Determinant and Invertibility

We use the Determinant Properties and Row Operations (multilinearity and alternating property of the determinant as a function of rows) together with induction on $n$.

Base case $n = 1$. Here $A = (a_{11})$ and $\det(A) = a_{11}$ by normalization, matching the product of the single diagonal entry.

Inductive step. Suppose the result holds for $(n-1)\times(n-1)$ upper triangular matrices. Let $A$ be $n \times n$ upper triangular. Expand $\det(A)$ along the first column using cofactor expansion (a standard consequence of multilinearity in the rows, established alongside the row-operation properties). Since $A$ is upper triangular, every entry in the first column below row $1$ is zero: $a_{21} = a_{31} = \cdots = a_{n1} = 0$. Cofactor expansion along the first column gives

where $A_{11}$ is the $(n-1)\times(n-1)$ matrix obtained by deleting row $1$ and column $1$ of $A$; all other terms vanish because $a_{i1} = 0$ for $i \geq 2$. The submatrix $A_{11}$ is again upper triangular, with diagonal entries $a_{22}, \dots, a_{nn}$. By the inductive hypothesis,

Combining, $\det(A) = a_{11}a_{22}\cdots a_{nn}$, completing the induction. $\blacksquare$

Proof of the main theorem. We prove the matrix version; the operator version follows immediately by choosing any basis $\mathcal{B}$ of $V$, setting $A = [T]_{\mathcal{B}}$, and using $\det(T) := \det(A)$ together with the matrix statement $A$ invertible $\iff T$ invertible (a linear operator is invertible iff its matrix in any fixed basis is invertible, since $[S]_{\mathcal B}[T]_{\mathcal B} = [ST]_{\mathcal B}$ and $[I]_{\mathcal B} = I_n$).

Step 1: Row reduce $A$ to echelon form. By Gaussian elimination, there is a finite sequence of elementary row operations transforming $A$ into an upper triangular matrix $U$. Each elementary row operation is one of three types:

• (Swap) interchange two rows,
• (Scale) multiply a row by a nonzero scalar $c \in \mathbb{F}$,
• (Add) add a scalar multiple of one row to another row.

By the row-operation properties of the determinant (prerequisite Determinant Properties and Row Operations):

• A swap multiplies the determinant by $-1$.
• A scale by $c \neq 0$ multiplies the determinant by $c$.
• An add operation leaves the determinant unchanged.

Each elementary row operation is itself realized by left-multiplication by an elementary matrix $E$, so that after $k$ operations,

and each $E_i$ is invertible (swap and add matrices are their own or explicit inverses; scale matrices have inverse scaling by $c^{-1}$). Consequently each $\det(E_i) \neq 0$: $\det(E_i) = -1$ for a swap, $\det(E_i) = c \neq 0$ for a scale, and $\det(E_i) = 1$ for an add operation.

Step 2: Relate $\det(A)$ to $\det(U)$ via multiplicativity. By the Multiplicativity of the Determinant, $\det(E_iM) = \det(E_i)\det(M)$ for any square matrix $M$. Applying this repeatedly,

Since each $\det(E_i) \neq 0$, the product $\det(E_k)\cdots\det(E_1)$ is a nonzero scalar, call it $c^\ast \in \mathbb{F}\setminus\{0\}$. Thus

In particular, $\det(A) = 0$ if and only if $\det(U) = 0$, since $c^\ast \neq 0$.

Step 3: Determinant of the echelon form controls its diagonal. By the Lemma above, since $U$ is upper triangular,

where $u_{11}, \dots, u_{nn}$ are the diagonal entries of $U$. This product is nonzero if and only if every $u_{ii} \neq 0$.

Step 4: Diagonal entries of $U$ all nonzero $\iff$ $A$ invertible. We now show these two conditions are equivalent, which will complete the proof via Steps 2–3. ($\Leftarrow$) Suppose $A$ is invertible. Since $U = E_k \cdots E_1 A$ is a product of invertible matrices ($E_1, \dots, E_k$ are invertible as noted, and $A$ is invertible by hypothesis), $U$ is invertible (a product of invertible matrices is invertible, with inverse the product of inverses in reverse order — this is immediate from the definition of invertibility and associativity of matrix multiplication). An upper triangular matrix $U$ is invertible if and only if all its diagonal entries are nonzero: if some $u_{ii} = 0$, consider the smallest such index $i$; then row $i$ of $U$ has zeros in positions $1,\dots,i$ (positions $1,\dots,i-1$ automatically zero by upper triangularity, position $i$ zero by assumption), so the first $i$ rows of $U$ lie in the span of the last $n-i$ standard coordinate directions restricted to columns $1,\ldots$ — more directly, the rows $1,\dots,i$ of $U$ all have zero entries in columns $1,\dots,i$ is false in general, so instead we argue via rank: if $u_{ii}=0$ for some $i$, then columns $1,\dots,i$ of $U$ have nonzero entries only in rows $1,\dots,i-1$ (since $U$ is upper triangular, entries below the diagonal are zero, and the diagonal entry $u_{ii}$ itself is zero). Hence the $i$ columns $1,\dots,i$ of $U$, each an $n$-dimensional vector, actually have all their possibly-nonzero coordinates confined to the $i-1$ positions $1,\dots,i-1$. That is, these $i$ column vectors lie in an $(i-1)$-dimensional subspace of $\mathbb{F}^n$ (spanned by the first $i-1$ standard basis vectors), so they must be linearly dependent (any $i$ vectors in an $(i-1)$-dimensional space are dependent). Linear dependence of columns of $U$ means $U$ is not invertible (a matrix with linearly dependent columns cannot have a left inverse, since an invertible matrix maps a basis, the standard basis, to a linearly independent set, namely its columns). This contradicts invertibility of $U$. So no diagonal entry of $U$ is zero. ($\Rightarrow$) Suppose all diagonal entries $u_{ii} \neq 0$. We construct an explicit inverse of $U$ by back-substitution, and hence of $A$. Since $U$ is upper triangular with nonzero diagonal, for each standard basis vector $e_j$ the system $Ux = e_j$ can be solved uniquely: the last equation reads $u_{nn}x_n = (e_j)_n$, solvable for $x_n$ since $u_{nn}\neq 0$; substituting back, each subsequent equation $u_{ii}x_i + (\text{known terms}) = (e_j)_i$ is solvable uniquely for $x_i$ since $u_{ii}\neq 0$. This produces, for each $j$, a unique vector $x^{(j)}$ with $Ux^{(j)} = e_j$; assembling these as columns of a matrix $C$ gives $UC = I_n$. A square matrix with a right inverse also has that same matrix as a two-sided inverse (standard fact for finite-dimensional square matrices, following from rank-nullity: $UC = I$ forces $U$ to be surjective as a linear map, hence injective since domain and codomain have equal finite dimension $n$, hence bijective, hence $C$ is also the left inverse). So $U$ is invertible. Then $A = E_1^{-1}\cdots E_k^{-1} U$ is a product of invertible matrices, hence invertible. This establishes: all diagonal entries of $U$ are nonzero $\iff$ $A$ is invertible.

Step 5: Assemble. Chaining Steps 2–4:

This is exactly the claimed equivalence. $\blacksquare$

Since $A$ is invertible, $AA^{-1} = I_n$. By Multiplicativity of the Determinant, $\det(A)\det(A^{-1}) = \det(I_n) = 1$. By the theorem, $\det(A) \neq 0$, so we may divide to get $\det(A^{-1}) = \det(A)^{-1}$. $\blacksquare$