Practice: Undoing a transformation

Self-check

Six short questions. Answer each one in your head (or on paper) before opening the collapsible. Trying to retrieve the answer is where the learning sticks; rereading feels productive but does much less.

1. What is the inverse M^-1, and what property defines it?

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The transformation that undoes M: apply M then M^-1 and you are back where you started. It is defined by M^-1 · M = M · M^-1 = I, the identity (do-nothing) transformation. It is not “one over the matrix”; there is no division involved.

2. When does the inverse of a matrix exist?

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Exactly when det(M) ≠ 0. A nonzero determinant means the transformation rearranged space without collapsing it, so every output came from one input and can be traced back. A zero determinant means a collapse, with information lost, so no undo transformation can exist.

3. What is the column space of a matrix, and what is the rank?

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The column space is the set of all possible outputs of M · v, which is the span of the matrix’s columns. The rank is the dimension of that column space: how many dimensions of output the transformation actually uses (rank 2 = fills the plane, rank 1 = collapses to a line).

4. What is the null space of a matrix?

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The set of all input vectors that the transformation sends to the zero vector (M · v = 0). For an invertible matrix it is just the origin {0}; for a collapsed matrix it is an entire line (or more) of inputs that get crushed to nothing, which is exactly the information that was destroyed.

5. State the rank-nullity conservation law and what it means.

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rank + nullity = input dimension. The dimensions the transformation keeps (rank) plus the dimensions it crushes (nullity) always equal the dimensions you started with. For a 2x2 matrix, rank 2 forces nullity 0; rank 1 forces nullity 1. Nothing leaks; the accounting is exact.

6. What are the three equivalent tests for invertibility?

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(1) det(M) ≠ 0; (2) full rank (the column space is the entire input dimension); (3) the null space is only the origin. If any one holds, all three hold; if any one fails, all three fail. They are the same fact seen three ways.

Try it yourself, part 1: test invertibility three ways

For each matrix, find (i) the determinant, (ii) the column space and rank, (iii) the null space and nullity, and (iv) whether it is invertible. Confirm all three tests agree. About 8 minutes, pen and paper.

a) [[3, 1], [2, 2]]
b) [[1, 2], [2, 4]]
c) [[1, 0], [0, 1]]

Check your work

a) det = (3)(2) - (1)(2) = 4 ≠ 0. Columns [3, 2] and [1, 2] point in different directions, so the column space is all of 2D, rank 2. The only input sent to the origin is the origin, so null space {0}, nullity 0. 2 + 0 = 2. Invertible (all three tests agree).
b) det = (1)(4) - (2)(2) = 0. Columns [1, 2] and [2, 4] are dependent ([2,4] = 2·[1,2]), so the column space is the line through [1, 2], rank 1. Null space: M · [x, y] = 0 gives x + 2y = 0, so x = -2y, the line through [-2, 1], nullity 1 (check: M · [-2, 1] = [-2+2, -4+4] = [0, 0]). 1 + 1 = 2. Not invertible.
c) det = 1. The identity: column space all of 2D (rank 2), null space {0} (nullity 0). Invertible.

Try it yourself, part 2: solve M·v = b

Each part is a system M · v = b. Decide how many solutions it has, and find them where you can. About 7 minutes.

a) [[3, 1], [2, 2]], target b = [4, 4]
b) [[1, 2], [2, 4]], target b = [1, 1]
c) [[1, 2], [2, 4]], target b = [1, 2]

Check your work

a) The matrix is invertible (part 1a), so there is exactly one solution. Solve 3x + y = 4 and 2x + 2y = 4 (the second gives x + y = 2): substituting y = 2 - x into the first, 3x + 2 - x = 4, so x = 1 and y = 1. Check: M · [1, 1] = [4, 4].
b) The matrix is not invertible; its column space is the line through [1, 2] (all points with second coordinate twice the first). The target [1, 1] is off that line (1 ≠ 2·1), so there is no solution: the transformation can never output [1, 1].
c) Same matrix, but [1, 2] is on the column-space line (2 = 2·1), so there are infinitely many solutions. One particular input is [1, 0] (M · [1, 0] = [1, 2]), and adding any null-space vector t · [-2, 1] still works, since the null-space part is crushed to zero. The full solution set is [1, 0] + t · [-2, 1] for every t.

Flashcards

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Q. What is the inverse M^-1, and what property defines it?

The transformation that undoes M. Defined by M^-1 · M = I, the do-nothing (identity) transformation. It is not a reciprocal or division; it is the move that brings every vector back to where it started.

Q. When does a matrix have an inverse?

Exactly when det(M) ≠ 0. Nonzero determinant means no collapse, so every output traces back to one input. Zero determinant means information was lost, so no undo transformation exists.

Q. What is the column space of a matrix?

The set of all possible outputs of M · v, which equals the span of the matrix’s columns. It is everything the transformation can reach on the output side.

Q. What is the rank of a matrix?

The dimension of the column space: how many dimensions of output the transformation actually uses. Rank 2 fills the plane; rank 1 collapses output to a line. A big matrix can still have small rank.

Q. What is the null space of a matrix?

The set of all inputs sent to the zero vector (M · v = 0). For an invertible matrix it is just the origin; for a collapsed one it is a line (or more) of inputs crushed to nothing, the destroyed information named.

Q. State the rank-nullity conservation law.

rank + nullity = input dimension. The dimensions kept (rank) plus the dimensions crushed (nullity) equal the dimensions you started with. For 2x2: rank 2 forces nullity 0; rank 1 forces nullity 1.

Q. What are the three equivalent tests for invertibility?

(1) det ≠ 0; (2) full rank (column space = entire input dimension); (3) null space is only the origin. All true together or all false together: one fact, three views.

Q. How many solutions does M·v = b have?

If M is invertible: exactly one (v = M^-1·b). If not invertible: none when b is off the column space, or infinitely many when b is on it (one solution plus any null-space vector).

Q. Column space vs null space: which side does each live on?

Column space lives on the output side (everything the transformation can produce). Null space lives on the input side (everything the transformation destroys). Different spaces, different questions.

Q. Why can a large matrix have small rank, and why does that matter for AI?

Rank counts output dimensions actually used, not the matrix’s row or column count. A thousand-by-thousand matrix can be rank 10, meaning it operates in far fewer dimensions, so it is highly compressible. Much model-shrinking work exploits low-rank weight matrices or updates.