Restrictive definition of diagonalizable matrix

Question

There is a theorem that says that every matrix of rank $r$ can be transformed by means of a finite number of elementary row and column operations into the matrix $$D=\begin{pmatrix} I_r & O_1 \\ O_2 & O_3 \\ \end{pmatrix}$$ where $O_1, O_2, O_3$ are zero matrices and $I_r$ is the identity matrix of size $r$.

A corollary of this theorem says that for every matrix of rank $r$ there exist inevertible matrices $B$ and $C$ of size $m$x$m$ and of size $n$x$n$ respectively such that $D=BAC$

So every matrix can be transformed to a diagonal matrix $D$ and in this sense every matrix can be diagonalized; but the definition of a diagonalizable matrix is that: $A$ is diagonalizable if there exist an invertible matrix $P$ such that $P^{-1}AP$ is a diagonal matrix

This definition is very similar to the corollary of the theorem but is more restrictive,so I would really appreciate if you can tell me why do we adopt this restrictive definition of diagonalizable matrix

Answer 1

This is a property that could motivate the restriction: let $k\in \Bbb N$,

• if $A=P^{-1}DP$, then $A^k = (P^{-1}DP)^{k} = P^{-1}D^kP$ with $(D^{k})_{i,i}=D_{i,i}^k$.

• if $B \neq C^{-1}$ then $C^{-1}B^{-1}\neq I$, and so $A^k = (B^{-1}DC^{-1})^k=\ldots=(B^{-1}DC^{-1})^k$...

Answer 2

Not every matrix can be diagonalized, in the sense that $D = P^{-1}AP$ (implicit in this statement, of course, is that $P$ is invertible). One of the classic examples is

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

(or you can have $0$'s on the diagonal, if you like). To show this is not diagonalizable, we first need to show "$A$ is diagonalizable if and only if it has $n$ linearly independent eigenvectors".

Let's just show one direction: assume $A$ is diagonalizable, we'll show it has $n$ linearly independent eigenvectors.

If we can write $D = P^{-1}AP$, then $e_1 = (1,0,0,\dots,0), e_2 = (0,1,0,\dots,0)$, etc. are eigenvectors of $D$, with eigenvalues corresponding to the entries of $D$. Then $A$ has the same eigenvalues (this is a fact about similarity). Further, $v_1 = Pe_1$ is an eigenvector of $A$:

$$ Av_1 = APe_1 = (PDP^{-1})Pe_1 = PDe_1 = \lambda_1 Pe_1 = \lambda_1 v_1 $$

and similarly for $v_2 = Pe_2$, etc. Finally, these eigenvectors $v_1,\dots,v_n$ of $A$ must be linearly independent (since the $e_j$'s were, and since $P$ is invertible).

Thus, if a matrix does not have a maximal linearly independent set of eigenvectors, then it cannot be diagonalizable.

We now just need to observe that the only eigenvalue of my original matrix is $\lambda = 1$, that

$$ A - \lambda I = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, $$ which only has a one dimensional kernel, and thus there cannot be two linearly independent eigenvectors.

So every matrix can be "quasi-diagonalized" (in the sense that $D = BAC$, as you write), but not necessarily "truly diagonalized" (i.e. $D = P^{-1}AP$, or in other words, that in the quasi-diagonalization, you can have $B = C^{-1}$).

Or said yet another way "not every matrix is similar to a diagonal matrix".

Answer 3

What you are confusing here is the notions of equivalence for rectangular matrices and similarity for square matrices. The former is relevant for classifying linear maps between distinct (finite dimensional) vector spaces, where one allows choosing two bases independently in both vector spaces. The latter relevant for classifying linear operators on one vector space (linear maps from the space to itself). In this situation it is natural to allow only one basis to be chosen, since this is what allows comparing vectors before and after application of the linear operator. The definition of eigenvectors is an example of such a comparison. Choosing just one basis is also clearly relevant when one wants to apply the operator multiple times, since for a composition of linear maps to correspond to a product of matrices, one must express the matrices using the same basis at the intermediate vector space.

The classification of rectangular matrices under equivalence says that apart from the shape, the rank is the only invariant: all matrices of the same shape and rank are equivalent, and the standard representative of the equivalence class with rank$~r$ has as only nonzero coefficients $r$ copies of $1$ on the initial entries of the main diagonal. So fo purposes of equivalence, every rectangular matrix can not only be brought to diagonal form, but even to one with all diagonal entries $1$ or $0$. The classification of square matrices under similarity is much more subtle; all non-leading coefficients of the characteristic polynomial are invariants (each can have an arbitrary field element as value, unlike the rank which has only very few possible values), and even these invariants do not characterise all similarity classes. The simplest classes are those of diagonal matrices, and the matrices of those classes are called diagonalisable. Here the set of diagonal entries of a diagonal matrix in a given class is fixed up to order, again showing how much more restricted the situation is for the question of similarity.