Linear algebra · Álgebra linear
| English | Português |
|---|---|
| rank/ræŋk/ | posto |
| eigenvalue/ˈaɪdʒənvæljuː/ | valor próprio |
A decision before an answer
- A square matrix may have no inverse even when all its entries are nonzero.
- Your goal: Relate rank, nullity and solutions of linear systems.
Read the relationship
- A real vector space is closed under its specified addition and scalar multiplication and satisfies the vector-space axioms. A basis is an independent spanning list: every vector has a unique coordinate representation in that list. To test independence of matrix columns, solve the homogeneous system Ac=0; only the zero coefficient vector means independence. A spanning set can contain redundant vectors, so spanning alone does not make a basis.
- Analyse vector spaces and linear transformations.
A 3-column matrix has rank 2. Its nullity is:
Rank-nullity gives 3−2=1.
Use the defining rule
- Row reduction exposes pivots and free variables without changing the solution set of a linear system. Rank is the dimension of the column image, equivalently the number of pivots. Nullity is the kernel dimension: for a map from an n-dimensional domain, rank+nullity=n. In a nonhomogeneous system, a zero coefficient row with nonzero right side means inconsistency; free variables give infinitely many solutions only after consistency has been established. The codomain dimension need not equal rank.
- Compute eigenvalues, determinants and diagonalisation conditions.
det A=0 implies:
A zero determinant means the square matrix is singular. It can have many nonzero entries; a single dependence is enough.
Check the conditions
- A square matrix is invertible exactly when its determinant is nonzero, its kernel is zero and its rank equals its size. These statements do not require each entry to be nonzero. Triangular determinants are products of diagonal entries, so a matrix depending on a complex variable can be singular at complex roots absent from a real-only calculation. Row swaps reverse determinant sign; adding a multiple of one row to another leaves it unchanged. Check singularity before applying an inverse formula.
- Compute eigenvalues, determinants and diagonalisation conditions.
A=[[1,1],[0,1]] has characteristic polynomial (1−λ)². Eigenvectors satisfy y=0, so its eigenspace has dimension one. Two independent eigenvectors are needed to diagonalise a 2×2 matrix; A is not diagonalizable.
The identity matrix has every eigenvalue equal to ____.
Iv=v for every vector.
Apply the task format
- An eigenvector is nonzero and satisfies Av=λv, so eigenvalues are roots of det(A−λI). Diagonalisation requires a full independent eigenvector basis over the chosen field. Distinct eigenvalues give independent eigenvectors, while repeated eigenvalues may have too small an eigenspace. A real symmetric matrix has a real orthonormal eigenbasis. A real odd-dimensional matrix has at least one real eigenvalue because its real characteristic polynomial has odd degree; that alone does not imply diagonalisation or all eigenvalues real.
- Compute eigenvalues, determinants and diagonalisation conditions.
The algebraic multiplicity of an eigenvalue is not automatically the dimension of its eigenspace.
Which answer fits this case? · Qual resposta se encaixa neste caso?
Relate rank, nullity and solutions of linear systems · Relacione posto, nuloidade e soluções de sistemas lineares
A repeated eigenvalue guarantees a full eigenvector basis.
A defective matrix is a counterexample.
Keep the distinctions
- rank 秩 — Dimension of the image of a linear map.
- eigenvalue 特征值 — A scalar satisfying Av=λv for a nonzero v.
- Relate rank, nullity and solutions of linear systems.
- Analyse vector spaces and linear transformations.
- Compute eigenvalues, determinants and diagonalisation conditions.
Match each term with its precise meaning in this lesson.
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
The order follows the stated process; check each stage before the next.