Worked examples — Change of basis matrix

If any symbol here feels unfamiliar, it was built in Change of basis matrix — this page assumes only that page.

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The scenario matrix

Every change-of-basis problem is one of these shapes. We enumerate them so nothing surprises you.

Cell	What makes it that case	Which direction	Example
A	Old = standard, new = non-standard	$E\to C$ (needs inverse)	Ex 1
B	Old = non-standard, new = standard	$B\to E$ (easy, no inverse)	Ex 2
C	Both non-standard	$B\to C$ (full $C^{-1}B$)	Ex 3
D	New basis is old scaled	$E\to C$	Ex 4
E	New basis is old rotated, $\det >0$ (geometry)	$E\to C$	Ex 5
F	Reverse an existing translator	use $P^{-1}$	Ex 6
G	Same basis in and out (identity)	$B\to B$	Ex 7
H	Bigger space, $3\times 3$	$E\to C$ in ${\mathbb{R}}^3$	Ex 8
I	Near-degenerate basis (limiting/breaking)	why $C^{-1}$ blows up	Ex 9
J	Real-world word problem	modelling	Ex 10
K	Reflected basis, $\det <0$ (orientation flip)	$E\to C$	Ex 11

The number line of "signs and quadrants" for this topic is really the sign of the determinant and whether $C^{-1}$ exists at all. Cell E has $\det >0$ (orientation preserved), cell K has $\det <0$ (orientation flipped, a mirror), and cell I is the degenerate boundary $\det \to 0$ where $C^{-1}$ ceases to exist.

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Example 1 — Cell A: standard → non-standard (needs inverse)

Forecast: Guess before computing — will the first $C$-coordinate be more or less than $4$? (The first $C$-vector is twice as long as $e_1$, so you should need fewer of it.)

1. Build the matrix of $C$. $C=\left[\begin{array}{cc}2& 1\\ 0& 3\end{array}\right].$ Why this step? Columns of $C$ are the new basis vectors written in standard coordinates — that's the standard-basis shortcut from the parent note.

2. We need $P_{C\leftarrow E}=C^{-1}$. Why this step? We are going into $C$-land from standard-land. Read the subscript: $P_{C\leftarrow E}$ eats $E$-coords, gives $C$-coords. The parent formula gives $P_{C\leftarrow E}=C^{-1}E=C^{-1}$ (since $E=I$).

3. Invert $C$. For $\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ the inverse is $\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$. Here $\det =2\cdot 3-1\cdot 0=6$: $C^{-1}=\frac{1}{6}\left[\begin{array}{cc}3& -1\\ 0& 2\end{array}\right].$ Why this step? Dividing by $\det =6$ and swapping/negating entries is the $2\times 2$ inversion rule — this is the translator that reads standard coords and reports them in $C$.

4. Apply to $(4,6{)}^T$. $[v{]}_C=\frac{1}{6}\left[\begin{array}{cc}3& -1\\ 0& 2\end{array}\right]\left[\begin{array}{c}4\\ 6\end{array}\right]=\frac{1}{6}\left[\begin{array}{c}12-6\\ 0+12\end{array}\right]=\left[\begin{array}{c}1\\ 2\end{array}\right].$ Why this step? Multiplying the translator by the input column is the translation — output is $[v{]}_C$.

Verify: Reconstruct the arrow: $1\cdot (2,0)+2\cdot (1,3)=(2+2,0+6)=(4,6)=v$. ✓ Forecast confirmed: first coordinate is $1$, much less than $4$, exactly because $(2,0)$ is a long ruler.

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Example 2 — Cell B: non-standard → standard (the easy direction)

Forecast: Do we need an inverse here? (No — decoding into standard is always the multiply-by-$B$ direction.)

1. Write $B$. $B=\left[\begin{array}{cc}3& -1\\ 1& 2\end{array}\right]$. Why this step? Its columns are already in standard coordinates, so $P_{E\leftarrow B}=B$ directly — no inverse.

2. Multiply. $[v{]}_E=B[v{]}_B=\left[\begin{array}{cc}3& -1\\ 1& 2\end{array}\right]\left[\begin{array}{c}2\\ 4\end{array}\right]=\left[\begin{array}{c}6-4\\ 2+8\end{array}\right]=\left[\begin{array}{c}2\\ 10\end{array}\right].$ Why this step? $[v{]}_B=(2,4)$ means $v=2b_1+4b_2$; matrix–column multiply is literally that linear combination.

Verify: $2(3,1)+4(-1,2)=(6-4,2+8)=(2,10)$. ✓

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Example 3 — Cell C: both bases non-standard (full $C^{-1}B$)

Forecast: The mnemonic is "C-inverse Bites B." Which matrix gets inverted — the input basis or the output basis? (The output, $C$.)

1. Assemble both matrices. $B=\left[\begin{array}{cc}1& 3\\ 2& 1\end{array}\right],C=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right].$ Why this step? Columns are the respective basis vectors in standard coordinates — the raw material for the formula.

2. Invert $C$. $\det C=1$, so $C^{-1}=\left[\begin{array}{cc}1& -1\\ 0& 1\end{array}\right].$ Why this step? Translating $B$-coords first to standard ($\times B$) then standard into $C$ ($\times C^{-1}$). The composite is $C^{-1}B$.

3. Multiply $C^{-1}B$. $P_{C\leftarrow B}=\left[\begin{array}{cc}1& -1\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& 3\\ 2& 1\end{array}\right]=\left[\begin{array}{cc}1-2& 3-1\\ 2& 1\end{array}\right]=\left[\begin{array}{cc}-1& 2\\ 2& 1\end{array}\right].$ Why this step? This product is the single translator card that eats $B$-coords and gives $C$-coords in one shot.

4. Apply to $[v{]}_B=(1,1{)}^T$. $[v{]}_C=\left[\begin{array}{cc}-1& 2\\ 2& 1\end{array}\right]\left[\begin{array}{c}1\\ 1\end{array}\right]=\left[\begin{array}{c}1\\ 3\end{array}\right].$ Why this step? Feeding the input column through the translator produces $[v{]}_C$.

Verify (Forecast-then-Verify): The actual arrow is $v=1(1,2)+1(3,1)=(4,3)$. In $C$: $1(1,0)+3(1,1)=(1+3,0+3)=(4,3)$. ✓

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Example 4 — Cell D: new basis is old, just scaled ($E\to C$)

Forecast: Stretching the ruler makes each measured number shrink. Predict the two coordinates before computing.

1. Matrix of $C$ is diagonal. $C=\left[\begin{array}{cc}5& 0\\ 0& 2\end{array}\right]$. Why this step? Diagonal because the two rulers don't interfere — pure scaling.

2. Inverse of a diagonal matrix = reciprocals on the diagonal. $C^{-1}=\left[\begin{array}{cc}1/5& 0\\ 0& 1/2\end{array}\right].$ Why this step? A diagonal matrix scales each axis independently, so undoing it just inverts each scale — no cross terms. Since $B=E$, $P_{C\leftarrow E}=C^{-1}$.

3. Apply. $[v{]}_C=\left[\begin{array}{cc}1/5& 0\\ 0& 1/2\end{array}\right]\left[\begin{array}{c}10\\ 6\end{array}\right]=\left[\begin{array}{c}2\\ 3\end{array}\right].$ Why this step? Running the standard-coord input through $C^{-1}$ reports it in the stretched rulers.

Verify: $2(5,0)+3(0,2)=(10,6)$. ✓ The numbers shrank ($10\to 2$, $6\to 3$) exactly as forecast — a longer ruler yields a smaller reading.

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Example 5 — Cell E: new basis is old, rotated, $\det >0$ (geometry)

Figure 1 — The faint white arrows are the old standard axes. The blue arrow is the new first basis vector $c_1$; the red arrow is $c_2$. Notice they are both length $1$ and meet at a right angle (an orientation-preserving rotation, $\det =+1$). The yellow arrow is the fixed vector $v=(1,1)$ — observe it lies flat along $c_1$, which is exactly why its second $C$-coordinate will come out zero. The arrow itself never moved; only the grid we measure it with turned.

Forecast: The arrow $(1,1)$ points exactly along the new $c_1$ direction. So one $C$-coordinate should be zero. Which one?

1. Build $C$ from its columns. $C=\left[\begin{array}{cc}\frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right].$ Why this step? Columns are the rotated basis vectors in standard coords.

2. For a rotation, the inverse is the transpose. $C^{-1}=C^T=\left[\begin{array}{cc}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right].$ Why this step? Rotation matrices are orthogonal — their columns are unit-length and perpendicular (see the blue and red arrows in Figure 1) — so $C^{T}C=I$, meaning $C^{-1}=C^T$. No fraction-heavy inversion needed.

3. Apply to $(1,1{)}^T$. $[v{]}_C=C^T\left[\begin{array}{c}1\\ 1\end{array}\right]=\left[\begin{array}{c}\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\end{array}\right]=\left[\begin{array}{c}\sqrt{2}\\ 0\end{array}\right].$ Why this step? Passing standard coords through $C^T=C^{-1}$ reports the same arrow in the rotated grid.

Verify: $\sqrt{2}\cdot c_1+0\cdot c_2=\sqrt{2}\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)=(1,1)=v$. ✓ Second coordinate is $0$, as forecast — the arrow lies along $c_1$ (the yellow arrow on the blue arrow in Figure 1).

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Example 6 — Cell F: reverse an existing translator ($P^{-1}$)

Forecast: You have $C$-coords but want $B$-coords. Which matrix moves you backwards? ($P_{B\leftarrow C}=P_{C\leftarrow B}^{-1}$ — flip the translator card over.)

1. Invert the given matrix. $\det =(-1)(1)-(2)(2)=-5$: $P_{B\leftarrow C}=(P_{C\leftarrow B}{)}^{-1}=\frac{1}{-5}\left[\begin{array}{cc}1& -2\\ -2& -1\end{array}\right]=\left[\begin{array}{cc}-1/5& 2/5\\ 2/5& 1/5\end{array}\right].$ Why this step? Reversing the direction of a change of basis is exactly taking the inverse — the parent note's inverse relation.

2. Apply to $(1,3{)}^T$. $[w{]}_B=\left[\begin{array}{cc}-1/5& 2/5\\ 2/5& 1/5\end{array}\right]\left[\begin{array}{c}1\\ 3\end{array}\right]=\left[\begin{array}{c}-1/5+6/5\\ 2/5+3/5\end{array}\right]=\left[\begin{array}{c}1\\ 1\end{array}\right].$ Why this step? Running the $C$-coord input through the reversed translator delivers the $B$-coords.

Verify: This should undo Ex 3, where $[v{]}_B=(1,1{)}^T$ gave $[v{]}_C=(1,3{)}^T$. Feeding $(1,3{)}^T$ back returns $(1,1{)}^T$. ✓

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Example 7 — Cell G: same basis in and out (the identity)

Forecast: Translating a language into itself changes nothing — so what matrix must this be?

1. Use the formula with $C=B$. $P_{B\leftarrow B}=B^{-1}B=I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right].$ Why this step? $C^{-1}B$ with $C=B$ collapses to the identity — the degenerate but essential "no change" case.

2. Apply. $I(2,4{)}^T=(2,4{)}^T$. Unchanged, as it must be. Why this step? The identity leaves any input column untouched — confirming "same basis in and out" does nothing.

Verify: Column $j$ of $P_{B\leftarrow B}$ should be $[b_j{]}_B=e_j$ — and indeed $[b_1{]}_B=(1,0{)}^T$, $[b_2{]}_B=(0,1{)}^T$, which stacks to $I$. ✓

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Example 8 — Cell H: bigger space, $3\times 3$ ($E\to C$)

Forecast: $C$ is "each vector adds one more $1$." Its inverse should be a simple difference pattern. Guess the top coordinate.

1. Matrix of $C$. $C=\left[\begin{array}{ccc}1& 1& 1\\ 0& 1& 1\\ 0& 0& 1\end{array}\right].$ Why this step? Columns = new basis vectors in standard coords.

2. We need $C^{-1}$. This upper-triangular "all ones above and on the diagonal" matrix inverts to the finite-difference matrix: $C^{-1}=\left[\begin{array}{ccc}1& -1& 0\\ 0& 1& -1\\ 0& 0& 1\end{array}\right].$ Why this step? Since $B=E$, $P_{C\leftarrow E}=C^{-1}$; and undoing "add the next component" is subtracting it, giving this difference pattern. You can check $C^{-1}C=I$ directly.

3. Apply to $(3,5,2{)}^T$. $[v{]}_C=\left[\begin{array}{ccc}1& -1& 0\\ 0& 1& -1\\ 0& 0& 1\end{array}\right]\left[\begin{array}{c}3\\ 5\\ 2\end{array}\right]=\left[\begin{array}{c}3-5\\ 5-2\\ 2\end{array}\right]=\left[\begin{array}{c}-2\\ 3\\ 2\end{array}\right].$ Why this step? Passing the standard-coord input through $C^{-1}$ reports it in the $C$-basis.

Verify: $-2(1,0,0)+3(1,1,0)+2(1,1,1)=(-2+3+2,3+2,2)=(3,5,2)$. ✓

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Example 9 — Cell I: near-degenerate basis (why it breaks)

Figure 2 — The blue arrow $c_1=(1,0)$ and the red arrow $c_2=(1,0.1)$ almost coincide: the two rulers point nearly the same way. The yellow arrow is the vector $v=(1,1)$ we want to measure. Because the rulers barely differ, reaching $v$ needs a huge push forward and a huge cancelling pull, so the coordinates $(-9,10)$ are enormous compared to the tiny arrow — the visual signature of a near-degenerate basis.

Forecast: As the two basis vectors become nearly parallel (Figure 2: red and blue almost overlap), does measuring a vector get easier or dangerously sensitive?

1. Determinant. $\det C_{\epsilon }=1\cdot \epsilon -1\cdot 0=\epsilon$. Why this step? The inverse divides by the determinant. As $\epsilon \to 0$, we divide by nearly zero — coordinates explode. This is the limiting/degenerate boundary of the whole topic.

2. The inverse. $C_{\epsilon }^{-1}=\frac{1}{\epsilon }\left[\begin{array}{cc}\epsilon & -1\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& -1/\epsilon \\ 0& 1/\epsilon \end{array}\right].$ Why this step? Applying the $2\times 2$ rule shows the $1/\epsilon$ entries — the translator's numbers blow up as the basis degenerates.

3. Apply with $\epsilon =0.1$ to $(1,1{)}^T$. $[v{]}_{C_{0.1}}=\left[\begin{array}{cc}1& -10\\ 0& 10\end{array}\right]\left[\begin{array}{c}1\\ 1\end{array}\right]=\left[\begin{array}{c}1-10\\ 10\end{array}\right]=\left[\begin{array}{c}-9\\ 10\end{array}\right].$ Why this step? Feeding the input through the exploded translator produces the enormous coordinates seen in Figure 2.

Verify: $-9(1,0)+10(1,0.1)=(-9+10,1.0)=(1,1)$. ✓

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Example 10 — Cell J: real-world word problem

Forecast: The speed magnitude is $\sqrt{8^2+6^2}=10$ m/s. Whatever coordinates we get, they must also have length $10$ (rotating axes can't change how fast the drone is). Keep that as your sanity anchor.

1. The pilot's basis $C=\{f,\ell \}$. $C=\left[\begin{array}{cc}0.6& -0.8\\ 0.8& 0.6\end{array}\right].$ Why this step? Columns are the runway axes in standard (East, North) coordinates.

2. This is a rotation ⇒ $C^{-1}=C^T$ (orthogonal, like Ex 5). $C^T=\left[\begin{array}{cc}0.6& 0.8\\ -0.8& 0.6\end{array}\right].$ Why this step? $f,\ell$ are perpendicular unit vectors, so no messy inverse — the transpose suffices, and $P_{C\leftarrow E}=C^{-1}=C^T$.

3. Apply. $[v{]}_C=C^T\left[\begin{array}{c}8\\ 6\end{array}\right]=\left[\begin{array}{c}0.6\cdot 8+0.8\cdot 6\\ -0.8\cdot 8+0.6\cdot 6\end{array}\right]=\left[\begin{array}{c}4.8+4.8\\ -6.4+3.6\end{array}\right]=\left[\begin{array}{c}9.6\\ -2.8\end{array}\right].$ Why this step? Running the ground-frame velocity through $C^T$ reports the same velocity in runway coordinates. So the screen shows $9.6$ m/s forward and $-2.8$ m/s left (i.e. $2.8$ m/s to the right).

Verify (units + magnitude): $\sqrt{9.6^2+(-2.8{)}^2}=\sqrt{92.16+7.84}=\sqrt{100}=10$ m/s. ✓ Speed preserved exactly as forecast — the drone doesn't fly faster just because the pilot uses runway axes.

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Example 11 — Cell K: reflected basis, $\det <0$ (orientation flip)

Forecast: A rotation (Ex 5) keeps $\det =+1$. Flipping one ruler reverses orientation, so predict: is $\det C$ positive or negative? And what should happen to the second coordinate of $v$?

1. Matrix of $C$. $C=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]$. Why this step? Columns are the mirror basis vectors in standard coords; the $-1$ encodes the downward $y$-ruler.

2. Determinant — check orientation. $\det C=(1)(-1)-(0)(0)=-1<0$. Why this step? A negative determinant is the algebraic fingerprint of an orientation-reversing (mirror) basis — the case cell K exists to cover. It is still non-zero, so $C^{-1}$ exists.

3. Inverse. This reflection is its own inverse: $C^{-1}=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]=C$. Why this step? Mirroring twice returns you to the start, so $C^2=I$ and $C^{-1}=C$. Since $B=E$, $P_{C\leftarrow E}=C^{-1}=C$.

4. Apply to $(3,4{)}^T$. $[v{]}_C=\left[\begin{array}{cc}1& 0\\ 0& -1\end{array}\right]\left[\begin{array}{c}3\\ 4\end{array}\right]=\left[\begin{array}{c}3\\ -4\end{array}\right].$ Why this step? Running the standard coords through $C^{-1}$ reports them in the flipped grid — the second coordinate changes sign, exactly as forecast.

Verify: $3(1,0)+(-4)(0,-1)=(3,0+4)=(3,4)=v$. ✓ The arrow $(3,4)$ is unchanged; only its description flipped its second component because the ruler points the other way.

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Recall

Recall In which direction do you NOT need an inverse?

Non-standard $\to$ standard ($B\to E$): just multiply by $B$, since its columns are already in standard coordinates.

Recall Why is

$C^{-1}=C^T$ for a rotated basis? A rotation's columns are perpendicular unit vectors (orthogonal matrix), so $C^{T}C=I$, hence $C^{-1}=C^T$ — no fractions needed.

Recall What does the sign of

$\det C$ tell you? $\det >0$: new axes are a rotation (orientation preserved). $\det <0$: a reflection (orientation flipped, a mirror). $\det \to 0$: the basis degenerates and $C^{-1}$ blows up / ceases to exist.

Recall What goes wrong as the basis becomes degenerate?

$\det C\to 0$, so $C^{-1}$ (which divides by $\det$) blows up: coordinates explode and errors amplify by $1/\det$.

Recall Same basis in and out — what is

$P_{B\leftarrow B}$? The identity $I$: translating a language into itself changes nothing.

Connections

• Change of basis matrix — the parent machine these examples exercise.
• Invertible Matrices — Ex 9's degenerate limit is exactly non-invertibility; Ex 11's $\det <0$ is still invertible.
• Coordinate Vectors — every input/output column here is one.
• Linear Transformations — Ex 5, 10 & 11 are the passive view (axes move, arrow fixed).
• Basis and Dimension — why an independent spanning set is required at all.
• Similar Matrices · Eigenvectors and Diagonalization — where these translators get reused on whole maps.