Format. Strang works through problems from old exams. The emphasis is on Chapter 3: rectangular matrices, the four subspaces, rank, and dimension.
Problem 1: Dimension of a span
Question. \(u, v, w\) are nonzero vectors in \(\mathbb{R}^7\). What are the possible dimensions of the subspace they span?
Answer: 1, 2, or 3.
- At most 3, because there are only three vectors.
- At least 1, because the vectors are nonzero. Dimension 0 would require all three to be the zero vector.
Problem 2: Block matrices built from \(U\)
\(U\) is a \(5 \times 3\) matrix in echelon form with \(r = 3\) pivots. Treat it as already in reduced form.
(a) What is \(N(U)\)? The rank equals the number of columns, so the 3 columns are independent:
(b) \(B = \begin{bmatrix} U \\ 2U \end{bmatrix}\) (\(10 \times 3\)). What are its reduced form and rank? Subtract 2 × the top block from the bottom block:
(c) \(C = \begin{bmatrix} U & U \\ U & 0 \end{bmatrix}\) (\(10 \times 6\)). What are its reduced form and rank?
- Subtract the top block row from the bottom: \(\begin{bmatrix} U & U \\ 0 & -U \end{bmatrix}\).
- Add the new bottom block row to the top: \(\begin{bmatrix} U & 0 \\ 0 & -U \end{bmatrix}\).
- Multiply the bottom block row by \(-1\): \(\begin{bmatrix} U & 0 \\ 0 & U \end{bmatrix}\).
The rank is \(\text{rank}(C) = 6\).
One last cosmetic step for true reduced form: each \(U\) has zero rows, and those would be moved to the bottom.
(d) What is \(\dim N(C^\top)\)? \(C\) is \(10 \times 6\), so \(m = 10\). Therefore
Problem 3: Reconstructing \(A\) from its complete solution
Given. The complete solution of \(Ax = (2,4,2)\) is
Rank and dimensions.
- Two independent vectors span the nullspace, so \(\dim N(A) = 2\).
- \(A\) has 3 columns, so \(r = 3 - 2 = 1\).
- The row space therefore has dimension 1.
Recovering \(A\), one column at a time.
- \(A(2,0,0) = (2,4,2)\) gives 2 × column 1 = \((2,4,2)\), so column 1 is \((1,2,1)\).
- \((0,0,1) \in N(A)\) means column 3 times 1 gives zero, so column 3 is \(0\).
- \((1,1,0) \in N(A)\) means column 1 + column 2 = 0, so column 2 is \(-(1,2,1)\).
For which \(b\) is \(Ax = b\) solvable? Exactly when \(b \in C(A)\), that is, when \(b\) is a multiple of \((1,2,1)\). The given \(b = (2,4,2)\) is such a multiple, which is consistent.
The lesson. A complete solution \(x_p + x_n\) carries a lot of information about \(A\).
Problem 4: Null space of a product
\(B\) is \(3 \times 4\), so \(N(B)\) is a subspace of \(\mathbb{R}^4\).
Key fact: an invertible left factor doesn't change the nullspace
If \(C\) is invertible, then \(N(CD) = N(D)\).
Proof, in both directions.
- If \(Dx = 0\), then \(CDx = C\,0 = 0\).
- If \(CDx = 0\), multiply on the left by \(C^{-1}\) to get \(Dx = 0\).
\(C\) is invertible here. The middle column is \((1,1,0)\). In any combination of the columns that gives zero, look at the second row: only the middle column has a nonzero entry there, so its coefficient must be 0. The remaining two columns, \((1,0,1)\) and \((0,0,1)\), are clearly independent. Alternatively, \(\det C = 1\).
Basis for \(N(B)\)
Read it from \(D\). \(D\) is already in reduced form, with pivot columns 1, 2 and free columns 3, 4. The rank is 2.
Special solutions:
- Set \(x_3 = 1\): \(s_1 = (1, -1, 1, 0)\).
- Set \(x_4 = 1\): \(s_2 = (-2, 1, 0, 1)\).
Check with \(-F\). The free block of \(D\) is \(F = \begin{bmatrix} -1 & 2 \\ 1 & -1 \end{bmatrix}\). The pivot entries of the special solutions are exactly \(-F\).
Complete solution of \(Bx = (1, 0, 1)\)
A particular solution. Column 1 of \(B\) is \(C \cdot (\text{column 1 of } D) = C\,(1,0,0)^\top = (1,0,1)\), which equals \(b\). So \(x_p = (1,0,0,0)\).
The complete solution:
Any solution can serve as \(x_p\). A student asked about this. "Particular" means one solution, not a specific required one. You pick the most convenient. The pattern "particular solution plus nullspace" holds for all linear systems, not just matrices.
True / False round
| Statement | Answer | Reason |
|---|---|---|
| \(A\) square and \(N(A) = \{0\}\) \(\Rightarrow\) \(N(A^\top) = \{0\}\) | True | Full rank: \(r = n = m\) |
| Invertible \(5\times5\) matrices form a subspace of the 25-dimensional space \(M\) | False | The zero matrix isn't invertible, and sums can be singular. Singular matrices don't form a subspace either, because sums can be invertible. |
| \(B^2 = 0 \Rightarrow B = 0\) | False | \(B = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\) has \(B^2 = 0\). Strang flags this matrix as a recurring source of trouble later in the course. |
| \(m \times m\) system with independent columns is solvable for every \(b\) | True | Rank \(m\), so the matrix is invertible |
| \(m = n\) \(\Rightarrow\) row space = column space | False | For \(\begin{bmatrix} 0&1\\0&0 \end{bmatrix}\), the row space is multiples of \((0,1)\) and the column space is multiples of \((1,0)\). It is true for symmetric matrices. |
| \(A\) and \(-A\) have the same four subspaces | True | Scaling by \(-1\) changes neither the spans nor the nullspaces |
| Same four subspaces \(\Rightarrow\) \(A\) is a multiple of \(B\) | False | Any two invertible \(6\times6\) matrices share the same subspaces: \(\mathbb{R}^6\), \(\mathbb{R}^6\), \(\{0\}\), \(\{0\}\). They do share the same rank. |
| Exchanging two rows of \(A\) preserves which subspaces? | Row space and nullspace | The column space generally changes |
Closing question: a preview of orthogonality
Question. Why can't \(v = (1,2,3)\) be both a row of \(A\) and in \(N(A)\)?
Answer. If \((1,2,3)\) were a row of \(A\), then that row's entry in \(Av\) would be
The general fact. The row space and the nullspace, both subspaces of \(\mathbb{R}^n\), share only the zero vector. More strongly, the nullspace is perpendicular to the row space. This is the topic of Lecture 14.
Review checklist
- Four dimensions: \(\dim C(A) = \dim C(A^\top) = r\), \(\dim N(A) = n - r\), \(\dim N(A^\top) = m - r\).
- Complete solution \(x = x_p + x_n\). This form can be used in reverse to reconstruct \(A\).
- Invertible factors: \(N(CD) = N(D)\) when \(C\) is invertible.
- Block elimination: the same row operations work on blocks.
- Sets that are not subspaces: invertible matrices, singular matrices, matrices of a fixed rank.
- Row operations preserve the row space and the nullspace, but not the column space.
- Full-rank cases: don't overlook \(r = m\) and \(r = n\).