Determinants and eigenvalue models

A determinant compresses a square matrix into a signed volume factor. Eigenvalues instead describe directions or modes of the transformation. Neither replaces row reduction, but each reveals structure that repeated arithmetic can conceal.

Build the determinant from smaller matrices

Definition

Submatrix, minor, and cofactor

Let $A=[a_{ij}]$ be an $n\times n$ matrix with $n\ge 2$. For indices $1\le i,j\le n$, delete row $i$ and column $j$ simultaneously. The resulting $(n-1)\times(n-1)$ matrix is the submatrix $A(i\mid j)$.

The minor belonging to $a_{ij}$ is the scalar

$$M_{ij}=\det\bigl(A(i\mid j)\bigr).$$

The cofactor belonging to $a_{ij}$ is the signed scalar

$$C_{ij}=(-1)^{i+j}M_{ij}.$$

The signs form a checkerboard whose upper-left sign is positive:

$$\begin{bmatrix} +&-&+&\cdots\\ -&+&-&\cdots\\ +&-&+&\cdots\\ \vdots&\vdots&\vdots&\ddots \end{bmatrix}.$$

The notation $A(i\mid j)$ names a matrix, $M_{ij}$ is a determinant of that matrix, and $C_{ij}$ is that scalar after the sign has been attached. Keeping these three objects distinct prevents a common double-sign error.

Definition

Recursive definition of the determinant

For a $1\times1$ matrix $[a_{11}]$, define

$$\det([a_{11}])=a_{11}.$$

For an $n\times n$ matrix with $n>1$, define its determinant by expansion along the first row:

$$\det(A)=\sum_{j=1}^{n}a_{1j}C_{1j} =\sum_{j=1}^{n}(-1)^{1+j}a_{1j}M_{1j}.$$

This is a definition, not yet a theorem about other rows or columns. Each minor in the sum has size $n-1$, so repeated use of the definition eventually reaches the $1\times1$ base case. No $0\times0$ determinant convention is needed: the submatrix and minor definitions are used only for $n\ge2$.

Arbitrary cofactor expansion from the recursive determinant

Let $A$ be real $n\times n$, $n\ge1$, with determinant defined by the one-by-one base case and first-row recursion. Let $S_n$ be the permutations of $1,\ldots,n$. Define the inversion number $\operatorname{inv}(\sigma)$ as the number of pairs $r<s$ with $\sigma(r)>\sigma(s)$, and put $\operatorname{sgn}(\sigma)=(-1)^{\operatorname{inv}(\sigma)}$.

Suppose $\sigma(i)=j$ and delete position $i$ and value $j$, keeping the relative orders of the remaining positions and values; the resulting permutation is $\tau\in S_{n-1}$. If $u$ entries before position $i$ are smaller than $j$, the inversions removed are $(i-1-u)+(j-1-u)=i+j-2-2u$. Consequently

$$\operatorname{sgn}(\sigma)=(-1)^{i+j}\operatorname{sgn}(\tau).$$

Define $P_n(A)=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{r=1}^n a_{r,\sigma(r)}$. For $n=1$, it is the recursive determinant. For $n>1$, grouping by $\sigma(1)=j$ and using the sign identity gives $P_n(A)=\sum_j(-1)^{1+j}a_{1j}P_{n-1}(A(1\mid j))$. Induction therefore proves $P_n(A)=\det A$.

Now fix any row $i$ and group the same finite permutation sum by $\sigma(i)=j$. Deletion is a bijection to all permutations of the corresponding minor, so

$$\det A=\sum_{j=1}^n(-1)^{i+j}a_{ij}\det(A(i\mid j)).$$

For a fixed column $j$, each permutation has a unique position $i$ with $\sigma(i)=j$. Grouping by that position gives

$$\det A=\sum_{i=1}^n(-1)^{i+j}a_{ij}\det(A(i\mid j)).$$

Both expansion formulas concern $n\ge2$, so no empty determinant convention was introduced. This proves the general expansion theorem without assuming alternating row rules or the determinant product theorem.

Why the structural rules follow

Zero rows and zero columns

Theorem

A zero row or zero column gives determinant zero

If every entry in one row or one column of a square matrix is zero, then its determinant is zero.

Theorem

A row or column with one possible nonzero entry

For a square matrix of size $n\ge2$, suppose all entries in row $k$ are zero except possibly $c_{k\ell}$. Then

$$\det(C)=c_{k\ell}(-1)^{k+\ell}\det\bigl(C(k\mid\ell)\bigr).$$

The same formula holds when all entries in column $\ell$ are zero except possibly $c_{k\ell}$. The formula includes the case $c_{k\ell}=0$.

For $n=1$, a matrix with a zero row is $[0]$, whose determinant is $0$ by the base definition. For $n\ge2$, expand along that row using the arbitrary-row theorem. Every term has a zero entry as its factor, so every term is zero and hence $\det(A)=0$.

The zero-column statement can be proved directly by induction, which makes the inheritance of the zero column explicit. For $n=1$, the only $1\times1$ matrix with a zero column is $[0]$, and its determinant is $0$. Assume the statement for $n=k$. Let $C$ be a $(k+1)\times(k+1)$ matrix with zero column $\ell$. Expand along the first row. The term from column $\ell$ is zero. In every other term, delete row $1$ and a column $j\ne\ell$; the original zero column survives as a zero column of the resulting $k\times k$ submatrix (with its index shifted when $j\lt\ell$). By the induction hypothesis each such submatrix has determinant zero. Hence every term is zero, proving $\det(C)=0$.

For the one-support theorem, the row case follows by expanding along row $k$: all terms except the $\ell$-term vanish. For the column case, expand along row $k$. The $\ell$-term remains. For each $j\ne\ell$, the submatrix $C(k\mid j)$ inherits a zero column from column $\ell$ of $C$, so its determinant is zero by the zero-column result. Therefore only $c_{k\ell}(-1)^{k+\ell}\det(C(k\mid\ell))$ remains. This is the precise reason that a sparse row or column collapses the calculation to one minor.

Triangular matrices

Theorem

Triangular determinant theorem

If $T$ is upper triangular or lower triangular, then

$$\det(T)=t_{11}t_{22}\cdots t_{nn}.$$

We prove the upper-triangular statement by induction on $n$. For $n=1$, it is the determinant definition. Suppose it holds for upper-triangular $k\times k$ matrices, and let $T$ be upper triangular of size $k+1$. Its last row has zero entries except possibly $t_{k+1,k+1}$. Expanding along that last row gives

$$\det(T)=t_{k+1,k+1}\det\bigl(T(k+1\mid k+1)\bigr),$$

because the sign is $(-1)^{2k+2}=1$. The retained submatrix is still upper triangular, with diagonal $t_{11},\ldots,t_{kk}$. By induction its determinant is $t_{11}\cdots t_{kk}$, giving the required product through $t_{k+1,k+1}$. For a lower-triangular matrix, expand along the first row, whose only possible nonzero entry is $t_{11}$; the retained submatrix is lower triangular with diagonal $t_{22},\ldots,t_{nn}$, and the same induction proves the result.

Multilinearity in one fixed row

Theorem

Multilinearity in one fixed row

Fix every row except row $p$. If the new row is $\alpha u+\beta v$, then

$$\det(\text{same rows except }\alpha u+\beta v) =\alpha\det(\text{same rows except }u) +\beta\det(\text{same rows except }v).$$

For $n=1$, the base definition gives

$$\det([\alpha a+\beta b])=\alpha a+\beta b =\alpha\det([a])+\beta\det([b]).$$

For $n\ge2$, fix every row except row $p$. Let matrices $A,B,C$ agree in all those fixed rows, and suppose their $p$th rows satisfy $c_p=\alpha a_p+\beta b_p$. For each column $j$, deleting row $p$ gives the same submatrix:

$$A(p\mid j)=B(p\mid j)=C(p\mid j).$$

Expanding $\det(C)$ along row $p$ and using the entrywise relation $c_{pj}=\alpha a_{pj}+\beta b_{pj}$ gives

$$\begin{aligned} \det(C) &=\sum_{j=1}^{n}(\alpha a_{pj}+\beta b_{pj})(-1)^{p+j}\det(C(p\mid j))\\ &=\alpha\det(A)+\beta\det(B). \end{aligned}$$

The equality works precisely because every other row is held fixed; changing two rows at once is a different statement.

Adjacent swaps and repeated rows

Theorem

Repeated rows force determinant zero

If a square matrix has two identical rows, then its determinant is $0$.

Let $A$ have adjacent rows $r_q,r_{q+1}$, and let $B$ be obtained by swapping those two physical rows. Expand $\det(A)$ along row $q$. Expand $\det(B)$ along row $q+1$, where the row $r_q$ has moved. For each column $j$, the retained submatrix in the two expansions is the same matrix: deleting the position occupied by $r_q$ leaves the other row in the same order. However, the signs differ:

$$(-1)^{q+1+j}=-(-1)^{q+j}.$$

The entries and minors therefore match term by term with the opposite sign, so $\det(B)=-\det(A)$. Swapping rows in positions $p\lt q$ can be realised by $2(q-p)-1$ adjacent swaps, an odd number, and hence also changes the sign.

If rows $p$ and $q$ of $A$ are identical, the matrix obtained by swapping them is still $A$. The swap result then gives $\det(A)=-\det(A)$. Since the entries are real, this implies $2\det(A)=0$ and therefore $\det(A)=0$. This proof does not rely on a later row-replacement rule.

Indexing and strategy

For each term, keep three objects visible: the selected entry, its submatrix after deleting one row and one column, and the minor obtained by taking that submatrix determinant. Write the checkerboard sign from the original row and column positions before evaluating the minor. When a minor is expanded again, its local indices guide the inner sign; they do not replace the outer sign.

The arbitrary-row and arbitrary-column theorem makes the choice of expansion line strategic. Choose a line with many zeros or simple surviving minors, and state that choice before calculating. A zero entry removes a branch immediately, whereas a zero minor must be established by its own calculation. The recursive first-row formula defines the determinant; arbitrary-line expansion is the theorem that permits this computational choice.

The determinant product follows from a finite multilinear expansion

Let $A,B$ be real $n\times n$ matrices with $n\ge1$. Use the already proved determinant permutation formula, multilinearity in the rows, zero determinant for repeated rows, and sign reversal for a row exchange. Row $i$ of $AB$ is $\sum_{k=1}^na_{ik}B_k$, where $B_k$ is row $k$ of $B$. Expanding the finite multilinear expression gives

$$\det(AB)=\sum_{k_1,\ldots,k_n=1}^n\left(\prod_{i=1}^na_{i k_i}\right)\det(B_{k_1},\ldots,B_{k_n}).$$

Tuples with a repeated index have zero determinant. Every remaining tuple is a permutation $\sigma$ of $1,\ldots,n$. Sort it by swapping adjacent inverted neighbours. If a list is not increasing, such neighbours exist. Each swap removes exactly their mutual inversion; comparisons with every other entry contribute the same total before and after. Thus exactly $\operatorname{inv}(\sigma)$ swaps sort the list. The row-exchange rule then gives $\det(B_{\sigma(1)},\ldots,B_{\sigma(n)})=(-1)^{\operatorname{inv}(\sigma)}\det B$. Substitution into the finite sum and the proved permutation formula for $A$ give $\det(AB)=(\det A)(\det B)$. No invertibility assumption or infinite expansion was used.

The adjugate identity and continuity of inversion

For $n\ge2$, let $C_{ij}=(-1)^{i+j}\det(A(i\mid j))$ and define $\operatorname{adj}(A)_{ji}=C_{ij}$. The $(i,j)$ entry of $A\operatorname{adj}(A)$ is $\sum_k a_{ik}C_{jk}$. If $i=j$, row expansion makes this $\det A$. If $i\ne j$, it is the row-$j$ expansion of the matrix obtained by replacing row $j$ by row $i$: its row-$j$ cofactors are unchanged, and its repeated rows make its determinant zero. Hence $A\operatorname{adj}(A)=(\det A)I$.

The permutation formula also gives $\det(A^T)=\det A$: replace a permutation by its inverse. Their inversion numbers agree, since an inversion $(r,s)$ corresponds to the inversion $(\sigma(s),\sigma(r))$ of the inverse, and their entry products agree after reindexing. This applies to every minor too, so $\operatorname{adj}(A^T)=(\operatorname{adj} A)^T$. Applying the first identity to $A^T$ and transposing yields $\operatorname{adj}(A)A=(\det A)I$. For $n=1$, define $\operatorname{adj}([a])=[1]$ and both identities are immediate.

Nonsingularity after the adjugate identity

The established cofactor expansions and repeated-row rule give $A\operatorname{adj}A=\operatorname{adj}A\,A=(\det A)I$, with $\operatorname{adj}[a]=[1]$ when $n=1$. If $\det A\ne0$, division supplies a two-sided inverse. Conversely, if $A$ has an inverse, the product formula gives $1=\det I=\det A\det(A^{-1})$, so $\det A\ne0$. The adjugate identity is an explicitly selected earlier proof here, not an assumption that every matrix is invertible.

1 The determinant and its consequences

For $A=(a_{ij})\in\mathbb R^{n\times n}$ define

$$\det A=\sum_{\sigma\in S_n}\operatorname{sgn}(\sigma)\prod_{i=1}^n a_{i,\sigma(i)}.$$

Here $S_n$ is the set of permutations and the sign is $+1$ or $-1$ according to parity. This formula makes the determinant linear in each row separately and alternating under exchange of two rows, with $\det I=1$. Two equal rows therefore give zero. Adding a multiple of one row to another does not change the determinant; scaling a row scales it; a row exchange reverses its sign. Triangular matrices have determinant equal to the product of diagonal entries.

Invertibility criterion. $A$ is invertible if and only if $\det A\ne0$.

Proof. Row reduction changes the determinant only by nonzero scale factors or sign. If there are $n$ pivots, reduction reaches $I$, whose determinant is 1, and the reversible operations also construct an inverse. Otherwise an echelon row is zero, so the determinant is zero and the homogeneous system has a free variable, making $A$ noninjective and hence noninvertible. $\square$

The product law $\det(AB)=\det A\det B$ follows by viewing the determinant of $AB$ as an alternating multilinear function of the rows of $A$: its value at $I$ is $\det B$, so expansion in standard basis rows gives $\det B$ times $\det A$. In particular, similarity preserves determinants.

For $\begin{pmatrix}2&1\\1&2\end{pmatrix}$ the determinant is 3. This is area scaling by 3, not length scaling by 3 in every direction.

2 Eigenvalues and independent modes

An eigenvector $v\ne0$ satisfies $Av=\lambda v$. The characteristic polynomial is $p_A(t)=\det(tI-A)$; its roots over the chosen scalar field are the eigenvalues. The eigenspace is $\ker(A-\lambda I)$.

Write $E_A(\lambda)=\ker(A-\lambda I)$ over the chosen scalar field.

Theorem

Eigenvectors of distinct eigenvalues are linearly independent

If $v_1,\dots,v_k$ are eigenvectors of $A$ corresponding to pairwise distinct eigenvalues $\lambda_1,\dots,\lambda_k$, then

$$v_1,\dots,v_k$$

are linearly independent.

Proof

Induction without assuming independence

Suppose $\lambda_1,\dots,\lambda_s$ are pairwise distinct and $w_j\in E_A(\lambda_j)$. We prove by induction on $s$ that $\sum_{j=1}^s w_j=0$ forces every $w_j=0$, allowing some $w_j$ to be zero. The case $s=1$ is immediate. For the induction step, apply $A-\lambda_s I$ to the relation. The last term vanishes and we obtain

$$\sum_{j=1}^{s-1}(\lambda_j-\lambda_s)w_j=0.$$

Each scaled vector remains in its own eigenspace. The induction hypothesis forces every $(\lambda_j-\lambda_s)w_j$ to be zero; distinctness makes each scalar nonzero, so $w_j=0$ for $j<s$, and the original relation gives $w_s=0$. Grouping a relation by eigenvalue then proves that the union of bases chosen inside the eigenspaces is independent: the grouped vector in each eigenspace is zero, and independence inside that eigenspace kills every coefficient.

A matrix is diagonalizable if an eigenbasis exists: $A=PDP^{-1}$. Then $A^k=PD^kP^{-1}$. Distinct $n$ eigenvalues guarantee diagonalizability, but repeated eigenvalues do not decide it. The identity has repeated eigenvalue 1 and is diagonalizable; $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ has only a one-dimensional eigenspace and is not.

3 Worked transition model

Consider the column-stochastic matrix

$$A=\begin{pmatrix}0.8&0.3\\0.2&0.7\end{pmatrix}.$$

It maps a two-category probability vector to the next period. Column sums are 1, so total mass is preserved. Its characteristic polynomial is $(t-1)(t-0.5)$. Eigenvectors can be chosen as $v=(3,2)$ for eigenvalue 1 and $w=(1,-1)$ for 0.5. Starting from $(1,0)$,

$$\binom10=\frac15v+\frac25w,\qquad A^k\binom10=\binom{3/5}{2/5}+\frac25(1/2)^k\binom1{-1}.$$

The distribution tends to $(3/5,2/5)$. This conclusion uses the specific decaying second eigenvalue. A matrix $\begin{pmatrix}0&1\\1&0\end{pmatrix}$ also preserves probabilities but alternates forever from $(1,0)$, since its second eigenvalue is $-1$.

4 Graded practice

E1 Foundation. Compute the determinant of $\begin{pmatrix}1&2&0\\0&3&1\\0&0&4\end{pmatrix}$. What happens after exchanging its first and third rows?

E2 Modes. Diagonalize $B=\begin{pmatrix}2&1\\0&3\end{pmatrix}$ and obtain $B^k$ for nonnegative integers $k$.

E3 Counterexample. Show $J=\begin{pmatrix}2&1\\0&2\end{pmatrix}$ is not diagonalizable and calculate $J^k$.

E4 Reasoning. Explain why a real plane rotation by $90$ degrees has no real eigenvector although its determinant is 1.

5 Full solutions

S1. Triangularity gives $1\cdot3\cdot4=12$. A single row exchange gives determinant $-12$.

S2. For 2 choose $(1,0)$; for 3 choose $(1,1)$. Thus $P=\begin{pmatrix}1&1\\0&1\end{pmatrix}$ and $D=\operatorname{diag}(2,3)$. Multiplication gives $B^k=\begin{pmatrix}2^k&3^k-2^k\\0&3^k\end{pmatrix}$. At $k=0$ this is $I$, as required.

S3. The only eigenvalue is 2 and its eigenspace has vectors $(x,0)$, so no eigenbasis exists. Write $J=2I+N$ with $N^2=0$. The binomial formula gives $J^k=2^kI+k2^{k-1}N$ for $k\ge1$, and $J^0=I$. The extra factor $k$ is missed by pretending the matrix were diagonal.

S4. A nonzero real eigenvector would have to remain on its original line, but a right-angle rotation sends it to a perpendicular line. Algebraically $t^2+1$ has no real root. Determinant nonzero guarantees invertibility, not real diagonalizability.

Summary

Determinants detect singularity and signed volume scaling. Eigenvalues describe invariant modes and power behavior. Always specify the scalar field and check eigenvector dimensions when eigenvalues repeat; determinant information alone cannot supply an eigenbasis.