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.
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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.
- $\lambda_1 = 0$
- $\dots$
- $\lambda_{p-1} = 0$
- $\lambda_{p} = ~ ?$
- $\dots$
- $\lambda_{n} = ~ ?$
Best Answer
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.