Diagonalization of Matrices

The prior subsection defines the relation of similarity and shows that, although similar matrices are necessarily matrix equivalent, the converse does not hold. Some matrix-equivalence classes break into two or more similarity classes (the nonsingular $n\!\times \!n$ matrices, for instance). This means that the canonical form for matrix equivalence, a block partial-identity, cannot be used as a canonical form for matrix similarity because the partial-identities cannot be in more than one similarity class, so there are similarity classes without one. This picture illustrates. As earlier in this book, class representatives are shown with stars.

We are developing a canonical form for representatives of the similarity classes. We naturally try to build on our previous work, meaning first that the partial identity matrices should represent the similarity classes into which they fall, and beyond that, that the representatives should be as simple as possible. The simplest extension of the partial-identity form is a diagonal form.

Definition 2.1:

A transformation is diagonalizable if it has a diagonal representation with respect to the same basis for the codomain as for the domain. A diagonalizable matrix is one that is similar to a diagonal matrix: $T$ is diagonalizable if there is a nonsingular $P$ such that $PTP^{-1}$ is diagonal.

Example 2.2: The matrix

${\begin{pmatrix}4&-2\\1&1\end{pmatrix}}$

is diagonalizable.

${\begin{pmatrix}2&0\\0&3\end{pmatrix}}={\begin{pmatrix}-1&2\\1&-1\end{pmatrix}}{\begin{pmatrix}4&-2\\1&1\end{pmatrix}}{\begin{pmatrix}-1&2\\1&-1\end{pmatrix}}^{-1}$

Example 2.3: Not every matrix is diagonalizable. The square of

$N={\begin{pmatrix}0&0\\1&0\end{pmatrix}}$

is the zero matrix. Thus, for any map $n$ that $N$ represents (with respect to the same basis for the domain as for the codomain), the composition $n\circ n$ is the zero map. This implies that no such map $n$ can be diagonally represented (with respect to any $B,B$) because no power of a nonzero diagonal matrix is zero. That is, there is no diagonal matrix in $N$'s similarity class.

That example shows that a diagonal form will not do for a canonical form— we cannot find a diagonal matrix in each matrix similarity class. However, the canonical form that we are developing has the property that if a matrix can be diagonalized then the diagonal matrix is the canonical representative of the similarity class. The next result characterizes which maps can be diagonalized.

In the next subsection, we will expand on that example by considering more closely the property of Corollary 2.4. This includes seeing another way, the way that we will routinely use, to find the $\lambda$'s.

Resources

• Diagonalizability, Wikibooks: Linear Algebra