3.1 Matrices as Geometric Grid Transformations: Stretching, Shearing, and Rotating

The fastest way to build geometric intuition in linear algebra is to stop viewing a matrix as a static block of numbers and start viewing it as a coordinate grid transformation. Every 2×22 \times 2 matrix AA is completely determined by where it sends just two arrows: the horizontal basis vector e1=[10]\mathbf{e}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix} and the vertical basis vector e2=[01]\mathbf{e}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}.

In this Article

1. Reading Matrix Columns as Transformed Basis Vectors

Multiply any 2×22 \times 2 matrix A=[abcd]A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} by the standard basis vectors e1=[10]\mathbf{e}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix} and e2=[01]\mathbf{e}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}:

Ae1=[abcd][10]=[ac],Ae2=[abcd][01]=[bd]A\mathbf{e}_1 = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} a \\ c \end{bmatrix}, \qquad A\mathbf{e}_2 = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} b \\ d \end{bmatrix}

Golden Rule of Matrix Geometry: Column 1 of AA is the new landing spot of e1=[10]\mathbf{e}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}. Column 2 of AA is the new landing spot of e2=[01]\mathbf{e}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}. Once you plot those two columns as arrows from the origin, you can immediately see how the entire plane is stretched, tilted, or flipped!

2. Linearity: Why Grid Lines Remain Parallel and Evenly Spaced

Any vector x=[x1x2]\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} is simply x1e1+x2e2x_1 \mathbf{e}_1 + x_2 \mathbf{e}_2. Because matrix multiplication is a linear map (A(x1e1+x2e2)=x1(Ae1)+x2(Ae2)A(x_1\mathbf{e}_1 + x_2\mathbf{e}_2) = x_1(A\mathbf{e}_1) + x_2(A\mathbf{e}_2)), a point that started x1x_1 steps along e1\mathbf{e}_1 and x2x_2 steps along e2\mathbf{e}_2 ends up x1x_1 steps along Column 1 and x2x_2 steps along Column 2. Geometrically, this means:

  • The origin 0\mathbf{0} stays fixed at 0\mathbf{0}.
  • Straight lines remain straight lines (never curving).
  • Parallel grid lines remain parallel and evenly spaced, turning unit squares into parallelograms and the unit circle into an ellipse.
   UNIT SQUARE BEFORE A:                PARALLELOGRAM AFTER A = [[2, 1], [0, 1]]:
         x2                                   x2
         ^                                    ^         (1,1) lands at [3,1]
   (0,1) +-------+ (1,1)                (0,1) .---------*-------* (3,1)
         |       |                           /         /       /
         |       |                          /         /       /
   ------+-------+---> x1               ---+---------+-------+---> x1
       (0,0)   (1,0)                     (0,0)             (2,0) = Ae_1
Transformation TypeMatrix AAWhere e1\mathbf{e}_1 LandsWhere e2\mathbf{e}_2 LandsGeometric Effect on Grid
Anisotropic Scaling[3002]\begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix}[30]\begin{bmatrix} 3 \\ 0 \end{bmatrix}[02]\begin{bmatrix} 0 \\ 2 \end{bmatrix}Stretches horizontal by 3×3\times and vertical by 2×2\times
Horizontal Shear[11.501]\begin{bmatrix} 1 & 1.5 \\ 0 & 1 \end{bmatrix}[10]\begin{bmatrix} 1 \\ 0 \end{bmatrix}[1.51]\begin{bmatrix} 1.5 \\ 1 \end{bmatrix}Slides horizontal layers rightward like a deck of cards
90∘90^\circ Counterclockwise Rotation[0−110]\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}[01]\begin{bmatrix} 0 \\ 1 \end{bmatrix}[−10]\begin{bmatrix} -1 \\ 0 \end{bmatrix}Rigidly rotates the entire grid 90∘90^\circ without distortion
Symmetric Stretch-Shear[2112]\begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}[21]\begin{bmatrix} 2 \\ 1 \end{bmatrix}[12]\begin{bmatrix} 1 \\ 2 \end{bmatrix}Pure stretch along 45∘45^\circ and −45∘-45^\circ diagonals

4. Tracing a Vector and Unit Circle Through a Matrix

Given A=[2101]A = \begin{bmatrix} 2 & 1 \\ 0 & 1 \end{bmatrix}, where does x=[−12]\mathbf{x} = \begin{bmatrix} -1 \\ 2 \end{bmatrix} land, and what is the area of the parallelogram formed by Ae1A\mathbf{e}_1 and Ae2A\mathbf{e}_2?

  1. Combine columns: Ax=(−1)[20]+2[11]=[−2+20+2]=[02]A\mathbf{x} = (-1)\begin{bmatrix} 2 \\ 0 \end{bmatrix} + 2\begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} -2+2 \\ 0+2 \end{bmatrix} = \begin{bmatrix} 0 \\ 2 \end{bmatrix}.
  2. Parallelogram area: The base along the horizontal axis has length 22 (from Ae1=[20]A\mathbf{e}_1 = \begin{bmatrix} 2 \\ 0 \end{bmatrix}), and the vertical height is 11 (from Ae2=[11]A\mathbf{e}_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}). Thus Area=2×1=2\text{Area} = 2 \times 1 = 2, which matches det⁡(A)=(2)(1)−(1)(0)=2\det(A) = (2)(1) - (1)(0) = 2.

5. Active Recall Check and Key Takeaways

Quick Check: Test Your Understanding

Question: A linear transformation in R2\mathbb{R}^2 reflects every vector across the 45∘45^\circ line x2=x1x_2 = x_1 (so horizontal and vertical axes swap places). Write down its 2×22 \times 2 matrix AA purely by tracking where e1\mathbf{e}_1 and e2\mathbf{e}_2 land.

Answer: Reflecting across x2=x1x_2 = x_1 sends e1=[10]\mathbf{e}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix} to [01]\begin{bmatrix} 0 \\ 1 \end{bmatrix} (Column 1) and sends e2=[01]\mathbf{e}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix} to [10]\begin{bmatrix} 1 \\ 0 \end{bmatrix} (Column 2). Therefore, A=[0110]A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}.

  • Columns are the destination of basis vectors: A=[Ae1Ae2]A = \begin{bmatrix} A\mathbf{e}_1 & A\mathbf{e}_2 \end{bmatrix}.
  • Grid coordinates are preserved: Ax=x1(Ae1)+x2(Ae2)A\mathbf{x} = x_1(A\mathbf{e}_1) + x_2(A\mathbf{e}_2) uses the original coordinates x1,x2x_1, x_2 on the warped grid.
  • Canonical References: Watch 3Blue1Brown on Linear Transformations and Matrices.

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linear-transformations
basis-vectors
matrix-multiplication
shear
rotation
linear-algebra