Try
Consider two real matrices A with rows (1,2),(0,3) and B with rows (2,0),(1,1). Their determinants are 3 and 2. Multiply them in that order: AB has rows (4,2),(3,3), whose determinant is 12−6=6. This agrees with the product 3·2, but a successful numerical example is not a proof for arbitrary matrices.
Now change B to have rows (2,4),(1,2). Its columns are dependent, so its determinant is zero. The product has rows (4,8),(3,6), also with determinant zero. A general proof must include this case. An argument that first divides by det(B) cannot cover it.
The preceding lesson defines the determinant as the normalized alternating multilinear function of columns and proves its uniqueness. Here alternating means the value is zero whenever two columns agree. That definition remains valid in characteristic two; a sign-change equation alone would not force zero there.
Predict
Before reading the proof, decide which predictions are valid: determinants multiply under matrix multiplication; determinants add under matrix addition; determinant one guarantees numerical stability; an invertible block matrix must have an invertible leading block. Only the first assertion is universally valid under the usual matching square sizes. We will give mechanisms or counterexamples for each.
For a fixed n×n matrix A over a field F, define a function of n column vectors by
Φ(v₁,…,vₙ)=det(Av₁,…,Avₙ).
Predict its value on the standard basis e₁,…,eₙ. Those images are exactly the columns of A, so the value is det(A). This is the constant that will appear in the product theorem, including when it is zero.
Experiment
Check the following small cases by multiplying the matrices and using the two-by-two determinant formula. The table uses real arithmetic. Then reduce the first case modulo 3 and explain why the same product identity still holds even though its determinant becomes zero.
| Calculation | Matrices or entries | Result to verify | Interpretation |
|---|
| Product | A=(1,2;0,3), B=(2,0;1,1) | det(AB)=6=3·2 | Multiplicativity includes reduction modulo a prime |
| Singular product | A as above, B=(2,4;1,2) | det(AB)=0 | A proof must permit zero determinants |
| Sum | A=B=I₂ | det(A+B)=4, det(A)+det(B)=2 | Not additive |
| Leading block | M=(0,1;1,0) | det(M)=−1 | An invertible full matrix can have a zero leading pivot |
The notation (a,b;c,d) lists the first row before the semicolon and the second row after it. For another action, scale just the first column of A by 2, then scale both columns by 2. The determinants become 6 and 12 respectively. Explain why whole-matrix scaling contributes one factor for every column, rather than a single factor overall.
Understand
The map Φ above is multilinear because A is linear and the determinant is multilinear in the resulting columns. It is alternating: if two input columns agree, their images agree, making the determinant zero. The preceding uniqueness theorem therefore gives
Φ(v₁,…,vₙ)=Φ(e₁,…,eₙ) det(v₁,…,vₙ)=det(A) det(v₁,…,vₙ).
Substitute the columns of B. Their images under A are precisely the columns of AB, so det(AB)=det(A)det(B). No determinant was divided out. If A is singular, Φ may be the zero form; uniqueness still applies. The empty dimension also agrees: the determinant of the empty matrix is the empty product 1.
We can now prove the invertibility criterion without assuming it in the argument. A singular square matrix has dependent columns by the independent rank/elimination theorem. In a nontrivial dependence, solve for a column whose coefficient is nonzero. Expanding the determinant in that column gives a sum of terms each repeating another column, hence zero. Thus det(A)≠0 forces independent columns and an inverse. Conversely, if the inverse exists, multiplicativity yields det(A)det(A⁻¹)=1. It follows that the determinant is nonzero and det(A⁻¹)=1/det(A). For invertible S, applying the same rule three times proves det(S⁻¹AS)=det(A): changing basis preserves this scalar.
Triangular matrices admit a second useful proof using permutation terms. In an upper triangular matrix, a nonzero term must select column σ(i)≥i in every row. But the sum of the selected column numbers equals the sum of all row numbers, so each inequality must be equality. Only the diagonal product survives. Zeros on that diagonal are allowed; they make the surviving product zero.
For an upper block triangular matrix with square diagonal blocks P and S and lower-left block zero, every surviving permutation sends all bottom rows to the last columns. Their counts are equal, leaving the first columns for the top rows. The permutation splits into two within-block permutations, with product sign. Summing their products independently gives det(P)det(S). The blocks themselves need not be triangular or invertible. The lower block triangular formula follows by the same argument with top rows, or by transposition.
If P is invertible, a more general block matrix M with blocks P,Q above R,S can be reduced by the block matrix E with blocks I,0 above −RP⁻¹,I. Direct multiplication gives EM with blocks P,Q above 0,S−RP⁻¹Q. Since det(E)=1, the product and block formulas prove
det(M)=det(P) det(S−RP⁻¹Q).
The expression S−RP⁻¹Q is called the Schur complement. Its order of multiplication matters. This formula assumes the inverse of P exists; it does not provide a way to divide by a singular block. In particular, the swap matrix in the table is invertible despite its zero leading scalar block.
Apply
Try a changed block calculation over the real numbers: take P=(1,1;0,1), Q=I₂, R=I₂, S=(3,0;1,2). First compute P⁻¹=(1,−1;0,1). Then S−P⁻¹=(2,1;1,1), whose determinant is 1. Since det(P)=1, the full four-by-four block determinant is 1. Verify at least one block product in E M to check the elimination rather than simply substituting into a memorized formula.
Does determinant one mean that solving a system is well-conditioned? For real M≥1 take A_M=diag(M,1/M). Its determinant is always 1, but its Euclidean operator norm and its inverse norm are both M. Indeed, the largest stretching factor of A_M occurs on the first coordinate, and that of its inverse occurs on the second. Their product, the condition number, is M². This can grow without bound while the determinant remains fixed.
Choose b=(M,0), whose solution is x=(1,0), and perturb it by δb=(0,εM). The relative right-hand-side change is ε, but the solution change is (0,εM²), of relative size εM². This is a computed sensitivity example, not a claim that a particular numerical algorithm always makes such an error. Nonzero determinant certifies exact solvability and uniqueness; it does not guarantee small perturbations stay small.
Continue
Retain the proof dependencies: alternating multilinear uniqueness gives multiplicativity; independent rank facts and multiplicativity give invertibility; permutation restrictions give triangular block formulas; legal block elimination gives the Schur complement. Do not use a result as its own premise or cancel a possibly zero determinant.
The cumulative problems combine these proofs with parameter-dependent rank, cofactors and field characteristics. For numerical interpretation, compare determinant tests with conditioning and sensitivity. The exact proofs here do not require floating-point computation.
