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There are many discussions of such type problems (comparison), for example:
Diagonalizable vs Normal

Today, I want to clearly understand the topic.
Suppose the matrix $A\in \mathbb{R}^{n\times n}$.

  1. Since the multiplication of all eigenvalues is equal to the determinant of the matrix, $A$ full rank is equivalent to $A$ nonsingular.
  2. The above also implies $A$ has linearly independent rows and columns. So $A$ is invertible.
  3. $A$ is diagonalizable iff $A$ has $n$ linearly independent eigenvectors. ($A$ is non-defective).
    Note: $A$ is defective if geo. multiplicity $<$ alge. multiplicity.
  4. A diagonalizable matrix does not imply full rank (or nonsingular).

My problem is

Does full rank matrix (nonsingular) imply it is diagonalizable?

Equivalently:

Does a matrix with all its columns or rows linear independent imply all its eigenvectors are linear independently?

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2 Answers 2

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Full rank and diagonalizability are independent from one another.

  1. A diagonalizable matrix with full rank: $$ \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} $$

  2. A non diagonalizable matrix with full rank: $$ \begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix} $$

  3. A diagonalizable matrix without full rank: $$ \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix} $$

  4. A non diagonalizable matrix without full rank: $$ \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix} $$

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  • $\begingroup$ how's the second matrix non-diagonalisable ? $\endgroup$
    – user1049492
    Commented Oct 17, 2023 at 14:55
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    $\begingroup$ @MahNeh A $2\times2$ matrix with coincident eigenvalues is diagonalizable if and only if it is diagonal. Anyway, the eigenspace relative to the unique eigenvalue $1$ has dimension $1$. $\endgroup$
    – egreg
    Commented Oct 17, 2023 at 15:13
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$1.$

No. A full rank matrix implies it's determinant is non-zero or the matrix is non-singular.

Speaking in terms of linear operator $T$ over a vector space $V$ is diagonalizable if and only if there exists an ordered basis consisting of eigenvectors of $T$, Furthermore if $T$ is diagonalizable and $\beta$ is an ordered basis of $T$ consisting eigen vectors of $T$, then $M=[T]_\beta$, the matrix representation of $T$, is a diagonal matrix.

One can check if a given matrix $M_{n\times n}$ is diagonalizable or not by-

  1. Characteristic polynomial splits or not,
  2. $n-rank(M-\lambda_iI)=$multiplicity of $\lambda_i$.

If this two criteria fulfills then a matrix is diagonalizable.

Counterexample: $\begin{bmatrix}3 & 1 &0 \\ 0&3&0\\ 0&0&4\end{bmatrix}$

$2$. Consider the identity matrix $I$, all it's rows and columns are linearly independent. What can you say about the eigenvectors?

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  • $\begingroup$ What do you mean 'splits'? And in your answer, how do you link between diagonalizable and nonsingular(full rank) after you say no? Or simply speaking, is there any example to say $A$ is full rank but not diagonalizable? $\endgroup$
    – sleeve chen
    Commented Jun 4, 2016 at 9:30
  • $\begingroup$ If $A$ is full rank, then there is no eigenvalues $=0$. Therefore, we have only to consider $U$ if $A = U^{-1}\Lambda U$ Then can we say $U$ is full rank by knowing that $A$ is full rank? $\endgroup$
    – sleeve chen
    Commented Jun 4, 2016 at 9:33
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    $\begingroup$ A polynomial splits if it can be written as a product of linear polynomials. This depends on the field you're working in, e.g. $X^2 - 2$ splits in $\mathbb{R}$, but not in $\mathbb{Q}$. $\endgroup$
    – Bib-lost
    Commented Jun 4, 2016 at 9:35
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    $\begingroup$ $\begin{bmatrix} 1& 1 \\ 0& 1 \end{bmatrix}$ is Full rank, but not diagonalizable. $\endgroup$
    – Bib-lost
    Commented Jun 4, 2016 at 9:38

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