13. Quiz 1 Review

Overview

The review connects linear-system solutions to rank, nullity, and the four fundamental subspaces, with emphasis on rectangular matrices from chapter three. It works through dimension counts, reconstructing a matrix from a complete solution, factorized-matrix null spaces, and quick tests about invertibility, row operations, and orthogonality.

Key takeaways

For an m-by-n matrix, rank-nullity gives dim N(A) = n - rank(A); a 5-by-3 matrix with three pivots therefore has a trivial null space.
A system Ax=b is solvable precisely when b lies in the column space of A; a rank-1 matrix with column space span{(1, 2, 1)} cannot solve for arbitrary right-hand sides.
Left multiplication by an invertible matrix does not change a product's null space: N(CD) = N(D) when C is invertible.
A complete solution separates into one particular solution plus the entire null space; for the example with two free variables, all solutions use two arbitrary coefficients.
The row space and null space occupy the same n-dimensional coordinate space but are orthogonal, so their only shared vector is zero.
Independent columns in a square m-by-m matrix imply rank m and invertibility, which guarantees a solution for every b.

Chapters

0:00 Quiz Review Focus: Rectangular Matrices and the Four Subspaces

The exam review centers on Ax=b and emphasizes chapter three, where rectangular matrices require reasoning about rank and null spaces.
The key framework is the four fundamental subspaces: the column space, row space, null space, and null space of the transpose.

1:18 Possible Dimensions of the Span of Three Vectors

Three nonzero vectors u, v, and w in R^7 can span a subspace of dimension 1, 2, or 3.
The span cannot have dimension above 3 because it has only three generators, or dimension 0 because each vector is nonzero.

2:29 Ranks and Nullities of Block Matrices Built from U

A 5-by-3 matrix U with three pivots has rank 3 and a trivial null space, since its three columns are independent.
Stacking U over 2U gives a 10-by-3 matrix of rank 3: duplicated row information does not add pivots.
For the block matrix formed from U and zero blocks, row reduction separates the independent blocks; with rank 6 and size 10-by-6, its transpose has a four-dimensional null space.
The examples reinforce that rank counts pivots, while nullity is the number of columns minus rank.

10:28 Recovering a Matrix from a Complete Solution to Ax=b

A complete solution reveals the null space: two independent null-space vectors imply nullity 2 and, by rank-nullity, rank 1 for a matrix with three columns.
The particular solution determines the column-space direction; the right-hand side shown is a multiple of (1, 2, 1), so the column space is its span.
A null-space vector such as (0, 0, 1) forces the third column of A to be zero; another null-space relation constrains the second column.
The system Ax=b is solvable exactly when b lies in the column space, not for every right-hand side.

18:14 Square-Matrix Tests: Null Spaces, Subspaces, and Nilpotence

For a square matrix, a trivial null space for A implies a trivial null space for A transpose, because full column rank also gives full row rank.
Invertible 5-by-5 matrices do not form a subspace: the zero matrix is excluded, and sums of invertible matrices need not be invertible.
Singular matrices are not a subspace either, since adding two singular matrices can produce an invertible matrix.
The claim that B squared equals zero forces B to equal zero is false; a nonzero nilpotent matrix can square to the zero matrix.

22:05 Independent Columns Make a Square System Solvable for Every Right-Hand Side

For an m-by-m matrix, independent columns mean rank m and therefore invertibility.
An invertible square matrix makes Ax=b solvable for every right-hand side b.
This full-rank case contrasts with rectangular systems, where independence or spanning alone may not guarantee both properties.

25:26 Finding a Null Space and Particular Solution from a Matrix Factorization

The example matrix B is presented as a product of a square invertible factor and a 3-by-4 factor, so its null space matches that of the 3-by-4 factor.
The 3-by-4 factor has rank 2, leaving two free variables and two special solutions that form a basis for its null space.
For Bx=(1, 0, 1), the right-hand side equals the first column of B, so the first standard basis vector is a convenient particular solution.
Every solution is the particular solution plus an arbitrary linear combination of the two null-space basis vectors.

36:43 What the Four Subspaces Reveal About Different Matrices

A square matrix need not have identical row and column spaces; a 2-by-2 example has row space along (0, 1) and column space along (1, 0).
A symmetric matrix does have equal row and column spaces because its rows and columns are the same vectors.
A and negative A share all four fundamental subspaces, but matrices with the same four subspaces need not be scalar multiples.
Any two invertible 6-by-6 matrices share row and column spaces equal to R^6 and both null spaces equal to {0}.

43:15 Row Operations Preserve Row Space and Null Space; Row Space Is Orthogonal to Null Space

Exchanging two rows preserves the row space and the null space, but generally changes the column space.
A nonzero vector cannot be both a row of A and a vector in the null space of A: its dot product with that row would have to be zero, but (1, 2, 3) dotted with itself is 14.
More generally, every vector in the row space is perpendicular to every vector in the null space, so their intersection contains only the zero vector.
The review closes by introducing orthogonality as the next topic in the course.

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