LINEAR-ALGEBRA · Interactive Practice | Unit 30 · Video 1

Your Photo Is a Vector: Why JPEG Throws Away the Pixel Grid

IKey Formulas

Formula	Name	Description
$\text{Image} = c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \dots + c_N\mathbf{v}_N$	Basis expansion	Any image is a weighted sum of basis vectors
$c_i = \mathbf{v}_i^\top \mathbf{x}$	Coefficient (orthonormal basis)	Projection of image $\mathbf{x}$ onto basis vector $\mathbf{v}_i$
$\mathbf{x} \in \mathbb{R}^{H \cdot W}$	Image as vector	$H \times W$ grayscale image flattened into a single column
$\hat{\mathbf{x}} = \sum_{i \in S} c_i \mathbf{v}_i$	Lossy reconstruction	Keep only large coefficients $S$ , discard the rest

IIVisualization 1 — An Image Is a Vector

Unroll a grayscale grid row by row and the picture becomes one point — a vector in $\mathbb{R}^{16}$ .

💡 The same idea scales: a $512 \times 512$ photo is a single point in $\mathbb{R}^{262{,}144}$ .

IIIVisualization 2 — Basis Vectors Can Be Pictures

Each basis vector is itself an $8 \times 8$ image — from a single spike to a uniform field to a fast oscillation.

💡 JPEG's basis is cosine waves at rising frequencies (the Discrete Cosine Transform); modern codecs use wavelets — choosing the basis is the whole game.

IVVisualization 3 — Compression by Discarding Coefficients

Keep only the largest few of 64 frequency coefficients and the reconstruction barely changes.

Keep top-

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💡 It works because real images are mostly smooth: they project heavily onto the low-frequency basis vectors and barely onto the high-frequency ones.

VQuiz Questions

Question 1

A grayscale photograph has resolution $512 \times 512$ . As a single vector, in which space does it live?

✅ Correct! Every pixel contributes one coordinate.

❌ Not quite. The dimension equals the total number of pixels, not the width or height alone.

Show solution

Solution:

To turn an image into a vector, we stack all the pixel values into one long column. For a $512 \times 512$ grayscale image, that's $512 \times 512 = 262{,}144$ pixel values, so the image is a single point in $\mathbb{R}^{262{,}144}$ .

(A color image would triple this, since each pixel has R, G, and B channels.)

Question 2

You photograph a perfectly uniform dark gray blackboard. Compare two bases:

Pixel basis $\{\mathbf{e}_1, \dots, \mathbf{e}_N\}$
Structured basis that includes the all-ones vector $\mathbf{1} = (1,1,\dots,1)$

In which basis does the image have a shorter description (fewer non-zero coefficients)?

✅ Correct! A uniform image is a scalar multiple of the all-ones vector, so almost every coefficient is zero.

❌ Not quite. Having every coefficient equal is not the same as having most coefficients zero. We want shortness, not uniformity.

Show solution

Solution:

In the pixel basis, every coefficient is the gray value $g$ . That's 262,144 equal, non-zero numbers — no short description.

In a structured basis that contains the all-ones vector $\mathbf{1}$ , the uniform image is simply $\text{blackboard} = g \cdot \mathbf{1} + 0 \cdot \mathbf{v}_2 + 0 \cdot \mathbf{v}_3 + \dots$ Almost all coefficients are zero. The whole image is captured by a single number $g$ .

Same vector, same information — just a much shorter description.

Question 3

True or False: When JPEG performs lossy compression, it removes information at random to make the file smaller.

✅ Correct! Lossy compression is a precise, structured discard of high-frequency coefficients — not random damage.

❌ Not quite. The whole point of compression-by-change-of-basis is that the discard is targeted.

Show solution

Solution: False.

JPEG does not discard information randomly. It changes basis (using the Discrete Cosine Transform) so that the image is expressed as a sum of low- to high-frequency components. The coefficients on high-frequency basis vectors are then aggressively quantized or zeroed out.

This is principled, not random:

Real photographs are mostly smooth, so high-frequency coefficients are already small.
The human visual system is much less sensitive to high-frequency detail than to large-scale structure.

So JPEG selectively discards the parts our eyes barely notice — that's why a 10× compressed photo can still look fine.

Question 4

Modern video codecs encode the difference between consecutive frames, rather than each frame on its own. What does this strategy have in common with JPEG?

✅ Correct! That's the deep idea: compression is choosing coordinates.

❌ Not quite. JPEG is lossy, and the DCT is spatial, not temporal. Look for the general principle the two share.

Show solution

Solution:

The unifying principle is: change coordinates until most numbers are zero.

JPEG changes the spatial basis (pixels → cosine waves). Smooth images become sparse in the new basis.
Video codecs change the temporal basis (current frame → current frame minus predicted frame). Since consecutive frames are highly correlated, the difference image is mostly black — a nearly-zero vector that's cheap to store.

Different axis, same philosophy: the right basis turns hard problems into trivial ones.

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