 & -2\\ 2 & 5 & 4\\ 0 & 0 & 5\end{bmatrix}.\) Note that \(p_A = (4-\lambda)(5-\lambda)^2\). So the eigenvalues are \(4\) and \(5\).
The geometric multiplicity of \(4\) is \(1\) since the geometric multiplicity cannot be \(0\) and cannot exceed the algebraic multiplity of \(4\), which is \(1\).
Now, note that \(A - 5I = \begin{bmatrix} -1 & 0 & -2 \\ 2 & 0 &4 \\ 0 & 0 & 0\end{bmatrix}\). This matrix has rank \(1\) since the second row is \(-2\) times the first row and the third row is a row of \(0\)'s. Hence, the nullity of \(A-5I\) is \(2\), implying that the geometric multiplicity of \(5\) is \(2\).
So \(A\) is diagonalizable as the sum of the geometric multiplities is \(3\).
Now that we know the matrix \(A\) in the second example above is diagonalizable, how do we actually find \(P\) and \(D\)?
First, observe that \(A= PDP^{-1}\) can be rewritten as \(AP = PD\). Column \(i\) of the left-hand side is given by \(AP_i\) where \(P_i\) denotes the \(i\) column of \(P\) and column \(i\) of the right-hand side is given by \(PD_i\). Since \(D\) is a diagonal matrix, \(PD_i = D_{i,i} P_i\). Thus \(P_i\) must be an eigenvector with eigenvalue \(D_{i,i}\). Since we need \(P\) to be invertible, we need \(n\) linearly independent eigenvectors. This is why we need the sum of the geometric multiplicities to equal \(n\).
To form \(D\), first list all the eigenvalues (including multiple appearances) in any order. (Hence, an eigenvalue with algebraic multiplicity \(k\) will appear \(k\) times in the list.) Then form \(D\) with this list of values as the diagonal.
For our example, the list could be \(4,5,5\). So we can set \(D = \begin{bmatrix} 4 & 0 & 0\\ 0 & 5 & 0\\0 & 0 & 5\end{bmatrix}\).
To form \(P\), first find a basis for each eigenspace. To form the first column of \(P\), take a vector from the basis for the eigenspace of the first diagonal entry of \(D\) and remove that vector from future inclusion into \(P\). For each subsequent column \(i\), take a vector from the basis for the eigenspace of the \(i\)th diagonal entry of \(D\) and remove it from future includsion into \(P\).
For our example, we first need to find a basis for the eigenspace of \(4\) and a basis for the eigenspace of \(5\).
For the eigenvalue \(4\), we want to find a basis for the nullspace of \(A-4I = \begin{bmatrix} 0 & 0 &-2 \\ 2 & 1 & 4 \\ 0 & 0 & 1 \end{bmatrix}\). Row-reducing this matrix to RREF gives the matrix \(\begin{bmatrix} 1 & \frac{1}{2} & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}\). Hence, all solutions to \(\begin{bmatrix} 1 & \frac{1}{2} & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0\end{bmatrix}\) are of the form \( \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -\frac{1}{2}s \\ s \\ 0\end{bmatrix} = s \begin{bmatrix} -\frac{1}{2} \\ 1 \\ 0 \end{bmatrix}\) since \(x_2\) is a free variable. Thus, the single vector \(\begin{bmatrix} -\frac{1}{2} \\ 1 \\ 0\end{bmatrix}\) forms a basis for the eigenspace.
For the eigenvalue \(5\), we want to find a basis for the nullspace of \(A - 5I = \begin{bmatrix} -1 & 0 & -2 \\ 2 & 0 &4 \\ 0 & 0 & 0\end{bmatrix}\). Row-reducing this matrix gives \(\begin{bmatrix} 1 & 0 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}\). Hence, all solutions to \(\begin{bmatrix} 1 & 0 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0\end{bmatrix}\) are of the form \( \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -2t \\ s \\ t\end{bmatrix} = s \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} + t \begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix}\) since \(x_2\) and \(x_3\) are free variables. Hence, a basis for the nullspace is \(\left\{\begin{bmatrix} 0 \\ 1 \\0\end{bmatrix}, \begin{bmatrix} -2 \\ 0 \\ 1\end{bmatrix}\right\}\).
To form the first column of \(P\), we take a vector from the basis for the eigenspace of \(4\). There is only one such vector. So \(P\) is now \(P = \begin{bmatrix} -\frac{1}{2} & * & *\\ 1 & * & *\\ 0 & * & *\end{bmatrix}\).
For the second column, we take a vector from the basis for the eigenspace of \(5\). There are two vectors to choose from. We choose \(\begin{bmatrix} 0 \\ 1 \\0\end{bmatrix}\). So \(P\) is now \(P = \begin{bmatrix} -\frac{1}{2} & 0 & *\\ 1 & 1 & *\\ 0 & 0 & *\end{bmatrix}\). And we have no choice but to take the remaining vector in the basis for eigenspace of \(5\) to fill the last column of \(P\). Hence, \(P = \begin{bmatrix} -\frac{1}{2} & 0 & -2\\ 1 & 1 & 0\\ 0 & 0 & 1\end{bmatrix}\).
Note that \(P^{-1} = \begin{bmatrix} -2 & 0 & -4\\ 2 & 1 & 4\\ 0 & 0 & 1\end{bmatrix}\). One can now verify that \(A = PDP^{-1}\).
Let \(A = \begin{bmatrix} 1 & -2 \\ 1 & 0\end{bmatrix}\), \(D = \begin{bmatrix} \frac{1+\sqrt{7}i}{2} & 0 \\ 0 & \frac{1-\sqrt{7}i}{2} \end{bmatrix}\), and \(P = \begin{bmatrix} \frac{1+\sqrt{7}i}{2} & \frac{1-\sqrt{7}i}{2} \\ 1 & 1 \end{bmatrix}.\)
Find \(P^{-1}\).
Check that \( A = PDP^{-1} \).
Let \(A = \begin{bmatrix} 8 & -9 \\ 4 & -4 \end{bmatrix}\). You are given that \(2\) is the only eigenvalue of \(A\). Give a basis for the eigenspace of this eigenvalue.