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The rank of a matrix is the maximum number of independent rows (or, the maximum number of independent columns).
A square matrix $A_{~ n ~ \times ~ n}$ is non-singular only if its rank is equal to n.
Source

Consider a quadratic matrix $A \in \mathbb{R}^{~ n ~ \times ~ n}$

  • with in total $p$ linearly dependent rows (or columns) and otherwise no zero-rows/-columns, which equals to $n-(p-1)$ linearly independent rows/columns, e.g.

$$A_1 = \begin{bmatrix} 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 & 4 \\ 5 & 0 & 6 & 7 & 8 \\ 9 & 8 & 0 & 7 & 6 \end{bmatrix} \quad \rightarrow \quad n = 5, ~ p = 2; ~ n-(p-1) = 4$$ $$ \hookrightarrow \lambda_1 = 0; ~ \lambda_2, \lambda_3, \lambda_4, \lambda_5 \neq 0 \quad \rightarrow ~ Rank(A_1) = 4$$

$$A_2 = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{bmatrix} \quad \rightarrow \quad n = 5, ~ p = 2; ~ n-(p-1) = 4$$ $$ \hookrightarrow \lambda_1 = 0; ~ \lambda_2, \lambda_3, \lambda_4, \lambda_5 \neq 0 \quad \rightarrow ~ Rank(A_2) = 4$$

  • or with exactly $p-1$ zero-rows/-columns and otherwise $n-(p-1)$ linearly independent rows/columns, e.g.

$$A_3 = \begin{bmatrix} 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 & 4 \\ 5 & 0 & 6 & 7 & 8 \\ 9 & 8 & 0 & 7 & 6 \end{bmatrix} \quad \rightarrow \quad n = 5, ~ (p-1) = 1; ~ n-(p-1) = 4$$ $$ \hookrightarrow \lambda_1 = 0; ~ \lambda_2, \lambda_3, \lambda_4, \lambda_5 \neq 0 \quad \rightarrow ~ Rank(A_3) = 4$$

$$A_4 = \begin{bmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{bmatrix} \quad \rightarrow \quad n = 5, ~ (p-1) = 1; ~ n-(p-1) = 4$$ $$ \hookrightarrow \lambda_1 = 0; ~ \lambda_2, \lambda_3, \lambda_4, \lambda_5 \neq 0 \quad \rightarrow ~ Rank(A_4) = 4$$

Are the following statements true?

  • $det(A) = 0$
  • The rank of $A$ is $rank(A) \leq n-(p-1)$, $rank(A) = n-m = k$,
    $m$ being the number/multiplicity of zero-eigenvalues and
    $k$ the number of non-zero-eigenvalues (counting multiplicity).
    Or should it be $rank(A) = n-(p-1) = n-m = k$?
  • The multiplicity of zero as an eigenvalue is at least $p-1$ (or should it be exactly $p-1$?), i.e.
    • $\lambda_1 = 0$
    • $\dots$
    • $\lambda_{p-1} = 0$
    • $\lambda_{p} = ~ ?$
    • $\dots$
    • $\lambda_{n} = ~ ?$
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  • $\begingroup$ What do you mean by "exactly $p$ linearly dependent rows"? If any $p$ of the rows are linearly dependent, then all $n$ rows are linearly dependent. Do you mean that $p$ is the least integer such that there are $p$ linearly dependent rows? $\endgroup$
    – joriki
    Commented May 29, 2016 at 12:03
  • $\begingroup$ @joriki Is the edited question clearer and are the formulations mathematically correct? Being an engineer, I did not understand your comment, specifically "If any $p$ of the rows are linearly dependent, then all n rows are linearly dependent.". $\endgroup$
    – Discbrake
    Commented May 29, 2016 at 13:17
  • $\begingroup$ @joriki It's easy to make an example of a three element linearly dependent set where any two elements for a linearly independent set: consider $(1,0,0)$, $(0,1,0)$ and $(1,1,0)$. $\endgroup$
    – egreg
    Commented May 29, 2016 at 13:21
  • $\begingroup$ @egreg: That's certainly true. Were you implying that this contradicts something I wrote? If so, what? $\endgroup$
    – joriki
    Commented May 29, 2016 at 16:09
  • $\begingroup$ @user5564832: Unfortunately it hasn't become any clearer to me -- "in total $p$ linearly dependent rows" makes as little sense to me as the previous formulation. Could you spell out what you mean? $\endgroup$
    – joriki
    Commented May 29, 2016 at 16:11

2 Answers 2

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The “maximum number of linearly dependent rows” doesn't really make sense. Consider the matrix $$ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix} $$ The three rows form a linearly dependent set, but no subset of rows is linearly dependent.

What you can define is the maximum number of linearly independent rows, which in the above case is $2$. Such number is the rank of the matrix.

By the rank-nullity theorem, if $k$ is the rank, $n-k$ is the dimension of the null space of the matrix, which is the geometric multiplicity of the eigenvalue $0$ (provided $0$ is an eigenvalue to begin with, that is, $k<n$).

The algebraic multiplicity of the zero eigenvalue is at least $n-k$. It can be bigger: consider $$ \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} $$ For this matrix, the algebraic multiplicity of the zero eigenvalue is $2$, but the geometric multiplicity is $1$.

A matrix is non-singular if and only if $0$ is not an eigenvalue, essentially by definition of eigenvalue (together with the rank-nullity theorem): $\lambda$ is an eigenvalue of the matrix $A$ if and only if $A-\lambda I$ is singular.

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$\DeclareMathOperator{\rank}{rank}$If I understand, you're asking:

Let $A$ be an $n \times n$ (real/complex) matrix with $k < n$ non-zero eigenvalues (counting multiplicity) and $m \geq 1$ the multiplicity of $0$ as an eigenvalue.

  • Is $\det A = 0$?
  • Is $\rank A = n - m = k$, the number of non-zero eigenvalues counted with multiplicity, or only $\rank A \geq n - m$?

To the first, Yes: $0$ is an eigenvalue.

For the second: If $A$ has $k$ non-zero eigenvalues, then the column space of $A$ clearly has dimension at least $k$, so $\rank A \geq n - m$. Equality, however, fails to hold in the "worst possible sense": An $n \times n$ matrix of rank $(n - 1)$ can have every eigenvalue equal to $0$; a Jordan block with $\lambda = 0$ has this property.

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  • $\begingroup$ Should those $\geq$ not rather be $\leq$ ? $\endgroup$
    – Discbrake
    Commented May 29, 2016 at 16:33
  • $\begingroup$ Actually not: The rank is bounded below by the number of non-zero eigenvalues. Again, a matrix of "large" rank can fail to have any non-zero eigenvalues at all. $\endgroup$ Commented May 29, 2016 at 16:57

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