ESPACES — L'Atelier

Basis & independence

How many arrows are needed — and from when on are you redundant?

A family is free when none of its arrows can be built from the others. It is generating when together they reach everything. A basis is both at once: just enough arrows, not one more. Add a third one in the plane and watch what happens: the dimension counter refuses to climb. Three arrows in a plane are always dependent — not by accident, but because the plane has only two directions to give.

The family

arrows—

dim of their span—

free?—

generating?—

verdict—

drag the arrowheads · the gold grid is what the family reaches

Free: $\lambda_1\vec v_1+\dots+\lambda_k\vec v_k=\vec0\Rightarrow$ every $\lambda_i=0$
Basis: free and generating — then every vector has exactly one set of coordinates
Dimension: every basis of the same space has the same number of vectors. That number is the dimension.

Rank

Rank does not count columns: it counts the directions that are genuinely different.

Here are three columns, so a $2\times3$ matrix. The rank is the dimension of what those columns reach together — their image. Three columns, but the rank will never exceed $2$: they live in the plane. Bring them all onto one line and the rank falls to $1$; the image is only a line now, and two dimensions of input were crushed on the way. Row rank always equals column rank — one of the most surprising facts in the subject, and nothing about it is obvious.

The three columns

——— ———

rank A—

dim image—

dim kernel—

the image is—

rank + dim kernel = — = number of columns

gold, marine, emeraude: the three columns · camellia: the image when it is a line

Kernel & image

What a map kills, and what it reaches — the two counts always add up.

The circle of unit vectors leaves, and comes back as an ellipse: that is the image, drawn in one stroke. When $\det A\neq0$ the ellipse is full and nothing was lost. Flatten $A$ and the ellipse closes into a segment: an entire direction has been sent to zero. That direction is the kernel. Drag the marine vector onto it and watch its image vanish. The rank–nullity theorem keeps the books: what is reached plus what is killed makes exactly the starting dimension, always, without exception.

The map

det A—

rank—

dim kernel—

‖Av‖—

v—

Av—

rank + dim kernel = — = 2

drag the gold and emeraude columns, then the marine vector

Kernel: $\ker A=\{\vec x:A\vec x=\vec0\}$ — everything the map sends to nothing
Image: $\operatorname{im}A=\{A\vec x\}$ — everything it can produce
Rank–nullity: $\operatorname{rank}A+\dim\ker A=n$, the number of columns

Change of basis

The point does not move. Only the numbers describing it change.

A vector has no coordinates: it receives them the moment a basis is chosen. Drag the camellia point, then drag the two basis vectors — the point does not move by a pixel, but its two numbers change completely. The matrix $P$ stacks the new basis as columns; it translates from new to old, and $P^{-1}$ translates the other way. All the rest of linear algebra is that one idea repeated: choose the frame in which the question becomes easy.

The two readings

————

P⁻¹

in the standard basis—

in the basis B—

det P—

the basis is—

pale grid: the standard basis · gold grid: the basis B · the point is the same

Inner product & projection

One number brings the angle, the length and the shadow.

$\vec u\cdot\vec v=u_xv_x+u_yv_y=\Vert u\Vert\Vert v\Vert\cos\theta$. That number measures how much of $\vec v$ goes in the direction of $\vec u$: it is the length of the shadow, to scale. The projection $\dfrac{\vec u\cdot\vec v}{\vec u\cdot\vec u}\vec u$ is that shadow itself, and what remains — $\vec v$ minus its shadow — is exactly perpendicular to $\vec u$. Press Gram–Schmidt and watch the procedure do precisely that: it does not manufacture orthogonality, it subtracts what was not orthogonal.

The shadow

u · v—

angle θ—

‖u‖—

‖v‖—

the projection—

verdict—

emeraude: the shadow of v on u · camellia: what remains, always perpendicular

The spectrum

The rare directions a matrix does not deflect — only stretch.

Walk the marine vector around the circle and watch its camellia image follow, almost always askew. Almost. At certain precise angles the two arrows lie down on one line: the image is nothing but a multiple of the input. That direction is an eigenvector, and the stretching factor is the eigenvalue $\lambda$. They are found by solving $\det(A-\lambda I)=0$, that is $\lambda^2-(\operatorname{tr}A)\lambda+\det A=0$. When the discriminant turns negative there is no fixed direction left at all — the matrix turns the plane, and nothing can stay put.

The spectrum

angle of v30°

trace—

det—

discriminant—

λ₁—

λ₂—

verdict—

vamp lines: the eigen-directions · marine: v · camellia: Av

Definition: $A\vec v=\lambda\vec v$ with $\vec v\neq\vec0$
Characteristic: $\det(A-\lambda I)=\lambda^2-(\operatorname{tr}A)\lambda+\det A=0$
Diagonal: $A=PDP^{-1}$ — in the eigenbasis the map is only a stretch along each axis
Powers: $A^n=PD^nP^{-1}$, and $D^n$ is just each $\lambda^n$ — which is why diagonalising pays

Épreuves the page checks by itself—

Six commissions in the geometer’s own algebra — and an endless examen. The page measures; the rank never lies.

Pitfalls

Six ways spaces catch the careless.

Generating is not freefour vectors can generate the plane without forming a basis. A basis must be both at once.

The zero vector kills any familyif $\vec0$ is on the list, the family is dependent, no matter what else is there.

The kernel is never emptyit always contains $\vec0$. “Zero kernel” means “only $\vec0$”, not “no kernel”.

u · v = 0 does not mean one of them is zeroperpendicularity is the interesting reason — but the zero vector kills the product too.

No real eigenvalue is not an anomalya rotation has none, and that is perfectly normal: no direction survives it.

Not every matrix diagonalisesyou need enough independent eigenvectors. A double eigenvalue does not always supply two.

Begun in MATRICES — La Transformation, where the matrix was still only a gesture on the plane.