Practice: Change of basis
Self-check
Section titled “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. Are a vector’s coordinates absolute or relative?
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Relative. Coordinates describe a vector against a chosen basis, not absolutely. The same arrow has different coordinates in different bases, and none is more correct than the others. The standard basis is just the default choice, not a privileged truth.
2. What is the basis matrix M, and which direction does it translate?
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M is the matrix whose columns are the other basis vectors written in your coordinates. Multiplying by M translates the other person’s coordinates into yours: M · [x', y'] = x'·b1 + y'·b2, expressed in your system.
3. What does M^-1 do?
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It translates your coordinates into the other basis (the reverse of M). This requires det(M) ≠ 0, that is, the other basis vectors must actually span the space; if they were dependent, M would collapse and there would be no clean way back.
4. What is the 2x2 inverse shortcut for M = [[a, b], [c, d]]?
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M^-1 = (1/det(M)) · [[d, -b], [-c, a]], where det(M) = ad - bc. Swap the main diagonal, negate the off-diagonal, divide by the determinant.
5. How do you express a transformation A in another basis, and how do you read the formula?
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A_their_basis = M^-1 · A · M, read right to left: M translates their coordinates into yours, A performs the transformation where you know how, and M^-1 translates the result back into their basis. Same physical operation, different matrix.
6. If a transformation’s matrix looks prettier in one basis than another, is it a different transformation?
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No. A transformation can look simple in one basis and ugly in another while being the exact same physical operation. The matrix is a basis-dependent description; the operation it describes does not change.
Try it yourself, part 1: translate coordinates both ways
Section titled “Try it yourself, part 1: translate coordinates both ways”Jennifer’s basis is b1 = [2, 1] and b2 = [1, 1], so M = [[2, 1], [1, 1]] with det(M) = 1 and M^-1 = [[1, -1], [-1, 2]]. About 7 minutes, pen and paper.
Step 1. Jennifer calls a vector [2, 3] in her basis. What do we call it (our coordinates)?
Step 2. We call a vector [4, 1]. What does Jennifer call it? Then translate your answer back with M to confirm the round trip returns [4, 1].
Check your work
Step 1. M · [2, 3] = 2 · [2, 1] + 3 · [1, 1] = [4, 2] + [3, 3] = [7, 5]. Her [2, 3] is our [7, 5].
Step 2. M^-1 · [4, 1] = [[1, -1], [-1, 2]] · [4, 1] = [4 - 1, -4 + 2] = [3, -2]. Our [4, 1] is her [3, -2]. Round trip: M · [3, -2] = 3 · [2, 1] + (-2) · [1, 1] = [6, 3] + [-2, -2] = [4, 1]. Back to the start, so M and M^-1 really do translate in opposite directions.
Try it yourself, part 2: transform in a new basis
Section titled “Try it yourself, part 2: transform in a new basis”Keep Jennifer’s basis (M = [[2, 1], [1, 1]], M^-1 = [[1, -1], [-1, 2]]). Let A = [[2, 0], [0, 1]] be a horizontal stretch by 2 in our basis. About 8 minutes.
Step 1. Compute A in Jennifer’s basis, M^-1 · A · M. (Do A · M first, then apply M^-1 to each column.)
Step 2. Verify by following her vector [1, 0] through the three steps (into our basis, stretch, back to hers) and confirming it matches your matrix applied to [1, 0].
Check your work
Step 1. A · M: A doubles the first component, so A · [2, 1] = [4, 1] and A · [1, 1] = [2, 1], giving A · M = [[4, 2], [1, 1]]. Then M^-1 · (A · M): M^-1 · [4, 1] = [4 - 1, -4 + 2] = [3, -2] and M^-1 · [2, 1] = [2 - 1, -2 + 2] = [1, 0]. So A_jennifer = [[3, 1], [-2, 0]].
Step 2. Her [1, 0] into our basis: M · [1, 0] = [2, 1]. Stretch: A · [2, 1] = [4, 1]. Back to hers: M^-1 · [4, 1] = [3, -2]. Now apply the matrix directly: A_jennifer · [1, 0] = [[3, 1], [-2, 0]] · [1, 0] = [3, -2]. Same answer, confirming the sandwich bundles the three steps into one matrix.
Flashcards
Section titled “Flashcards”Ten cards. Click any card to reveal the answer. Use the Print flashcards button to lay out the full set as one card per page, ready to print or save as a PDF for offline review.
Q. Are a vector's coordinates absolute or relative?
Relative. They describe the vector against a chosen basis, not absolutely. The same arrow has different coordinates in different bases; none is more correct. The standard basis is the default, not a privileged truth.
Q. What is the basis matrix M, and which way does it translate?
The matrix whose columns are the other basis vectors written in your coordinates. M · [x', y'] translates their coordinates into yours (it equals x'·b1 + y'·b2 in your system).
Q. What does M^-1 do in change of basis?
It translates your coordinates into the other basis, the reverse of M. It requires det(M) ≠ 0, meaning the other basis vectors actually span the space.
Q. What is the 2x2 inverse shortcut?
For M = [[a, b], [c, d]], M^-1 = (1/det) · [[d, -b], [-c, a]] with det = ad - bc. Swap the main diagonal, negate the off-diagonal, divide by the determinant.
Q. How do you express a transformation A in another basis?
A_their_basis = M^-1 · A · M, read right to left: translate into your basis (M), apply A, translate back (M^-1). Same operation, a different basis-dependent matrix.
Q. Does a prettier matrix in a new basis mean a different transformation?
No. The same physical operation can look simple in one basis and ugly in another. The matrix is a basis-dependent description; the operation itself does not change with the basis.
Q. Why must det(M) be nonzero for a change of basis?
Because M^-1 must exist to translate back. A nonzero determinant means the proposed basis vectors are independent and actually span the space; if they were dependent, M would collapse and the translation would not be reversible.
Q. Which way does M translate versus M^-1?
M (columns are the other basis in your coordinates) takes their coordinates to yours. M^-1 takes your coordinates to theirs. Mixing them up translates in the wrong direction.
Q. How does change of basis connect to dimensionality reduction?
PCA picks a basis aligned with the data’s directions of greatest variation, so keeping the first few coordinates compresses with minimal loss. Whitening picks a basis where every direction has unit variance. SVD finds bases that expose a matrix’s structure.
Q. What does the next lesson chase about basis choice?
The best basis for a given transformation, in which its matrix becomes a clean diagonal of pure stretch factors. The vectors defining that basis are eigenvectors, and finding them is the next lesson.