When Is A Matrix Not Diagonalizable

April 10, 2024

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Throughout this post, as always, $M_n(F)$ denotes the ring of $n times n$ matrices with entries from a field $F,$ and $I$ denotes the identity matrix.

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Recall that $A in M_n(F)$ is said to be diagonalizable if it is similar to a diagonal matrix, i.e. if there exist a diagonal matrix $D in M_n(F)$ and an invertible matrix $P in M_n(F)$ such that $A=PDP^{-1}.$ In other words, $A$ is diagonalizable if there exists a basis $mathcal{B}={v_1, cdots , v_n}$ of $F^n$ such that the matrix of the linear map $T: F^n to F^n, T bold{x}=Abold{x},$ with respect to $mathcal{B},$ call it $D,$ is diagonal. So if $lambda_1, cdots , lambda_n$ are the diagonal entries of $D,$ then $lambda_kv_k=Tv_k=Av_k,$ i.e. each $v_k$ is an eigenvector of $A.$ So $A$ is diagonalizable if and only if $F^n$ has a basis consisting of eigenvectors of $A.$

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In practice, it is not easy in general to prove whether or not a given matrix is diagonalizable. There is however the following well-known result which is quite useful.

Theorem. Let $A in M_n(F).$

i) If $A$ has $n$ distinct eigenvalues in $F,$ then $A$ is diagonalizable.

i) $A$ is diagonalizable if and only if the minimal polynomial of $A$ has the form $(x-lambda_1) cdots (x-lambda_k)$ for some distinct elements $lambda_1, cdots lambda_k in F.$

Proof. See, for example, Theorem 3.2 and Theorem 4.11 here, or any decent linear algebra textbook $Box$

Note that the first part of the Theorem only gives a sufficient condition for a matrix to be diagonalizable. The condition is not necessary because, for example, the identity matrix $I$ is diagonal hence diagonalizable but its eigenvalues, which are all $1,$ are not distinct.

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Remark 1. Diagonalizability depends on the base field $F.$ For example consider the matrix

$A=begin{pmatrix}0 & -1 1 & 0end{pmatrix} in M_2(mathbb{C}).$

The eigenvalues of $A$ are $pm i$ and so $A$ is diagonalizable, by the first part of the above Theorem. However, if we look at $A$ as an element of $M_2(mathbb{R}),$ then $A$ is no longer diagonalizable because its eigenvalues are not real.

Corollary (characterizing non-diagonalizable $2 times 2$ matrices). Let

$A=begin{pmatrix} a & b c & d end{pmatrix} in M_2(mathbb{C}).$

Then $A$ is not diagonalizable if and only if $(a-d)^2+4bc=0$ and $A ne lambda I$ for all $lambda in mathbb{C}.$

Proof. Let $p(x)$ be the characteristic polynomial of $A,$ i.e. $p(x)= x^2-(a+d)x+ad-bc.$ Suppose first that $A$ is not diagonalizable. Then $A$ is not diagonal and so it’s not a multiple scalar of the identity matrix. Also, by the first part of the above Theorem, the two eigenvalues of $A$ must be equal, i.e. the two roots of $p(x)$ must be equal. Thus $(a+d)^2-4(ad-bc)=0,$ which gives $(a-d)^2+4bc=0.$ Conversely, the condition $(a-d)^2+4bc=0$ means that $p(x)$ has two equal roots, i.e. $p(x)=(x-lambda)^2$ for some $lambda in mathbb{C}.$ Thus the minimal polynomial of $A$ is either $x-lambda$ or $(x-lambda)^2.$ If the minimal polynomial of $A$ is $x-lambda,$ then $A=lambda I,$ contradiction. Thus the minimal polynomial of $A$ is $(x-lambda)^2$ and hence, by the second part of the above Theorem, $A$ is not diagonalizable. $Box$

Remark 2. The above Corollary holds for any algebraically closed field $F$ not just $mathbb{C}$ but it does not always hold if $F$ is not algebraically closed. For example, if we choose $F=mathbb{R},$ then the $2 times 2$ matrix given in Remark 1 is not diagonalizable but it does not satisfy the identity $(a-d)^2+4bc=0$ given in the Corollary.

Example. By the above Corollary, the following matrices, as elements of $M_2(mathbb{C}),$ are not diagonalizable

$begin{pmatrix}0 & 1 0 & 0end{pmatrix}, begin{pmatrix}0 & 0 1 & 0end{pmatrix}, begin{pmatrix}1 & 1 0 & 1end{pmatrix}, begin{pmatrix}3 & -1 1 & 1end{pmatrix}, begin{pmatrix}3 & -2 2 & -1end{pmatrix}.$