Projections & Idempotent Matrices

Shine a light straight down on a vector and look at the shadow it casts on the floor. That shadow is a projection — the vector squashed onto a lower-dim subspace. The neat thing about shadows is obvious once you say it out loud: the shadow of a shadow is the same shadow. A point already on the floor projects to itself. Apply the projection twice and nothing extra happens.

That “applying it twice changes nothing” idea is exactly the equation P² = P. Matrices with this property get a name — idempotent — and from that one fact the rank, the nullity, and the allowed eigenvalues all fall out. Which is why GATE keeps reaching for it.

Projection onto a subspace

A projection P sends each vector to the closest point in some subspace U and leaves vectors that already live in U untouched. Picture dropping a perpendicular onto a line:

Three properties follow, and they are the whole exam syllabus here:

• Idempotent: P² = P. (And therefore P³ = P·P² = P·P = P, and so on.)
• Eigenvalues in {0, 1}: if Pv = λv, then Pv = P²v = λ²v, so λ² = λ, giving λ = 0 or λ = 1. Vectors in U have eigenvalue 1 (kept); vectors perpendicular to U have eigenvalue 0 (killed).
• Rank and nullity: the range of P is U, so rank(P) = dim(U). By rank–nullity, nullity(P) = n − dim(U).

An orthogonal projection (the perpendicular kind, as drawn) has the extra property that it is symmetric: P = Pᵀ. Not every idempotent matrix is symmetric — oblique projections drop the shadow at a slant — but the orthogonal ones are.

Want to feel the squash directly? In the playground below, set the matrix to [[1, 0], [0, 0]] — the orthogonal projection onto the x-axis. The whole unit shape collapses down to a line segment, and any point already on that line is left untouched. Apply the matrix again and the result is identical: that’s P² = P in pictures.

The centering matrix — PCA’s engine

The single most important projection in data analysis subtracts the mean from a data vector. With the all-ones vector 1, the centering matrix is

C = I − (1/n)·11ᵀ

Here 11ᵀ is the n × n matrix of all ones, so (1/n)·11ᵀ averages the entries, and Cx = x − mean(x)·1 removes that average. C is symmetric and idempotent — it is the orthogonal projection onto the subspace of mean-zero vectors. PCA begins by applying C to every feature, which is why this matrix shows up in DA papers.

How GATE asks this

Two flavours. As an MSQ: a matrix M is described as a projection onto a k-dimensional subspace, and you select the true statements — M² = M, M³ = M, rank(M) = k, eigenvalues in {0, 1}. As a NAT: you are given the subspace dimension and asked for the nullity (n − k) or the rank. GATE DA 2024 used a projection onto a 2-D subspace of R³; GATE DA 2026 returned to the theme via the centering matrix and its C² = C property.

Worked example — the 2024 and 2026 questions

Let M be the orthogonal projection of R³ onto a 2-dimensional subspace U. Evaluate M², M³, rank(M), and nullity(M).

Idempotence. A projection satisfies M² = M. Then M³ = M·M² = M·M = M as well — every positive power of a projection equals itself.

Rank. The output of M fills exactly U, so rank(M) = dim(U) = 2.

Nullity. Rank–nullity in R³: nullity(M) = 3 − rank(M) = 3 − 2 = 1. The null space is the 1-D direction perpendicular to U (the part the shadow discards).

Now the centering matrix part used in 2026. Show C² = C for C = I − (1/n)11ᵀ. The key fact is 1ᵀ1 = n (summing n ones):

C² = (I − (1/n)11ᵀ)(I − (1/n)11ᵀ)
   = I − (1/n)11ᵀ − (1/n)11ᵀ + (1/n²)·1(1ᵀ1)1ᵀ
   = I − (2/n)11ᵀ + (1/n²)·1·(n)·1ᵀ          [since 1ᵀ1 = n]
   = I − (2/n)11ᵀ + (1/n)11ᵀ
   = I − (1/n)11ᵀ  =  C   ✓

The two −(1/n)11ᵀ terms and the recovered +(1/n)11ᵀ collapse back to a single −(1/n)11ᵀ, proving C is idempotent. Concretely for n = 3, (1/3)11ᵀ is the 3-by-3 matrix of all 1/3s; squaring it returns itself, so C² = C numerically too.

Quick check