4.5.44 · D5Linear Algebra (Full)
Question bank — Subspaces — four fundamental subspaces of a matrix
Before we start, two words we lean on constantly:
- is — that means rows (one per output coordinate) and columns (one per input coordinate). So inputs have entries, outputs have entries.
- = number of pivots after row reduction — see Rank of a Matrix and RREF and Pivots.
True or false — justify
Each claim is either right or subtly broken. State which, and why.
The null space is a subspace of .
True — means , and has entries (one per column), so it lives in the input space .
The column space is a subspace of .
False — outputs have entries, so . Only the row space and null space live in .
If has only the solution , then is invertible.
Only if is square. Trivial null space means the columns are independent; for a square matrix that forces invertibility, but a tall () matrix can have without being invertible.
The row space and column space of any matrix are the same set.
False — they have the same dimension , but while , so unless they can't even be compared. Equal dimension, different homes.
For a matrix of rank , the null space has dimension .
True — . Count from columns (), never from rows.
Adding a row of zeros to leaves all four subspace dimensions unchanged.
True for , , ; but a zero row adds nothing to the row space and doesn't change rank, so dimension stays too. Only grows by 1 because increased.
Every vector orthogonal to all rows of lies in .
True — orthogonality to every row means each row-dot-, i.e. for all , i.e. . That's exactly the definition of the null space, and it is — see Orthogonal Complements.
If then has exactly one solution.
False — guarantees at least one solution. Uniqueness needs ; otherwise you can add any null-space vector and still land on . See Solving Ax=b.
The left null space contains inputs that get sent to zero.
False — ; these are output-space vectors orthogonal to every column, marking the directions that make unsolvable.
Spot the error
Each line states a plausible-sounding claim with a hidden mistake. Name it.
" is a subspace of because it's a straight line."
Wrong — it misses the origin: at you get . A line must pass through to satisfy the zero-vector rule.
"Since row rank = column rank, the row space equals the column space."
Confuses equal dimension with equal set. Row rank column rank is a statement about counts; the spaces sit in different dimensions unless is square (and even then need not coincide).
" is with rank , so ."
Wrong subtraction — the left null space uses : . The plain null space uses : .
"To find I take the pivot columns of the RREF, since those are independent."
Take pivot columns of the original , not the RREF. Row operations change the column space, so RREF columns span the wrong set; only the pivot positions transfer.
" and are orthogonal complements."
They live in different spaces ( vs ) so can't be complements. The correct pairs are (in ) and (in ).
"Rank of equals the number of nonzero rows in ."
No — count nonzero rows after row reduction to echelon form, not in the raw matrix. Original rows can be dependent (e.g. row 2 row 1) and still both nonzero.
"If two columns of are identical, the rank drops but the null space stays trivial."
Contradiction — identical columns are dependent, so a nonzero combination gives (put on those two columns). That vector is in , so .
Why questions
Answer with the mechanism, not a restatement.
Why must a subspace always contain the zero vector?
Closure under scalar multiplication forces it: pick any and scalar , then . A nonempty closed set can't avoid the origin.
Why does always hold?
The pivot columns give independent row-space directions and the free variables give independent null directions; together they partition all input coordinates — this is the Rank–Nullity Theorem.
Why is automatically a combination of 's columns?
Because : each input entry scales column . So the reachable outputs are exactly .
Why does come "for free" from ?
Each entry of is a row dotted with . Setting says every row-dot-, so is perpendicular to every row and hence to their whole span, the row space.
Why do the four dimensions determine solvability and uniqueness of ?
Solvable (dimension tells how big the reachable set is); unique (dimension tells how many free directions you can add). Solving Ax=b ties them together.
Why is the left null space useful in least squares?
The residual of a projection lands in ; forcing it there is exactly the normal equation — see Least Squares & Projections.
Edge cases
Boundary and degenerate inputs — never let the reader hit an unshown scenario.
The zero matrix : what are its four subspaces?
, so , , (everything is crushed), (nothing is reachable).
An invertible matrix: describe all four.
, so and both null spaces are . Nothing is crushed, every output is reachable — the machine is a perfect bijection.
A single row vector (so ): what does each subspace look like?
; (a full line in output space), , is the plane (dim ), since .
A single column vector (so ): the four subspaces?
; (a line in ), (), , is the plane in orthogonal to , dim .
Can the null space and left null space both be for a non-square matrix?
No — and can't both be unless , which forces a square (invertible) matrix. A rectangular matrix always has at least one nontrivial null space.
A wide matrix with full row rank : which subspace is forced nontrivial?
has dimension , so infinitely many inputs map to any reachable ; but and , so every output is reachable.
A tall matrix with full column rank : which subspace is forced nontrivial?
has dimension , so some outputs are unreachable (least-squares territory); but , so any solvable is unique.
If , are the orthogonality pairings still valid?
Yes — trivially and they sum to ; the Fundamental Theorem holds even in this degenerate case with one partner being all of the space.
Recall One-line self-test
Cover the answers and race through the "Spot the error" block: if you can name the mistake in each in under ten seconds, you've internalised the input/output ( vs ) and dimension (, , ) split.
Connections
- Subspaces — four fundamental subspaces of a matrix — the parent this bank drills.
- Rank of a Matrix — the number every trap hinges on.
- Rank–Nullity Theorem — the bookkeeping.
- Orthogonal Complements — why the pairs are perpendicular.
- Solving Ax=b — solvability and uniqueness traps.
- RREF and Pivots — where pivots (and the pivot-column error) come from.
- Least Squares & Projections — the left-null-space application.