Basic Procedure for Diagonalization

Changing to an Eigenvector Basis

Diagonalization does not change the linear transformation itself. It changes the coordinate basis so that the transformation acts independently along eigenvector directions. In the diagonal representation, matrix multiplication reduces to scaling each eigenvector coordinate.

To diagonalize a matrix $A \in \mathbb{K}^{n \times n}$, compute its eigenvalues and eigenspaces. If there are $n$ linearly independent eigenvectors, place those vectors in a change-of-basis matrix.

Diagonalization Steps

The diagonalization procedure consists of three main steps that must be performed sequentially:

1. Compute the characteristic polynomial to find all eigenvalues $\lambda_1, \ldots, \lambda_k \in \mathbb{K}$ along with their algebraic multiplicities $\mu_A(\lambda_1), \ldots, \mu_A(\lambda_k)$.

   Over the chosen field $\mathbb{K}$, the characteristic polynomial $\chi_A(t)$ must split completely into linear factors:

   $$\chi_A(t) = \prod_{i=1}^k (\lambda_i - t)^{\mu_A(\lambda_i)}$$

   $$\sum_{i=1}^k \mu_A(\lambda_i) = n$$

   If it does not split, then the matrix is not diagonalizable over that field.

2. Compute the eigenspaces for each eigenvalue by solving the homogeneous linear system:

   $$(A - \lambda_i \cdot I) \cdot v = 0$$

   Here we find every vector that matrix $A$ maps to a scalar multiple of itself. The factor is the eigenvalue $\lambda_i$. A negative factor reverses the vector's direction but keeps it on the same line.

3. Check the diagonalization conditions by verifying whether the algebraic multiplicity equals the geometric multiplicity for all eigenvalues. Mathematically, for all $i = 1, \ldots, k$ we must have $\mu_A(\lambda_i) = \dim E_A(\lambda_i)$.

   This condition ensures that we have enough independent eigenvectors to form a complete basis. If satisfied, the basis vectors from all eigenspaces form the columns of the transformation matrix $S$, yielding:

   $$\Lambda = S^{-1} \cdot A \cdot S$$

Example Application of the Procedure

Consider the matrix:

Apply the diagonalization procedure to this example:

1. We compute the characteristic polynomial and factor it:

   $$\begin{aligned} \chi_A(t) &= \det(A - t \cdot I) \\ &= -t^3 + t^2 + t - 1 \end{aligned}$$

   After factoring, we obtain:

   $$\chi_A(t) = (1 - t)^2 \cdot (-1 - t)$$

   The factors give the eigenvalues $\lambda_1 = 1$ with algebraic multiplicity $\mu_A(1) = 2$ and $\lambda_2 = -1$ with algebraic multiplicity $\mu_A(-1) = 1$. Their multiplicities sum to $2 + 1 = 3$, which equals the matrix dimension.

2. We determine the eigenspace for each eigenvalue:

   $$\begin{aligned} E_A(1) &= \ker(A - I) \\ &= \operatorname{span}\left\{\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\right\} \end{aligned}$$

   $$\begin{aligned} E_A(-1) &= \ker(A + I) \\ &= \operatorname{span}\left\{\begin{pmatrix} 1 \\ 3 \\ 2 \end{pmatrix}\right\} \end{aligned}$$

   For eigenvalue 1, every row of $A-I$ gives the same plane equation $z=x+y$. For eigenvalue −1, solve the first two equations in $(A+I)v=0$:

   $$\begin{aligned}x-y+z&=0\Rightarrow y=x+z\\-3x-y+3z&=0\Rightarrow -4x+2z=0\\z&=2x,\qquad y=3x\end{aligned}$$

   Here, the kernel or null space contains every vector that the matrix maps to the zero vector, while the span of a collection of vectors contains all of their linear combinations.

3. We check whether the diagonalization conditions are satisfied. For eigenvalue $\lambda_1 = 1$, its algebraic multiplicity is $2$ and its eigenspace has dimension $2$ (two independent basis vectors). For eigenvalue $\lambda_2 = -1$, its algebraic multiplicity is $1$ and its eigenspace has dimension $1$.

   Since $\mu_A(1) = 2 = \dim E_A(1)$ and $\mu_A(-1) = 1 = \dim E_A(-1)$, the diagonalization conditions are satisfied.

Now we can form the transformation matrix $S$ by arranging all eigenvectors as columns, and the diagonal matrix $\Lambda$ with eigenvalues on the main diagonal:

The columns of $S$ and the corresponding diagonal entries of $\Lambda$ must remain in the same order. Here the construction is valid because:

The matrix $A$ is successfully diagonalized to $\Lambda$ through the transformation $\Lambda = S^{-1} \cdot A \cdot S$.

The procedure therefore proves both requirements: the eigenvectors form a basis, and the matrix representation in that basis is diagonal.

Rotate the model to separate the two eigenspaces. Every point of the green plane $z=x+y$ stays fixed because its eigenvalue is 1. The orange vector $v_3$ lies outside that plane and maps to the red vector $Av_3=-v_3$. Together, these directions span the whole space.