LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generated

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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A square matrix with zero determinant has a nonzero kernel vector

Statement

If $M \in F^{n \times n}$ , $n \geq 1$ , $F \in {R, C}$ , and $\det M = 0$ , then there is a nonzero vector $v \in F^{n}$ with $M v = 0$ .

Facts & Assumptions

Given: A square real or complex matrix $M$ of positive order.

[F1]

For an $m \times n$ matrix $A$ , $rank (A) + \dim N (A) = n$ : $rank (M) + \dim \ker M = n$ .

[F2]

The rank of a matrix equals the rank of the linear map $x \mapsto A x$ : matrix rank is the dimension of the image of the associated linear map.

[F3]

A finite-dimensional linear operator over a field is invertible if and only if its determinant is nonzero: a finite-dimensional operator over a field is invertible exactly when its determinant is nonzero.

Proof

1.1givenF1F2algebraassume-contra

Suppose $\ker M = {0}$ . By [F1], $rank (M) = n$ ; by [F2] the image of $M$ has dimension $n$ in the $n$ -dimensional space $F^{n}$ , so $M$ is surjective. It is also injective by the supposition, hence invertible.

2.1step 1.1F3contradictiondischarge-contradiction∎

By [F3], invertibility implies $\det M \neq 0$ , contradicting the given $\det M = 0$ . Therefore $\ker M \neq {0}$ and a nonzero $v$ with $M v = 0$ exists.

Depends on

Used by

Every regular real or complex pencil has a generalized Schur form Theorem

Dependency tree · two levels

25 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

LAPACK Users' Guide, Generalized Nonsymmetric Eigenproblems (standard reference, not scraped)