I have studied before that the rank of a matrix = number of non zero Eigen values. But recently i came across a problem and i dont think it is valid there. I know i am going wrong somewhere.

$$A= \begin{bmatrix} 0 & 4 & 0 \\ 0 & 0 & 4\\ 0 & 0 & 0 \\ \end{bmatrix} \quad $$ The Rank of this matrix is 2. So there should be 2 non zero eigen values. But I only get 0 as the eigen value(λ) using $$[A-λI]=0$$

Can anybody explain? Thanks


4 Answers 4


Unfortunately, the answer is no in general, though the claim will hold true for diagonalizable matrices. Not all matrices are diagonalizable, including the matrix that you gave in your example. If your matrix is $n \times n$, then diagonalizability is equivalent to having a set of $n$ linearly independent eigenvectors, and those eigenvectors corresponding to non-zero eigenvalues will form a basis for the range of the matrix; hence rank is obtained (including multiplicities).

However, if you look at $A^T A$, then you can use the eigenvalues of that matrix to obtain the rank, regardless of what $A$ is. This is because $A^T A$ is symmetric, and thus must be diagonalizable, and furthermore one can show that $\mathrm{rank}(A^T A) = \mathrm{rank}(A)$.

  • 4
    $\begingroup$ “Most matrices are not diagonalizable” is a probably not true, under any reasonable definition of “most.” For example, “most” matrices have a characteristic polynomial with no repeated roots, and thus are diagonalizable, under any definition of “most” that I have heard. $\endgroup$ Commented Jun 7, 2020 at 19:33
  • 1
    $\begingroup$ You don't really need diagonalizability either. You just need the algebraic and geometric multiplicity of the eigenvalue $0$ to agree, not all of them. $\endgroup$
    – Arthur
    Commented Jun 7, 2020 at 19:37
  • $\begingroup$ @ThomasAndrews According to Devadas et al. (The proportion of diagonalizable matrices over a finite field), when $n\ge3$, most $n\times n$ matrices are indeed non-diagonalisable over $GF(q)$ when $q$ is large. The proportion of the diagonalisable ones approaches $1/n!$ as $q\to\infty$. The story about diagonalisability over an algebraic closure may be different, however. $\endgroup$
    – user1551
    Commented Jun 7, 2020 at 20:08
  • $\begingroup$ @ThomasAndrews Indeed, I was not precise here. $\endgroup$ Commented Jun 7, 2020 at 20:24
  • $\begingroup$ @Arthur We would need algebraic and geometric multiplicities of non-zero eigenvalues to be equal, since any Jordan for a non-zero eigenvalue is full-rank, but only has a one-dimensional eigenspace. $\endgroup$ Commented Jun 7, 2020 at 20:26

rank of a matrix = number of non zero Eigen values

is not true, as you have witnessed.

Consider that $A^3=0$, so if $A$ has an eigenvalue $\lambda$ and $v\neq0$ is a corresponding eigenvector, then $$ 0=A^3v=\lambda^3v $$ meaning $\lambda^3=0$, so $\lambda$ must be $0$.

The rank is, however, equal to the dimension of the image. Which is to say, the size of the largest possible set of linearly independent vectors of the form $Av$.

It is also the case that nilpotency (or more specifically the fact that the image may contain elements of the kernel) is in some sense the only thing that can go wrong with your statement.

  • $\begingroup$ actually i got this theory from this article which seems to be misleading: personal.utdallas.edu/~herve/Abdi-EVD2007-pretty.pdf Please refer page 8. So is it true the rank is the maximum number of non zero eigen values of a matrix? thank you $\endgroup$
    – Shubham
    Commented Jun 7, 2020 at 19:31
  • 1
    $\begingroup$ @Subham You have yourself found a matrix where the rank and the number of non-zero eigenvalues are different. You don't need to ask me whether it's true. It's not. $\endgroup$
    – Arthur
    Commented Jun 7, 2020 at 19:35
  • $\begingroup$ Note that you can connect the rank of the matrix to the number of Jordan blocks corresponding to the eigenvalue $0$, i.e. the geometric multiplicity of ev $0$. Notice that OPs matrix has eigenvalue $0$ with algebraic multiplicity $3$ but geometric multiplicity $1$. $\endgroup$ Commented Jun 8, 2020 at 3:55

This statement holds true when the Jordan blocks of A associated with eigenvalue 0 are all 1 by 1; Otherwise, there would be extra rank entailed by such Jordan blocks. For example, $rank(J_2(0)) = 1$.

Equivalently you can say the geometric and algebraic multiplicity of eigenvalue 0 agrees with each other.

Or, the minimal polynomial of A is $q_A(t)=t*\Pi_{i=1}^{d-1}(t-\lambda_i)^{r_i}$, where we assumed there are d distinct eigenvalues of A and the maxiaml sizes of their corresponding Jordan blocks are $r_i$.

Note that I took nonzero eigenvalue of algebraic mulitplicty k>1 as k nonzereo eigenvalues instead of 1; Otherwise we would need the equivalence between algebraic and geometric multiplicity of not only zero eigenvalues but also that of the nonzero eigenvalues of A. And therefore the diagonalizability would be necessary.

I could be wrong, feel free to correct me.


The short answer is you are right, the statement "Rank= number of non zero eigen values " is NOT correct One example: A Nilpotent matrix

Now $A^3$ is $0$ matrix,hence

if $A^3 x=λ^3 x$ where $x$ is the eigen vector, λ =0 (only one value exists)

so number of non zero eigen values is 0 therefore Rank is 0, but the matrix $A$ suggests the rank is $1$

I hope this counter example helps

  • 1
    $\begingroup$ I think there is already a well-written answer to this question, so your answer is a bit redundant(besides, the question's matrix is already nilpotent). $\endgroup$
    – abacaba
    Commented Sep 22, 2023 at 4:25
  • $\begingroup$ This countetexample was already given by Arthur. $\endgroup$ Commented Sep 22, 2023 at 5:40
  • $\begingroup$ Okay,I actually read @Arthur 's answer and took idea from it, just added a numerical counter example , I am new here, should I delete my answer? $\endgroup$ Commented Sep 22, 2023 at 5:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .