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my question is about diagonalizable matrices and similar matrices.

I have a trouble proving a matrix is diagonalizable.

I know some options to do that:

Matrix $A$ $(n \times n)$, is diagonalizable if:

  • Number of eigenvectors equals to number of eigenvalues.
  • There exists an invertible matrix $B$ and a diagonal matrix $D$ such that: $D=B^{-1}AB$.

But i have a trouble to determine it according the second option, Do i really need to search if there exists an invertible matrix $B$ and a diagonal matrix $D$ such that: $D=B^{-1}AB?$

I really sorry to ask an additional question here: If a matrix has a row of $0$'s (one of its eigenvalues is $0$), That matrix is diagonalizable?

in general, given a matrix, how do i know if is a diagonalizable matrix? Are there some additional formulas to do that?

Thanks for help!!

José Carlos Santos
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D.Rotnemer
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    The second option is, more or less, the definition of diagonalization, and diagonalizability. As for the additional question, the zero matrix is trivially diagonalizable. – Arthur Jul 26 '18 at 08:53
  • @Arthur: The OP didn't ask about the zero matrix, he asked about a matrix with a row of zeros. That's insufficient to be diagonalizable. – Meni Rosenfeld Jul 26 '18 at 09:57
  • @MeniRosenfeld That is one way to interpret the question. I read it as "can a matrix with a row of zeroes be diagonalizable?" I might've been wrong, but there it is. – Arthur Jul 26 '18 at 10:52
  • If the roots of the minimal polynomial are all simple roots, then the matrix is diagonalizable. Rmk: all the roots of the minimal polynomial are eigenvalues of the matrix and if $\chi (t)=(t-\lambda_1)^{\mu_1}...(t-\lambda_k)^{\mu_k}$ the minimal polynomial is $\varphi (t)=(t-\lambda_1)^{\eta_1}...(t-\lambda_k)^{\eta_k}$, where $\eta_i \le \mu_i$ $\forall i=1,...,k$ (I'm assuming that the matrix is $\mathrm{n}\times \mathrm{n}$). I wrote you this property because is really useful for problems and exercises. – Vajra Jul 05 '20 at 10:05
  • If a matrix $A \in \mathrm{M}(n,\mathbb{R})$ is diagonalizable, then $\exists$ $C \in \mathrm{Gl}(n,\mathbb{R})$ such that $D=C^{-1}AC$, where $C^{-1}$ is a matrix whose columns are eigenvectors and they appear with the same order of the eigenvalues on the diagonal of $D$. – Vajra Jul 05 '20 at 10:31

3 Answers3

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First a comment

The wording Number of eigenvectors equals to number of eigenvalues... is confusing. If $A$ has a non zero eigenvector then $A$ has an infinite number of eigenvectors (providing you work in $\mathbb R$ or $\mathbb C$ for example). A proper wording would be $A$ has a basis of eigenvectors.

If a matrix has a row of $0$'s (one of its eigenvalues is $0$), That matrix is diagonalizable?

The implication "If a matrix has a row of $0$'s" then "that matrix is diagonalizable" is not true. The matrix $$A=\begin{pmatrix} 0 & 0\\ 1 & 0 \end{pmatrix}$$ is an example. The only eigenspace is $\mathbb F e_2$ were $e_2$ is the second vector of the canonical basis (and $\mathbb F$ the field of the vector space).

Some equivalent conditions for a matrix $A$ to be diagonalizable

  • The sum of the dimensions of its eigenspaces is equal to the dimension $n$ of the space.
  • $A$ is similar to a diagonal matrix.
  • Its minimal polynomial is a product of distinct linear factors.
mathcounterexamples.net
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    The statement that a matrix with a row of zeroes is never diagonalisable is clearly wrong, just consider $B=\begin{pmatrix}0&0\0&1\end{pmatrix}$. – Toffomat Jul 26 '18 at 11:42
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    @Toffomat I haven't say that it is never diagonalizable! I meant that the implication $P \implies Q$ is not true where $P$ is "a matrix has a row of $0$'s and $Q$ is "it is diagonalizable". I'll precise that is my answer. – mathcounterexamples.net Jul 26 '18 at 12:09
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If a $n\times n$ matrix has $n$ distinct eigenvalues, then it is automatically diagonalizable. Otherwise, compute the dimension of each eigenspace. The matrix is diagonalizable if and only if the sum of these dimensions is $n$.

Concerning the sentence “There exists a matrix $B$ and a diagonal matrix $D$ such that: $D=B^{−1}AB$”, well… That's basically what being a diagonalizable matrix means.

José Carlos Santos
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The statement "matrix $A$ ($n×n$), is diagonalizable if: number of eigenvectors equals to number of eigenvalues" is not correct.

We would say that a matrix $A$ ($n×n$), is diagonalizable if and only if the sum of the dimension of eigenspaces is equal to $n$, that is if and only if for any eigenvalues the algebraic multiplicity is equal to the geometric multiplicity.

When a matrix is diagonalizable, of course, by definition the diagonal form is similar to the original matrix. Note that similarity holds, more in general, also with the Jordan normal form when the matrix is not diagonalizable.

user
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    Existence of Jordan normal form supposes that the characteristic polynomial of $A$ splits into linear factors in the field $\mathbb F$ over which the vector space is defined. – mathcounterexamples.net Jul 26 '18 at 09:25
  • @mathcounterexamples.net Thanks for the pointing out, even if I can't relate that to the OP. Bye! – user Jul 26 '18 at 09:27
  • Just a comment as the OP has not precised the field over which the vector space is defined. Jordan normal form may not exist. – mathcounterexamples.net Jul 26 '18 at 09:29
  • @mathcounterexamples.net Ah ok, thanks again! – user Jul 26 '18 at 09:30
  • The Jordan Normal Form may not exist. The correct test for the similarity of two matrices (over a field $\mathbb{F}$) is that they have the same Rational Canonical Form. – ancient mathematician Jul 26 '18 at 09:34
  • @ancientmathematician I'm not aware about those kind of generalization but I believe to your words in faith. Bye :) – user Jul 26 '18 at 09:59