The First Isomorphism Theorem states that the coimage is isomorphic to the image (ℝ³/ker(T) ≌ ℝ².

• Every single vector in the image (a point (x₀, x₀) in ℝ²) corresponds to exactly one element in the coimage (the entire vertical line { (x₀, x₀, z) }).
• Any vector chosen along that vertical line will map to the exact same point in the image.

• The kernel consists of all vectors x such that A x = 0. The kernel space is spanned on one vector: ker(T) = span{[1, 1, −1]^T}. With Mathematica, we find the kernel (Null Space) of T/matrix A:
  A = {{1, 2, 3}, {1, 0, 1}, {2, 2, 4}, {-1, 2, 1}}; NullSpace[A]
  {{-1, -1, 1}}
• The dimension of the kernel is called the nullity and it is 1.
• The image (column space) is the span of the columns of the original matrix A. Looking back at our REF, the pivots are located in Column 1 and Column 2. Therefore, the first two columns of our original matrix form a basis for the image.
  \[ \mbox{im}(\mathbf{A}) = \mbox{span} \left\{ \begin{pmatrix} 1 \\ 1 \\ 2 \\ -1 \end{pmatrix}, \ \begin{pmatrix} 2 \\ 0 \\ 2 \\ 2 \end{pmatrix} \right\} . \]
• Rank is 2 because it is the dimension of the image. There are 2 pivot columns (Columns 1 and 2).
  \[ \mbox{rank}(T) = \mbox{rank}(\mathbf{A}) = 2. \]
• Corank (Cokernel dimension): The matrix maps into a 4-dimensional space (ℝ⁴). The corank represents the dimension of the left null space (the geometric constraints on the codomain):
  \[ \mbox{corank}(\mathbf{A}) = \dim (\mbox{codomain}) - \mbox{rank}(\mathbf{A}) = 4 -2 =2 . \]
• Index: Using the standard linear algebra operator index formula (dim(kerA) - corank(A)):
  \[ \mbox{ind}(\mathbf{A}) = 1 - 2 = -1 . \]

The Rank-Nullity Theorem states:

We will show how this abstract space is isomorphic to a concrete 2-dimensional subspace of the codomain called the left null space, ker(A^T).

The left null space consists of all vectors y ∈ ℝ^4×1 ≌ ℝ⁴ such that A^Ty = 0. These are vectors orthogonal to every column in the image.

We set up the transpose matrix A^T:

1. R₂ ← R₂ − 2R₁     ⇒     [0, −2, −2, 4]
2. R₃ ← R₃ − 3R₁     ⇒     [0, −2, −2, 4]
3. Subtract R₂ from R₃ to get a row of zeros, then normalize R₂:
   \[ \mbox{RREF}\left( \mathbf{A}^{\mathrm T} \right) = \begin{bmatrix} 1 & 1 & 2 & -1 \\ 0 & 1 & 1 & -2 \\ 0 & 0 & 0 & 0 \end{bmatrix} . \]

• From row 2:    y₂ + y₃ − 2y₄ = 0     ⇒     y₂ = − y₃ + 2y₄.
• From row 1:    y₁ + y₂ + 2y₃ − y₄ = 0     ⇒     y₁ + (−y₃ + 2y₄) + 2 y₃ − y₄ = 0     ⇒     y₁ = − y₃ − y₄.

We define our linear mapping ϕ : ℝ^4×1 ⇾ ker(A^T) by sending a coset to its unique orthogonal representative:

1. Well-defined: If we pick a different representative for the same coset, say y + v (where v ∈ im(A)), then
   \[ \left( \mathbf{y} + \mathbf{v} \right)_{\perp} = \mathbf{y}_{\perp} + \mathbf{v}_{\perp} = \mathbf{y}_{\perp} + \mathbf{0} = \mathbf{y}_{\perp} . \]
   Changing the representative doesn't change the output.
2. Injective (One-to-One): If ϕ([y]) = 0, it means y_⊥ = 0, so y = y_∥ ∈ im(A). This is exactly the zero coset [0].
3. Surjective (Onto): For any vector y sitting inside our physical subspace ker(A^T), we can just feed the coset [y] into the function, and it maps directly to y.

Here is a numerical example showing how a specific vector w ∈ ℝ^4×1 is split into its image component (w_∥) and its cokernel/left null space component (w_⊥), along with the Mathematica code to compute it automatically.

Let's pick an arbitrary vector w in the codomain ℝ^4×1:

The image space is spanned by the columns of A. Because the third column is just a combination of the first two (col₃ = col₁ + col₂), we strip it out to avoid redundancy. We form a trimmed matrix A_col containing only the basis vectors:

The perpendicular leftover vector is w_⊥ = w − A_colx. For this error to be as small as possible, it must hit the plane at a perfect 90-degree angle. This means it must be completely orthogonal to the columns of A_col:

You can copy and paste the following script into Wolfram Mathematica to run the exact calculation above, find the structural bases, and perform the vector splitting automatically:

• Domain (X) and Codomain (Y): Both are ℝ^3×1.
• Kernel (ker(T): Solving A x = 0 yields a 1D line spanned by:
  \[ \mathbf{k} = \begin{pmatrix} 1 \\ -2 \\ 1 \end{pmatrix} , \]
• Image (im(T): Spanned by the first two linearly independent columns of A:
  \[ \mathbf{c}_1 = \begin{pmatrix} 1 \\ 4 \\ 7 \end{pmatrix} , \qquad \mathbf{c}_2 = \begin{pmatrix} 2 \\ 5 \\ 8 \end{pmatrix} . \]
• Coimage (doim(T) = X/ker(T)): Since ker(T) is 1-dimensional, the coimage is 2-dimensional (ℝ²). We represent a coset by eliminating the kernel's direction.
• Cokernel (coker(T) = Y/im(T)): Since im(T) is 2-dimensional, the cokernel is 1-dimensional (ℝ). It is spanned by the left null space of A, which happens to be m^T = [1, −2, 1].

Step 1: \( \displaystyle \quad \mbox{ker}(T) \, \stackrel{i}{\rightarrow} \, X \ \) (inclusion). Maps the 1D kernel into ℝ^3×1.

Step 2: \( \displaystyle \quad X \, \stackrel{\sigma}{\rightarrow} \, \mbox{coim}(T) \ \) Canonical Projection). Maps ℝ^3×1 ⇾ ℝ^2×1 by collapsing the kernel direction k. A valid algebraic projection matrix is:

Step 3: \( \displaystyle \quad \mbox{coim}(T) \, \stackrel{h}{\rightarrow} \, \mbox{im}(T) \ \) (The Unique Isomorphism). Maps the 2D coimage coordinates into the 3D landing plane. By solving A = H · Σ, we extract:

Step 4: \( \displaystyle \quad \mbox{im}(T) \, \stackrel{j}{\rightarrow} \, Y \ \) (Inclusion). Since H already outputs full 3D vectors in Y, the inclusion map j is implicitly handled by H mapping into ℝ^3×1.

Step 5: \( \displaystyle \quad Y \, \stackrel{f}{\rightarrow} \, \mbox{coker}(T) \ \) (Canonical Projection). Maps ℝ³ ⇾ ℝ¹ by squashing everything inside im(T) to zero. This is computed by the left null vector:

For verification, let's trace a non-trivial vector v = [1, 1, 1]^T through the system:

1. Direct Transformation:
   \[ T(\mathbf{v}) = \mathbf{A\,v} = \begin{pmatrix} 6 \\ 15 \\ 24 \end{pmatrix} . \]
2. Factored Path (h ⚬ σ):
   1. \[ \sigma (\mathbf{v}) = \Sigma \bullet \mathbf{v} = \begin{pmatrix} 0 \\ 3 \end{pmatrix} . \]
      This is the vector's coordinate inside the Coimage.
   2. \[ h \left( \sigma (\mathbf{v}) \right) = \mathbf{H} \cdot \begin{pmatrix} 0 \\ 3 \end{pmatrix} = \begin{pmatrix} 6 \\ 15 \\ 24 \end{pmatrix} \]
   3. Both paths match perfectly, proving the commutative diagram holds: A = H · Σ.
3. Cokernel Check:
   \[ \mathbf{F} \cdot T(\mathbf{v}) = \begin{bmatrix} 1&-2&1 \end{bmatrix} \cdot \begin{pmatrix} 6 \\ 15 \\ 24 \end{pmatrix} = 0 , \]
   proving the sequence is exact at Y.

Let's look at an explicit example where the condition holds, demonstrate how to build h, and look at the uniqueness criteria. Let our domain be X = ℝ^3×1 ≌ ℝ³, and let the target spaces be U = ℝ^2×1 ≌ ℝ² and V = ℝ^2×1 ≌ ℝ². We define two linear mappings f and g via their standard matrices:

• Map f : ℝ^3×1 ⇾ ℝ^2×1:
  \[ f = \begin{bmatrix} 1&0&0 \\ 0&1&0 \end{bmatrix} . \]
  This map strips away the z-coordinate.
  \[ \ker (f) = \left\{ \begin{pmatrix} 0 \\ 0 \\ z \end{pmatrix} \ : \ z \in \mathbb{R} \right\} . \]
• Map g : ℝ^3×1 ⇾ ℝ^2×1:
  \[ g = \begin{bmatrix} 2&3&0 \\ 4&5&0 \end{bmatrix} . \]
  This map transforms the x and y coordinates but also completely flattens the z -coordinate.
  \[ \ker (g) = \left\{ \begin{pmatrix} 0 \\ 0 \\ z \end{pmatrix} \ : \ z \in \mathbb{R} \right\} . \]

• Direct path (g):
  \[ g(\mathbf{v}) = \begin{pmatrix} 2x + 3y \\ 4x + 5y \end{pmatrix} . \]
• Factored path (h ∘ f):
  \[ f(\mathbf{v}) = \begin{pmatrix} x \\ y \end{pmatrix} \qquad \Longrightarrow \qquad h \left( f(\mathbf{v})\right) = \begin{pmatrix} 2x + 3y \\ 4x + 5y \end{pmatrix} . \]

Uniqueness check: The lemma states: If im(f) = U, then h is unique.

In our setup,\( \displaystyle \quad f = \begin{bmatrix} 1&0&0 \\ 0&1&0 \end{bmatrix} . \quad \) The columns span all of ℝ^2×1 ≌ ℝ², meaning f is surjective (onto), so im(f) = U = ℝ^2×1. . Because f completely covers the target space U, the behavior of h is entirely locked down for every single element in U. There is exactly one matrix h that works.

What if h wasn't unique? (Counter-example for uniqueness) Imagine if we enlarged U to ℝ^3×1 ≌ ℝ³ by adding a dead row to f, making\( \displaystyle \quad f = \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 0&0&0 \end{bmatrix} . \quad \)

We have spaces:    X = ℝ^3×1 ≌ ℝ³, U = ℝ³, and V = ℝ^2×1 ≌ ℝ². The linear map f : ℝ^3×1 ⇾ ℝ^3×1 acts as a projection, while g : ℝ^3×1 ⇾ ℝ^2×1 acts as a standard linear transformation.

For a mapping h : U ⇾ V to exist, the lemma states we must have ker(f) ⊆ ker(g). Let's check the kernels of both mappings:

• Kernel of f: Looking at your matrix f, any vector where the first two components are zero maps to zero:
  \[ \ker (f) = \mbox{span}\left\{ \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \right\} . \]
• Kernel of g: Multiplying matrix g by the same basis vector yields:
  \[ \begin{bmatrix} 2&3&0 \\ 4&5&0 \end{bmatrix} \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} . \]
  This shows that [0, 0, 1]^T also lives inside ker(g).

The second clause of the lemma states that h is unique if and only if im(f) = U. In our example, the image of f is only a 2D plane spanned by the first two standard basis vectors:

Let h : ℝ³ ⇾ ℝ² be represented by a general 2 × 3 matrix:

• \[ \mathbf{H}_1 = \begin{bmatrix} 2 & 3 & 0 \\ 4 & 5 & 0 \end{bmatrix} . \]
• \[ \mathbf{H}_2 = \begin{bmatrix} 2 & 3 & 1 \\ 4 & 5 & -1 \end{bmatrix} . \]

1. By looking at the columns, the second column is zero, and the third column is identical to the first. Therefore, the column space (image) is a 1-dimensional subspace in ℂ² spanned by the first column:
   \[ \mbox{im}(\mathbf{A}) = \mbox{span}\left\{ \begin{pmatrix} 1 \\ \mathbf{j} \end{pmatrix} \right\} . \]
2. The standard transpose of A flips the rows and columns without changing the signs of ⅉ:
   \[ \mathbf{A}^{\mathrm T} = \begin{bmatrix} 1& \mathbf{j} \\ 0&0 \\ 1& \mathbf{j} \end{bmatrix} . \]
   To find ker(A^T), we solve A^Ty = 0 for a vector y = [y₁, y₂]^T ∈ ℂ^2×1:
   \[ \begin{bmatrix} 1& \mathbf{j} \\ 0&0 \\ 1& \mathbf{j} \end{bmatrix} \begin{pmatrix} y_1 \\ y_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} . \]
   Using Mathematica, we get
       A = {{1, 0, 1}, {I, 0, I}};
       NullSpace[Transpose[A]]
      {-I, 1}}
   \[ \ker\left( \mathbf{A}^{\mathrm T} \right) = \mbox{span}\left\{ \begin{pmatrix} -\mathbf{j} \\ 1 \end{pmatrix} \right\} . \]
   Note that if we multiply this kernel's basis vector by the scalar ⅉ, we get the exact basis vector of the image:
   \[ \mathbf{j}\,\begin{pmatrix} -\mathbf{j} \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ \mathbf{j} \end{pmatrix} . \]
   Because im(A) = ker(A^T), the standard transpose kernel contains the exact same vectors as the image. It completely fails to provide a complementary, orthogonal subspace to im(A) inside ℂ².
3. Now we show what happens when we use the conjugate transpose, which changes the sign of ⅉ:
   \[ \mathbf{A} = \begin{bmatrix} 1 & -\mathbf{j} \\ 0&0 \\ 1 & -\mathbf{j}\end{bmatrix} . \]
   Soolving A^✶y = 0, we get
   \[ y_1 - \mathbf{j}\,y_2 = 0 \qquad \Longrightarrow \qquad y_1 = \mathbf{j}\,y_2 . \]
   Setting y₂ = 1 gives y₁ = ⅉ, so
   \[ \ker\left( \mathbf{A}^{\ast} \right) = \mbox{span}\left\{ \begin{pmatrix} \mathbf{j} \\ 1 \end{pmatrix} \right\} . \]
   Take the standard complex inner product between the basis vector of the image and this new kernel:
   \[ \left( \mbox{image vector} . \mbox{kernel vector} \right) = \begin{pmatrix} \mathbf{j} \\ 1 \end{pmatrix}^{\ast} \begin{pmatrix} 1 \\ \mathbf{j} \end{pmatrix} = 0. \]
   They are perfectly perpendicular! The space decomposes cleanly as ℂ² = im(A) ⊕ ker(A^✶), proving that ker(A^✶) is the correct geometric representation of the cokernel.

Metric / Property	Standard Transpose \((\mathbf{A}^{\mathrm{T}})\)	Conjugate Transpose \((\mathbf{A}^{\mathrm{H}})\)
Matrix Operation	\(\mathbf{A}^{\mathrm{T}} = \begin{bmatrix} 1 & i \\ 0 & 0 \\ 1 & i \end{bmatrix}\)	\(\mathbf{A}^{\mathrm{H}} = \begin{bmatrix} 1 & -i \\ 0 & 0 \\ 1 & -i \end{bmatrix}\)
Subspace Kernel	\(\ker(\mathbf{A}^{\mathrm{T}}) = \text{span}\left(\begin{bmatrix} -i \\ 1 \end{bmatrix}\right)\)	\(\ker(\mathbf{A}^{\mathrm{H}}) = \text{span}\left(\begin{bmatrix} i \\ 1 \end{bmatrix}\right)\)
Geometric Relationship to \(\text{im}(\mathbf{A})\)	Identical Subspace \(\ker(\mathbf{A}^{\mathrm{T}}) = \text{im}(\mathbf{A})\)	Orthogonal Complement \(\ker(\mathbf{A}^{\mathrm{H}}) = \text{im}(\mathbf{A})^\perp\)
Inner Product Result \((\langle \mathbf{v}_{\text{img}}, \mathbf{v}_{\text{ker}} \rangle)\)	\(\begin{bmatrix} -i & 1 \end{bmatrix}\begin{bmatrix} -i \\ 1 \end{bmatrix} = -1 + 1 = \mathbf{0}\) (Self-orthogonal / Isotropic)	\(\begin{bmatrix} -i & 1 \end{bmatrix}\begin{bmatrix} 1 \\ i \end{bmatrix} = -i + i = \mathbf{0}\) (Genuinely Perpendicular)
Direct Sum Decomposition	Fails \((\text{im}(\mathbf{A}) \oplus \ker(\mathbf{A}^{\mathrm{T}}) \neq \mathbb{C}^2)\)	Succeeds \((\text{im}(\mathbf{A}) \oplus \ker(\mathbf{A}^{\mathrm{H}}) = \mathbb{C}^2)\)

The following Mathematica script defines the matrices F and G, builds a general matrix H with symbolic entries, and solves the system H F = G to show exactly how the free parameters (variables) arise due to the violation of uniqueness.
(* 1. Define the matrices from your unique map violation example *) F = {{1, 0, 0}, {0, 1, 0}, {0, 0, 0}}; G = {{2, 3, 0}, {4, 5, 0}}; Print["--- Commutative Diagram Setup ---"]; Print["Matrix F (f: X -> U) = ", MatrixForm[F]]; Print["Matrix G (g: X -> V) = ", MatrixForm[G]]; (* 2. Construct a general 2x3 matrix H with symbolic entries *) H = Array[c, {2, 3}]; Print["\nGeneral Symbolic Matrix H = ", MatrixForm[H]]; (* 3. Compute the composition H . F *) composition = H . F; Print["Calculated H . F = ", MatrixForm[composition]]; (* 4. Solve the matrix equation H . F == G *) (* Thread is used to equate the matrices element-by-element *) system = Thread[Flatten[composition] == Flatten[G]]; solutions = Solve[system, Flatten[H]]; Print["\n--- Uniqueness Solutions Output ---"]; Print["Matrix H Solutions: ", solutions]; (* 5. Display the parameterized matrix H to show the free variables *) parameterizedH = H /. solutions[[1]]; Print["Final Parameterized Matrix H = ", MatrixForm[parameterizedH]];
   ■
End of Example 56

Let's construct a non-square, complex matrix A ∈ ℂ^3×4 where the columns are linearly dependent. This ensures that the adjoint matrix A^✶ has a nontrivial kernel. We will then use Mathematica to find a complex vector y that satisfies A^✶y = 0 and show how it serves as an orthogonal filter for any target vector b.

To find the orthogonal complement to this image, take the conjugate transpose A^✶:

On the other hand, the kernel of matrix A is

The second row is 3 times the first, so rank(A) = 1. The image is \[ \operatorname{im} (\mathbf{A}) = \operatorname{span} \left\{ \left[ \begin{matrix}1\\ 3\end{matrix}\right] \right\} . \] Thus, \[ \operatorname{coker} (\mathbf{A}) = \mathbb{R}^{2}/\operatorname{im} (\mathbf{A}) \] is 1‑dimensional. A convenient representative of a nonzero coset is, say, \[ [\mathbf{v}]=\left[ \left[ \begin{matrix}1\\ 0\end{matrix}\right] \right] . \] Construct the left null space, ker(A^T):

Compute \[ \mathbf{A}^T = \left[ \begin{matrix}1&3\\ 2&6\\ \end{matrix}\right] . \] We want A^Ty = 0, i.e., \[ \left[ \begin{matrix}1&3\\ 2&6\\ \end{matrix}\right] \left[ \begin{matrix}y_1\\ y_2\end{matrix}\right] =\left[ \begin{matrix}0\\ 0\end{matrix}\right] . \] This gives \[ \left\{ \, \begin{array}{l}\textstyle y_1+3y_2=0,\\ \textstyle 2y_1+6y_2=0,\end{array}\right. \] and the second equation is redundant. So \[ y_1=-3y_2. \] A basis vector for ker(A^T) is \[ \mathbf{y} = \left[ \begin{matrix}-3\\ 1\end{matrix}\right] . \] So \( \displaystyle \quad \ker (\mathbf{A}^T) = \operatorname{span} \left\{ \left[ \begin{matrix}-3\\ 1\end{matrix}\right] \right\} , \quad \) a 1‑dimensional subspace.

The functional on the cokernel:

Take \( \displaystyle \quad y=\left[ \begin{matrix}-3\\ 1\end{matrix}\right] . \quad \) Define \( \displaystyle \quad \Phi (\mathbf{y})([\mathbf{v}]) = \langle \mathbf{y}, \mathbf{v}\rangle = \mathbf{y}^T\mathbf{v}. \)

Check that this is well-defined on cosets:

If v' = v + A w, then \[ \mathbf{y}^T \mathbf{v}' = \mathbf{y}^T \mathbf{v} + \mathbf{y}^T {\bf A}\,{\bf w} = \mathbf{y}^T \mathbf{v} + (\mathbf{A}^T \mathbf{y})^T\mathbf{w} = \mathbf{y}^T\mathbf{v}+0, \] since A^Ty = 0.

So Φ(y) is a linear functional on coker(A).

Evaluate it on our chosen coset: \[ \Phi (y)\left( \left[ \left[ \begin{matrix}1\\ 0\end{matrix}\right] \right] \right) =y^T\left[ \begin{matrix}1\\ 0\end{matrix}\right] =-3. \] This is nonzero, so Φ(y) is a nontrivial functional on the 1‑dimensional space coker(A); hence it spans (coker(A))′.

Thus, in this example:

• ker(A^T) is the line spanned by \( \displaystyle \quad \left[ \begin{matrix}-3\\ 1\end{matrix}\right] .\)
• (coker(A))′ is the line spanned by the functional [v] ↦ y • v.
• The map y ↦ ([v] ↦ y • v) is a linear isomorphism between these two 1‑dimensional spaces.

1. Find index of the following matrices and determine which of them defines an injective (one-to-one) or surjective (onto) linear mapping.
   1. \( \displaystyle \quad \begin{bmatrix} 2&1&0 \\ 4&3&2 \\ 1&2&3 \\ 0&2&1 \end{bmatrix} ; \)
   2. \( \displaystyle \quad \begin{bmatrix} 1&0&0 \\ 2&2&2 \\ 1&2&0 \\ 0&1&0 \end{bmatrix} ; \)
   3. \( \displaystyle \quad \begin{bmatrix} 0&3&0&2&0&1&0 \\ 0&2&3&1&1&8&6 \\ 0&1&1&5&0&2&3 \\ 0&2&1&7&0&4&3 \end{bmatrix} ; \)
   4. \( \displaystyle \quad \begin{bmatrix} 0&1&0&2&0&1&0 \\ 0&3&2&6&0&5&4 \\ 0&1&1&2&0&2&2 \\ 0&2&1&4&0&3&2 \end{bmatrix} ; \)
   5. \( \displaystyle \quad \begin{bmatrix} 0&1&0&2&1&1&2 \\ 0&3&2&6&1&5&1 \\ 0&1&1&2&0&2&1 \\ 0&2&1&4&0&3&1 \end{bmatrix} . \)
2. Suppose A is an m × n matrix and that m ≤ n. Suppose also that the rank A equals m. Show that T_A maps 𝔽^n×1 onto 𝔽^m×1. Hint: vectors e₁, e₂, … , e_{m occur as columns in the row reduced echelon form for A}
3. Suppose A is an m × n matrix and that m ≥ n. Suppose also that the rank A equals n. Show that T_A is one-to-one. Hint: If not, there exists a vector x ≠ 0 such that A x = 0, and this implies at least one column of A is a linear combination of the others. Show that this would require the column rank to be less than n.
4. Suppose A is an m × n matrix and that m < n. Show that A is not one to one.

1. Beezer, R., A Second Course in Linear Algebra, 2013.
2. Fitzpatrick, S., Orthogonal sets of vectors, Linear Algebra.