Multiply any number by 1 and nothing happens: 1 \times 7 = 7, 1 \times (-3) = -3. The number 1 is multiplication's "do nothing" button. Matrices multiply vectors and other matrices — so is there a matrix that does nothing, the way 1 does nothing to a number?

Yes. It's called the identity matrix, written I. It has 1s down the main diagonal and 0s everywhere else:

Multiply any vector by I and it comes back unchanged, I\vec{x} = \vec{x}; multiply any matrix by I and you get that matrix back, IA = AI = A. The identity plays the role for matrices that 1 plays for numbers: the neutral element of multiplication — and, as you'll see later, it's the target every "undo" operation aims for.

Building it: the diagonal-of-1s pattern

The pattern scales to any size. The identity matrix I_n is the n\times n matrix whose entry in row i, column j follows one simple rule:

In words: put a 1 wherever the row number matches the column number, and a 0 everywhere else. I_2 is 2\times 2, I_3 is 3\times 3, I_4 is 4\times 4 — one identity matrix for every size, always square, always built from the same rule.

Try applying the rule by hand to build I_4. Row 2, column 2: row number equals column number, so that entry is 1. Row 2, column 3: row number does not equal column number, so that entry is 0. Fill in all sixteen entries this way and you get:

Twelve of those sixteen entries are 0; only the four where the row matches the column are 1. That ratio gets more lopsided the bigger the matrix gets — a 10\times 10 identity has a hundred entries, but only ten of them are 1.

Nothing moves

The columns of I are exactly the standard unit vectors \hat{\imath} and \hat{\jmath} — so, read as a weighted sum of columns, I\vec{x} just rebuilds \vec{x}. Move the input below; the output arrow sits exactly on top of it, every time.

Worked example: why each entry survives

Let's see exactly why, using the row-dot-product recipe from matrix-times-vector: each output entry is the dot product of one row of the matrix with the whole input vector. Take

The first output entry is row 1 of I_3 dotted with \vec{x}:

The 1 lands exactly on the entry it's "pointing at" — the first entry of \vec{x} — and every other term is multiplied by 0, so it vanishes. The second and third rows do the same trick, picking out -2 and 7 respectively:

Nothing was computed, really — every row's single 1 just copies one coordinate straight through. That's the whole mechanism behind "I does nothing."

Worked example: I is a universal "do nothing"

The same trick works when I multiplies an entire matrix, not just a single vector — because multiplying two matrices just means doing the row-times-column recipe once per column of the second matrix. Take

Column 1 of A is (3, 8) — treat it as a vector and multiply by I_2 exactly as before, and it comes back untouched: (3, 8). Column 2, (-5, 2), comes back the same way. Stack the two surviving columns back together:

I doesn't just leave one special vector alone — it leaves every vector alone, so it leaves every column of every matrix alone, so it leaves every matrix alone. That's what "universal do-nothing" means.

Here's a detail worth pausing on: matrix multiplication usually cares deeply about order — for most matrices, AB and BA are completely different matrices, or one of them might not even be defined. Try multiplying the same A by I_2 the other way round, AI_2, and run the identical row-times-column argument on the rows of A instead of its columns:

Same answer either way: I_2 A = A I_2 = A. The identity matrix is one of the very few matrices for which order never matters — another way of saying it truly does nothing, no matter which side it sits on.

Why a "do nothing" is so useful

Because it's the goal of undoing. The inverse of a matrix A is the matrix A^{-1} that cancels it: A^{-1}A = I and AA^{-1} = I. "Get back to where you started" is the identity — every time you'll solve for an inverse matrix later, I is the target you're aiming the answer at.

See it explained