Determinants, trace, block matrices, rank inequalities

In one sentence

The determinant is multiplicative and violently non-additive, block structure is the fastest route through any large matrix, and the adjugate's rank takes only three values.

Why the exam asks it

Because these are the questions that can be made hard without being deep: a $4\times4$ or $5\times5$ matrix with structure, where the intended route takes ten seconds and brute force takes ten minutes. Recognising the structure is the skill being tested.

The idea

What the determinant does and does not respect

$$\det(AB) = \det A \det B, \qquad \det(A^T) = \det A, \qquad \det(cA) = c^{n}\det A .$$

The third catches people constantly: scaling a matrix scales the determinant by $c^n$, not by $c$, because every one of the $n$ rows is scaled.

$\det(A + B)$ has no formula. $\begin{pmatrix}1&0\\0&0\end{pmatrix}$ and $\begin{pmatrix}0&0\\0&1\end{pmatrix}$ are both singular and sum to $I$.

Also standard: $\det(A^{-1}) = 1/\det A$, and $\det$ of a triangular matrix is the product of the diagonal.

Block matrices

$$\det\begin{pmatrix}A & B\\ 0 & D\end{pmatrix} = \det A \cdot \det D$$

for square $A$ and $D$, whatever $B$ is. Same for block lower triangular.

The general formula $\det\begin{pmatrix}A&B\\C&D\end{pmatrix} = \det(AD - BC)$ is not a theorem. It holds when $C$ and $D$ commute, and fails otherwise, for the obvious reason that the "entries" here are matrices. When a question hands you a block matrix, the reliable move is to clear one block by row operations and reduce to triangular form.

Two identities worth carrying:

• Sylvester: $\det(I_m + AB) = \det(I_n + BA)$ for $A$ of size $m\times n$ and $B$ of $n \times m$. It converts a large determinant into a small one, and is the intended route whenever $AB$ is big and $BA$ is $1\times1$.
• Rank-one update: $\det(I + uv^{T}) = 1 + v^{T}u$, which is Sylvester with $n = 1$.

Trace

$\operatorname{tr}(A + B) = \operatorname{tr}A + \operatorname{tr}B$ and $\operatorname{tr}(AB) = \operatorname{tr}(BA)$, always. Trace is additive; determinant is multiplicative; neither is both, and options swap them.

A consequence used often: $AB - BA = I$ is impossible over a field of characteristic $0$, since the left side has trace $0$ and the right has trace $n$.

Rank inequalities

• $\operatorname{rank}(A + B) \le \operatorname{rank}A + \operatorname{rank}B$.
• $\operatorname{rank}(AB) \le \min(\operatorname{rank}A, \operatorname{rank}B)$.
• Sylvester: $\operatorname{rank}A + \operatorname{rank}B - n \le \operatorname{rank}(AB)$.
• Frobenius: $\operatorname{rank}(AB) + \operatorname{rank}(BC) \le \operatorname{rank}(B) + \operatorname{rank}(ABC)$.

Sylvester is the one that gets used: it is what proves $AB = 0$ forces $\operatorname{rank}A + \operatorname{rank}B \le n$.

The adjugate

$\operatorname{adj}(A)$ has $(i,j)$ entry the $(j,i)$ cofactor, and

$$A \cdot \operatorname{adj}(A) = \operatorname{adj}(A)\cdot A = (\det A)\,I .$$

$$\operatorname{rank}\operatorname{adj}(A) = \begin{cases} n & \operatorname{rank}A = n\\ 1 & \operatorname{rank}A = n-1 \\ 0 & \operatorname{rank}A \le n-2\end{cases}$$

The middle case is the interesting one: if $\operatorname{rank}A = n-1$ then some $(n-1)$-minor is nonzero so $\operatorname{adj}A \ne 0$, while $A\operatorname{adj}A = 0$ forces every column of $\operatorname{adj}A$ into the one-dimensional null space of $A$. Also $\det(\operatorname{adj}A) = (\det A)^{n-1}$.

Special determinants

• Vandermonde: $\det(x_j^{i-1}) = \prod_{i<j}(x_j - x_i)$. Nonzero exactly when the nodes are distinct, which is why polynomial interpolation is uniquely solvable.
• Skew-symmetric of odd order: determinant $0$, since $\det A = \det A^T = \det(-A) = (-1)^n \det A$.
• Circulant: eigenvalues are $\sum_j c_j \omega^{jk}$ over the $n$-th roots of unity.

Where intuition breaks

"$\det(cA) = c\det A$." It is $c^n \det A$. The most common arithmetic slip in the topic.

"$\det(A + B) = \det A + \det B$." No formula exists. Two singular matrices can sum to the identity.

"$\det\begin{pmatrix}A&B\\C&D\end{pmatrix} = \det(AD-BC)$." Only when $C$ and $D$ commute. Tempting because it looks like the $2\times2$ formula, and it is that formula with the wrong kind of entries.

"$\operatorname{rank}(\operatorname{adj}A) = 2$." Impossible for any $A$ of any size — the rank is $n$, $1$ or $0$. Options offering an intermediate value can be struck without computing anything.

"$AB = 0$, so $A = 0$ or $B = 0$." Matrices have zero divisors. What is true is $\operatorname{rank}A + \operatorname{rank}B \le n$, by Sylvester, and that is usually what the question wants.

"$\operatorname{tr}(AB) = \operatorname{tr}A\operatorname{tr}B$." Trace is additive, not multiplicative. Determinant is the multiplicative one.

"A skew-symmetric matrix has determinant zero." Only in odd order. In even order it is the square of the Pfaffian, and generally nonzero: $\begin{pmatrix}0&1\\-1&0\end{pmatrix}$ has determinant $1$.

"$\det A = 0$, so $A$ has a zero row or column." Singularity is a statement about dependence among rows, not about any row being zero.

The exam's angle

1. Look for block structure before computing anything. A $4\times4$ with a zero block is two $2\times2$ determinants.
2. Look for rank-one structure. $\det(I + uv^T) = 1 + v^Tu$ turns an $n\times n$ determinant into a dot product.
3. Use trace to kill impossibilities. Anything of the form $AB - BA = cI$ with $c \ne 0$ is impossible.
4. For adjugate questions, get $\operatorname{rank}A$ and read the table. No computation needed.
5. Recognise Vandermonde on sight — rows or columns in geometric progression in the entries.

The night before

• $\det(AB) = \det A\det B$; $\det(cA) = c^n \det A$; $\det(A+B)$ has no formula.
• Block triangular: multiply the diagonal blocks. The general block formula needs $CD = DC$.
• Sylvester: $\det(I + AB) = \det(I + BA)$. Rank one: $\det(I + uv^T) = 1 + v^Tu$.
• $\operatorname{tr}$ additive and cyclic; $\det$ multiplicative. $AB - BA = I$ is impossible.
• $\operatorname{rank}A + \operatorname{rank}B - n \le \operatorname{rank}(AB) \le \min$.
• $A\operatorname{adj}A = (\det A)I$; $\operatorname{rank}\operatorname{adj}A \in \{n, 1, 0\}$; $\det(\operatorname{adj}A) = (\det A)^{n-1}$.
• Vandermonde $= \prod_{i<j}(x_j - x_i)$, nonzero iff nodes distinct.
• Skew-symmetric: determinant $0$ in odd order only.