This lecture starts the second half of the course. The focus moves from rectangular matrices to square ones. The two big topics ahead are determinants and eigenvalues. Determinants are needed mainly because eigenvalues depend on them.
Notation. The determinant of \(A\) is written \(\det A\) or \(|A|\). It is a single number that packs in as much information about the matrix as possible.
The famous fact. \(A\) is invertible exactly when \(\det A \neq 0\).
Strang's approach. He does not start with the big formula for the determinant, because it has too much packed into it. Instead he starts from three defining properties and derives everything else from them.
The three defining properties
Property 1. \(\det I = 1\).
Property 2. Exchanging two rows reverses the sign of the determinant.
Consequence: every permutation matrix has \(\det P = \pm 1\). It is \(+1\) if \(P\) comes from an even number of row exchanges and \(-1\) if from an odd number.
Two-by-two check:
Property 3. The determinant is linear in each row separately, with all other rows held fixed.
-
3A, scaling a row:
\[\begin{vmatrix} ta & tb \\ c & d \end{vmatrix} = t\begin{vmatrix} a & b \\ c & d \end{vmatrix}\] -
3B, adding inside a row:
\[\begin{vmatrix} a + a' & b + b' \\ c & d \end{vmatrix} = \begin{vmatrix} a & b \\ c & d \end{vmatrix} + \begin{vmatrix} a' & b' \\ c & d \end{vmatrix}\]
Warning: this does not say \(\det(A + B) = \det A + \det B\). That is false. Linearity holds one row at a time, with the other rows unchanged.
Why it holds for every row, not just row 1. Swap the row you care about into position 1, apply the property, and swap back. The two sign changes from Property 2 cancel.
Throughout, Strang checks each property against the familiar \(2\times2\) formula, \(ad - bc\).
Properties 4–10 (all derived from 1–3)
Property 4: two equal rows give \(\det A = 0\)
Proof.
- Exchange the two equal rows.
- The matrix is unchanged, so the determinant is unchanged.
- But Property 2 says the sign flips.
- The only number equal to its own negative is 0.
This matches what we already know: two equal rows mean rank less than \(n\), so the matrix is singular.
Property 5: subtracting a multiple of one row from another leaves \(\det\) unchanged
Proof (2×2 version; the same argument works for any size). Split the changed row using 3B, then factor out \(-\ell\) using 3A:
Why this matters. Elimination steps don't change the determinant. So
Property 6: a row of zeros gives \(\det A = 0\)
Proof. Use 3A with \(t = 0\). Multiplying the zero row by 0 doesn't change the matrix, but it multiplies the determinant by 0.
Property 7: a triangular matrix has \(\det\) equal to the product of its diagonal entries
The entries above the diagonal don't matter.
Proof, when every \(d_i \neq 0\):
- Eliminate upward (Property 5) to clear the entries above the diagonal. This leaves the diagonal matrix \(\text{diag}(d_1, \dots, d_n)\) with the same determinant.
- Factor each \(d_i\) out of its row (3A). This leaves the identity.
- By Property 1, \(\det I = 1\), so \(\det U = d_1 \cdots d_n\).
If some \(d_i = 0\). Elimination produces a row of zeros, so \(\det = 0\) by Property 6.
How software computes determinants. Eliminate to \(U\), then multiply the pivots, adjusting the sign for row exchanges:
\[\det A = \pm(\text{product of the pivots})\]
Recovering \(ad - bc\) from the properties. For \(\begin{bmatrix} a & b \\ c & d\end{bmatrix}\) with \(a \neq 0\):
- The pivots are \(a\) and \(d - \tfrac{c}{a}b\).
- Their product is \(a\left(d - \tfrac{c}{a}b\right) = ad - bc\).
- If \(a = 0\), a row exchange is needed first. If no exchange helps, the matrix is singular.
Property 8: \(\det A = 0\) exactly when \(A\) is singular
If \(A\) is singular: elimination produces a row of zeros, so \(\det A = 0\).
If \(A\) is invertible: elimination produces \(n\) nonzero pivots, so \(\det A = \pm d_1 \cdots d_n \neq 0\).
Property 9: \(\det(AB) = (\det A)(\det B)\)
Strang calls this very valuable and somewhat surprising: determinants are not additive, yet they are multiplicative.
Consequences:
Inverse. From \(A^{-1}A = I\), taking determinants gives
Squares.
Scaling the whole matrix. For \(A\) of size \(n\times n\):
This connects to volume: doubling every side of a box in \(\mathbb{R}^3\) multiplies its volume by \(2^3 = 8\).
Property 10: \(\det A^\top = \det A\)
2×2 check. \(\begin{vmatrix} a & c \\ b & d \end{vmatrix} = ad - bc\).
Proof. Strang covers the typical case where \(A = LU\) with no row exchanges.
- By Property 9:
\[\det A = \det L \cdot \det U, \qquad \det A^\top = \det(U^\top L^\top) = \det U^\top \cdot \det L^\top\]
- \(L\) and \(L^\top\) are triangular with 1s on the diagonal, so both determinants equal 1 (Property 7).
- \(U\) and \(U^\top\) are triangular with the same diagonal, so \(\det U^\top = \det U\) (Property 7).
- Therefore \(\det A^\top = \det A\).
When row exchanges are needed. Write \(PA = LU\). Then \(\det P^\top = \det P\), because \(P^\top = P^{-1}\) and \(\det P = \pm1\).
Consequence. Every statement about rows now holds for columns too:
- A zero column gives \(\det = 0\).
- Exchanging two columns flips the sign.
- The determinant is linear in each column separately.
A hidden fact: every permutation is either even or odd
Why it matters. Property 2 would be contradictory if some permutation could be reached by both an odd and an even number of row exchanges.
- An odd count would force \(\det = -1\).
- An even count would force \(\det = +1\).
The fact. Every permutation is either even or odd, never both. If one sequence of exchanges producing it has 7 steps, every sequence has an odd number of steps (21, 23, 101, …).
Consequence. This guarantees the determinant is well-defined by Properties 1–3. Strang says the proof takes one more algebraic trick, which he doesn't give here.
Summary table
| # | Property | Source |
|---|---|---|
| 1 | \(\det I = 1\) | definition |
| 2 | Row exchange flips the sign | definition |
| 3 | Linear in each row separately | definition |
| 4 | Two equal rows give \(\det = 0\) | 2 |
| 5 | Elimination steps don't change \(\det\) | 3, 4 |
| 6 | A zero row gives \(\det = 0\) | 3A |
| 7 | Triangular: \(\det\) = product of the diagonal | 5, 3A, 1 |
| 8 | \(\det A = 0\) exactly when \(A\) is singular | 5, 6, 7 |
| 9 | \(\det(AB) = \det A \det B\) | stated, proof sketched |
| 10 | \(\det A^\top = \det A\) | 9, 7 |
Big ideas from Lecture 18
- Three properties define the determinant: \(\det I = 1\), sign flips under row exchange, and linearity in each row.
- Elimination preserves the determinant, up to sign for row exchanges. So \(\det A = \pm(\text{product of the pivots})\), which is how software computes it.
- \(\det A = 0\) exactly when \(A\) is singular.
- Multiplicative, not additive: \(\det(AB) = \det A\det B\), which gives \(\det A^{-1} = 1/\det A\). But \(\det(A+B) \neq \det A + \det B\).
- Scaling the whole matrix: \(\det(tA) = t^n \det A\), consistent with volume.
- \(\det A^\top = \det A\), so every row property is also a column property.