A matrix has always been a grid of coefficients for solving systems — a bookkeeping device. This section adds a second, more powerful way to see the same object. A matrix A is a function. It sends each vector \mathbf{x} to A\mathbf{x}. A linear map is any function between vector spaces that respects addition and scaling: f(\mathbf{x}+\mathbf{y})=f(\mathbf{x})+f(\mathbf{y}) and f(c\mathbf{x})=cf(\mathbf{x}). A matrix is exactly how such a map gets written down, once you fix coordinates. It stretches, rotates, reflects, and shears the plane, or space, all at once, everywhere, by the same rule. Chapter 02 builds the entire multivariable derivative on top of exactly this idea. That is why it earns a close look now, on numbers small enough to compute by hand.
Formula
Eigenvalues of a Symmetric 2×2 Matrix
also known as: the characteristic equation, real spectral case
For a symmetric matrix A = \begin{pmatrix} a & b \\ b & d \end{pmatrix}, the eigenvalues are the roots of the characteristic equation \det(A - \lambda I) = \lambda^2 - (a+d)\lambda + (ad - b^2) = 0, namely \lambda = \frac{(a+d) \pm \sqrt{(a-d)^2 + 4b^2}}{2}. Because the discriminant (a-d)^2 + 4b^2 is a sum of squares, it is never negative — a symmetric matrix always has two real eigenvalues (counted with multiplicity), never a complex pair.
\det(A - \lambda I) = (a-\lambda)(d-\lambda) - b^2 = \lambda^2 - (a+d)\lambda + (ad - b^2), a quadratic in \lambda. Apply the quadratic formula with A=1, B=-(a+d), C=ad-b^2: \lambda = \frac{(a+d) \pm \sqrt{(a+d)^2 - 4(ad-b^2)}}{2}. Expanding (a+d)^2 - 4(ad-b^2) = a^2 + 2ad + d^2 - 4ad + 4b^2 = (a-d)^2 + 4b^2 gives the stated discriminant. It is a sum of two squares, so it is \geq 0, and the two roots are always real.
Picture an image printed on a stretchy rubber sheet, pinned flat, then pulled harder in some spots than others. Most straight lines through the center tilt as the sheet stretches. But there are always one or two special directions that don't tilt at all — they only get longer or shorter. Which directions those are depends on how the pulling was done. Those are the eigenvector directions. How much longer or shorter each one gets is its eigenvalue.
A linear map is a function between vector spaces that respects both vector addition and scalar multiplication. A matrix is exactly how such a map is written down, once coordinates are fixed. A matrix's action reveals three facts that its coefficients alone don't.
First, the determinant is the area-scaling factor. The unit square's area, 1, becomes |\det A| after the map. A NEGATIVE determinant means the map also flips orientation: clockwise becomes counterclockwise.
Second, a matrix's eigenvectors are special directions. The linear map doesn't rotate them at all — it only stretches them, by a factor \lambda, the eigenvalue.
Third, for a symmetric matrix, those special directions are always perpendicular to each other. The same theorem also shows why the eigenvalues themselves are always real, never a complex pair — that follows from a discriminant argument.
Formula
2×2 Determinant
Recalled from College Algebra · Computing determinants — first taught there.
A matrix as a shape-transformer
Set a=2, b=0, c=0, d=0.5 -- the square stretches horizontally and squashes vertically, and the circle becomes an ellipse. Now set b=1, c=0 with a=d=1 -- pure shear, area unchanged (det still 1).
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Negative determinant flips orientation
Set a=-1, d=1, b=c=0 -- the square's image is congruent to the original, but det is negative: a reflection, not a rotation.
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Worked example
Eigenvalues of a symmetric matrix
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1
Find the eigenvalues of A=\begin{pmatrix}3 & 1\\ 1 & 3\end{pmatrix}.
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2
Here a=3, d=3, b=1. The discriminant is (a-d)^2+4b^2 = 0+4=4.
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3
\lambda = \dfrac{(a+d)\pm\sqrt{4}}{2} = \dfrac{6\pm 2}{2}.
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4
\boxed{\lambda_1 = 4,\ \lambda_2 = 2} — both real, as guaranteed for a symmetric matrix. The eigenvector directions are \langle 1,1\rangle (stretched by 4) and \langle 1,-1\rangle (stretched by 2).
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Where this goes next
Chapter 2 of this course
The Jacobian matrix, which chapter 02 builds, IS this section's "matrix as a function" idea. The only change is where it applies. Here, A acts on the whole plane at once. There, Df(\mathbf{a}) acts only near one point \mathbf{a}. It acts on nearby vectors, in exactly the way A acts on \mathbf{x} here.
The Hessian shows up in that same chapter. It is a symmetric matrix, and its eigenvalues are always real, by this section's theorem. They sort a critical point into one of three shapes: a bowl, a dome, or a saddle.
Check your understanding (counts toward your progress)
A matrix has determinant -3. What does the sign tell you?
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A determinant of exactly 0 signals a non-invertible map — a nonzero determinant, positive or negative, always means the map is invertible.
The determinant's sign says nothing about which numbers a matrix is allowed to act on — it acts on every vector in the space either way.
The determinant scales AREA (a 2D quantity), not individual vector lengths one at a time — a shear, for instance, can leave every length far from uniformly rescaled while still having a clean determinant.
Why
The determinant's magnitude is the area-scaling factor; its sign tracks orientation. Negative means a region's boundary, traced counterclockwise, becomes clockwise after the map — a reflection is baked in.
|det A| is always the area-scaling factor, regardless of sign — the sign carries a separate piece of information.
Find the larger eigenvalue of A=\begin{pmatrix}5 & 2\\ 2 & 5\end{pmatrix}.
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Why
(a+d)=10, (a-d)^2+4b^2 = 0+16=16, sqrt(16)=4. λ = (10±4)/2 gives 7 and 3 — the larger is 7.
Correct answer
Also accepted:
Why
Use λ = [(a+d) ± sqrt((a-d)^2 + 4b^2)] / 2 with a=d=5, b=2.
Let A=\begin{pmatrix}2 & 1\\ 0 & 3\end{pmatrix}. Compute the second component of A\langle 1, 2\rangle.
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Why
Row 2 is (0, 3). (0)(1) + (3)(2) = 6.
Correct answer
Also accepted:
Why
Row 2 of A dotted with the vector gives the second output component.
A matrix sends the unit square to a parallelogram with the exact same area, but tilted. What must be true of its determinant?
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A zero determinant collapses the square to a line segment or a point — no area survives at all, let alone the same area.
A negative determinant is possible here (it would mean the tilt includes a flip), but nothing in the problem forces the sign — only the magnitude is pinned down by "same area."
The identity matrix leaves the square completely unchanged, un-tilted — this map visibly tilts it, so it can't be the identity, even though it preserves area.
Why
Area scales by |det A|. Same area as the original unit square (area 1) means |det A| = 1 exactly — the determinant could be +1 (orientation preserved) or -1 (orientation reversed, a flip hidden in the tilt).
'Same area' pins down |det A| directly — think about what area-scaling factor leaves area 1 at area 1.
Practice: Matrices, determinants, and linear maps (5 problems · doesn’t affect your progress)
Which of these transformations of the plane is NOT a linear map — not expressible as A\mathbf{x} for any matrix A?
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A rotation about the origin fixes the origin (it maps \mathbf{0} to \mathbf{0}) and respects addition and scaling — it's a genuine linear map, written down by a rotation matrix.
A reflection across the x-axis fixes the origin too, and it's exactly the map \begin{pmatrix}1&0\\0&-1\end{pmatrix}\mathbf{x} — a linear map.
Uniform scaling by 2 is 2I\mathbf{x}, a linear map through and through — every vector doubles in length, direction unchanged.
Why
Every linear map sends the origin to the origin — f(\mathbf{0}) = f(0\cdot\mathbf{0}) = 0\cdot f(\mathbf{0}) = \mathbf{0}. A translation moves the origin away from itself, so it can never be written as A\mathbf{x} for any matrix A, while rotation, reflection, and uniform scaling all fix the origin and are genuine linear maps.
Shift every point 3 units to the right
Let A=\begin{pmatrix}1 & 2\\ 3 & 0\end{pmatrix}. Compute the first component of A\langle -1, 2\rangle.
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Why
Row 1 is (1, 2). (1)(-1) + (2)(2) = -1 + 4 = 3.
Correct answer
Also accepted:
Why
Row 1 of A dotted with the vector gives the first output component.
By what factor does area scale under A=\begin{pmatrix}3 & 1\\ 1 & 2\end{pmatrix}?
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Why
|det A| = |(3)(2) - (1)(1)| = |6 - 1| = 5.
Correct answer
Also accepted:
Why
|det A| = |ad - bc| is the area-scaling factor, regardless of what else the matrix does.
For A=\begin{pmatrix}6 & 4\\ 4 & 0\end{pmatrix}, which of its two eigenvalues is the smaller one?
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Why
(a+d)=6, (a-d)^2+4b^2 = 36+64=100, sqrt(100)=10. λ = (6±10)/2 gives 8 and -2 — the smaller is -2. Eigenvalues can be negative even though the matrix's entries aren't.
Correct answer
Also accepted:
Why
Use λ = [(a+d) ± sqrt((a-d)^2 + 4b^2)] / 2 with a=6, d=0, b=4.
A matrix has determinant +1. Can you conclude it's the identity map, or even that it leaves every shape's appearance unchanged?
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A shear like \begin{pmatrix}1&1\\0&1\end{pmatrix} has determinant (1)(1)-(1)(0)=1 but visibly tilts the unit square — it shares its determinant with the identity without being the identity.
Determinant tracks how AREA scales, not individual vector lengths one at a time — that same shear stretches some vectors while leaving others unchanged, yet still has determinant 1.
A positive determinant means orientation IS preserved, not reversed — that's the opposite of what a flip would give.
Why
det A = +1 only pins down two things: area is preserved exactly, and orientation is preserved (no flip). It says nothing about rotation or shearing — infinitely many non-identity matrices, like shears and rotations, all have determinant 1.
Think of a matrix that visibly tilts the unit square while still keeping its area exactly 1 — does such a matrix exist?
A shear matrix like \begin{pmatrix}1&1\\0&1\end{pmatrix} has determinant (1)(1)-(1)(0)=1 — work out what it does to the unit square's shape.
No — determinant +1 only pins down that area and orientation are both preserved; shapes can still be tilted, sheared, or rotated