Invertibility of Linear Maps

Invertibility of linear maps is the gateway to the concept of an isomorphism, developed next as Isomorphisms and Isomorphic Vector Spaces: two vector spaces are "the same" precisely when an invertible linear map connects them. In matrix language this becomes the theory of Invertible Matrices, the computational backbone of solving linear systems by matrix inversion, computing determinants, and diagonalizing operators for eigenvalue problems. Beyond pure linear algebra, invertibility governs whether a linear system of equations $Ax=b$ has a unique solution for every $b$ — the central question in numerical linear algebra and in engineering applications like circuit analysis and structural mechanics. In quantum mechanics, unitary (hence invertible) linear operators model time evolution, and invertibility of the underlying transformation is what guarantees information is never lost. In cryptography, block ciphers are built from invertible linear (or affine) maps over finite fields, since encryption must always be reversible by the holder of the key. In statistics and machine learning, the invertibility of a design matrix (or covariance matrix) determines whether a least-squares or regression problem has a unique solution, tying this article's abstract criterion directly to everyday computation.

Ask the tutor: Why does an injective linear map between spaces of the same finite dimension have to be surjective, and where exactly does rank–nullity enter that argument?

Think of a linear map $T: V \to W$ as a machine that eats vectors from $V$ and produces vectors in $W$. To "undo" $T$, you need a machine $S$ that, run after $T$, gets you back exactly where you started ($ST = I_V$), and also, run before $T$, produces exactly what $T$ would have needed to hit any target in $W$ ($TS = I_W$). For this undoing to be possible at all, two failures must be ruled out. First, $T$ must not collapse distinct vectors onto the same image — if $T(v_1) = T(v_2)$ with $v_1 \ne v_2$, no machine $S$ can know whether to send that common output back to $v_1$ or to $v_2$. This is exactly injectivity, equivalently $\operatorname{null}(T) = \{0\}$. Second, $T$ must not miss any vectors in $W$ — if some $w \in W$ is never hit, then $S(w)$ has nothing sensible to map to under the requirement $TS = I_W$. This is exactly surjectivity, equivalently $\operatorname{range}(T) = W$. The picture below shows the two failure modes side by side: a non-injective map (two inputs collapse to one output — no way back) and a non-surjective map (a target point is never reached — no consistent value to assign there).

Once both failure modes are excluded — that is, once $T$ is bijective — the ordinary set-theoretic inverse function $T^{-1}: W \to V$ exists automatically, and the entire content of the theorem below is that this set-theoretic inverse is forced to be linear.

Example 1 (An explicit isomorphism between polynomials and a coordinate space). Let $V = \mathcal{P}_2(\mathbb{R})$, the space of real polynomials of degree at most $2$, and let $W = \mathbb{R}^3$. Define $T: V \to W$ by

This $T$ is linear (each coordinate is a linear combination of $a_0, a_1, a_2$, and coefficient extraction $a_0+a_1x+a_2x^2 \mapsto (a_0,a_1,a_2)$ is linear, so $T$ is a linear combination of linear coordinate functionals into $\mathbb{R}^3$). Check injectivity via the null space. Suppose $T(a_0+a_1x+a_2x^2) = (0,0,0)$. Then

From the first equation $a_1 = -a_0$. Substituting into the second, $-a_0 - 2a_2 = 0$, so $a_0 = -2a_2$. From the third, $a_0 = -a_2$. Combining $a_0 = -2a_2$ and $a_0 = -a_2$ gives $-2a_2 = -a_2$, so $a_2 = 0$, hence $a_0 = 0$ and $a_1 = 0$. So the only polynomial mapping to $0$ is the zero polynomial: $\operatorname{null}(T) = \{0\}$, and $T$ is injective. Check surjectivity. Since $\dim V = 3 = \dim W$ (a fact we use only to motivate the computation, not to substitute for the proof — the theorem above requires an actual verification, and here we verify surjectivity directly by exhibiting a preimage of an arbitrary target). Let $(b_1, b_2, b_3) \in \mathbb{R}^3$ be arbitrary; we solve

for $a_0, a_1, a_2$. From equation 1, $a_1 = b_1 - a_0$. Substitute into equation 2: $b_1 - a_0 - 2a_2 = b_2$, so $a_0 + 2a_2 = b_1 - b_2$. Combined with equation 3, $a_0 + a_2 = b_3$, subtract: $a_2 = (b_1 - b_2) - b_3$. Then $a_0 = b_3 - a_2 = b_3 - (b_1 - b_2 - b_3) = 2b_3 - b_1 + b_2$. And $a_1 = b_1 - a_0 = b_1 - 2b_3 + b_1 - b_2 = 2b_1 - b_2 - 2b_3$. So an explicit preimage exists for every $(b_1,b_2,b_3)$, namely

Thus $T$ is surjective. By the theorem, $T$ is invertible, and its inverse $S = T^{-1}: \mathbb{R}^3 \to V$ is given exactly by this formula:

Sanity check. Take $(b_1,b_2,b_3) = (3,-1,2)$. Then $a_0 = -3-1+4=0$, $a_1 = 6+1-4=3$, $a_2=3+1-2=2$. So $T^{-1}(3,-1,2) = 3x+2x^2$. Check: $T(3x+2x^2) = (0+3,\, 3-4,\, 0+2) = (3,-1,2)$. ✓.

Example 2 (A matrix that fails to be invertible, diagnosed via the theorem). Let $A$ be the $3\times 3$ real matrix

viewed as the linear map $T_A: \mathbb{R}^3 \to \mathbb{R}^3$, $T_A(x) = Ax$, composition-compatible with Matrix Multiplication as Composition. Check injectivity. Solve $Ax = 0$ for $x = (x_1,x_2,x_3)$:

From equation 2, $x_2 = -3x_3$. Substitute into equation 1: $x_1 - 6x_3 - x_3 = 0$, so $x_1 = 7x_3$. Check equation 3: $7x_3 + 3(-3x_3) + 2x_3 = 7x_3 - 9x_3+2x_3 = 0$. This holds automatically for every $x_3$, so equation 3 gives no new constraint. Hence, for any $x_3 = t \in \mathbb{R}$, the vector $(7t, -3t, t)$ solves $Ax=0$. Taking $t=1$: $(7,-3,1) \ne 0$ is in $\operatorname{null}(T_A)$. So $\operatorname{null}(T_A) \ne \{0\}$, and $T_A$ is not injective. By the theorem (contrapositive of $(1)\Rightarrow(2)$: if $T$ fails to be injective, $T$ cannot be invertible), $T_A$ has no two-sided inverse — as a linear map, and hence $A$ has no inverse matrix. (One can also check directly that $T_A$ is not surjective here: since $\operatorname{null}(T_A)$ is one-dimensional, the Rank–Nullity-type dimension count forces $\operatorname{range}(T_A)$ to be $2$-dimensional, a proper subspace of $\mathbb{R}^3$ — consistent with the theorem's equivalence, since a linear map between finite-dimensional spaces of equal dimension is injective iff it is surjective, though we do not need that fact here since non-injectivity alone already rules out invertibility.) This illustrates the theorem's practical power: to disprove invertibility, exhibiting a single nonzero vector in the null space suffices — no need to separately analyze the range.