Linear-Algebra-Done-Right · Unit 7 · Video 5 · Interactive Practice

Change of Basis for an Operator: A = C⁻¹BC

IKey Formulas

Formula	Name	Setting
$\mathcal{M}(I) = I$ , $\ AI = IA = A$	Identity operator and identity matrix	Any basis of $V$ ; any square $A$ of the same size
$(A^{-1})^{-1} = A$ , $\ (AD)^{-1} = D^{-1}A^{-1}$	Inverse rules	$A$ , $D$ invertible square matrices of the same size
$\mathcal{M}\big(I, (v), (u)\big) = \big(\mathcal{M}(I, (u), (v))\big)^{-1}$	Change-of-basis matrices	$(u)$ , $(v)$ bases of $V$ ; column $k$ of $\mathcal{M}(I, (u), (v))$ writes $u_k$ in the basis $(v)$
$A = C^{-1}BC$	Change-of-basis formula	$T \in \mathcal{L}(V)$ , $A = \mathcal{M}(T, (u))$ , $B = \mathcal{M}(T, (v))$ , $C = \mathcal{M}(I, (u), (v))$

Key Insight: Read $C^{-1}BC$ from right to left: $C$ turns $u$ -coordinates into $v$ -coordinates, $B$ applies $T$ , and $C^{-1}$ turns the result back into $u$ -coordinates. The same bookkeeping gives $\mathcal{M}(T^{-1}) = \big(\mathcal{M}(T)\big)^{-1}$ for an invertible $T$ .

IIThe Change-of-Basis Matrix and Its Inverse

Columns of $C$ write each $u_k$ in the standard basis; columns of $C^{-1}$ write each standard vector in the $u$ -basis.

IIIOne Vector Through C, B and C⁻¹

For the video's operator and basis, $C$ , then $B$ , then $C^{-1}$ carry the $u$ -coordinates of $x$ to those of $Tx$ , exactly as $A$ does.

p (u₁-coordinate of x) 1/2 q (u₂-coordinate of x) 1/2

IVSame Operator, Another Basis

Column $k$ of $A = C^{-1}BC$ holds the $u$ -coordinates of $Tu_k$ , so $A$ changes with the basis while $B$ stays fixed.

💡 Challenge: find a basis in which $A = \begin{pmatrix} 3 & 0 \\ 0 & 2 \end{pmatrix}$ , and another one, with $u_1 \ne (4, 2)$ , in which $A = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}$ .

VQuiz Questions

Problem 1 · Which Way Does the Matrix Go?

Given: the basis $u_1 = (1, 2)$ , $u_2 = (1, 3)$ of $\mathbf{F}^2$ , and the standard basis $(v)$ : $(1, 0), (0, 1)$ . Find $\mathcal{M}\big(I, (v), (u)\big)$ .

✅ Correct! Its first column says

(1, 0) = 3u_1 - 2u_2

and its second says

(0, 1) = -u_1 + u_2

❌ Wrong direction. That is

\mathcal{M}\big(I, (u), (v)\big)

: its columns write

u_1

u_2

in the standard basis. With

(v)

in the domain, the columns must write

(1,0)

and

(0,1)

in the basis

u_1, u_2

, which gives the inverse matrix.

❌ Rows and columns swapped. Since

(1, 0) = 3u_1 - 2u_2

, the coefficients

3, -2

form the first column, not the first row.

❌ Not quite. This writes

u_1

and

u_2

as rows. Even written as columns they would give

\mathcal{M}\big(I, (u), (v)\big)

, and the matrix asked for is its inverse.

Show solution

The identity sends $u_k$ to itself, so the columns of $C = \mathcal{M}\big(I, (u), (v)\big)$ are $u_1$ and $u_2$ in standard coordinates:

C = \begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix}.

By the theorem, $\mathcal{M}\big(I, (v), (u)\big) = C^{-1}$ . The candidate below works, since

\begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix}\begin{pmatrix} 3 & -1 \\ -2 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 3 & -1 \\ -2 & 1 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix}.

Check the columns directly: $3(1,2) - 2(1,3) = (1, 0)$ and $-(1,2) + (1,3) = (0, 1)$ . So

\mathcal{M}\big(I, (v), (u)\big) = \begin{pmatrix} 3 & -1 \\ -2 & 1 \end{pmatrix}.

Problem 2 · The Order of the Factors

Given: $T \in \mathcal{L}(\mathbf{F}^2)$ with $B = \mathcal{M}(T) = \begin{pmatrix} 2 & 1 \\ 1 & 0 \end{pmatrix}$ in the standard basis $(v)$ , and the basis $u_1 = (1, 0)$ , $u_2 = (1, 1)$ . Find $A = \mathcal{M}\big(T, (u)\big)$ .

✅ Correct!

Tu_1 = (2, 1) = u_1 + u_2

and

Tu_2 = (3, 1) = 2u_1 + u_2

, which are the two columns.

❌ Factors in the wrong order. This is

CBC^{-1}

. The formula is

A = C^{-1}BC

: the

C

that acts first, on the right, converts

u

-coordinates into

v

-coordinates.

❌ One step short. This is

BC = \mathcal{M}\big(T, (u), (v)\big)

: its columns are

Tu_1

Tu_2

in standard coordinates. Multiply by

C^{-1}

on the left to write them in the

u

-basis.

❌ Not quite. This is

C^{-1}B = \mathcal{M}\big(T, (v), (u)\big)

, which still takes standard coordinates as input. The factor

C

on the right is missing.

Show solution

The columns of $C = \mathcal{M}\big(I, (u), (v)\big)$ are $u_1$ , $u_2$ , and $C^{-1}$ can be checked by multiplying:

C = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}, \qquad C^{-1} = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}.

First $BC$ , whose columns are $Tu_1$ and $Tu_2$ in standard coordinates:

BC = \begin{pmatrix} 2 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 3 \\ 1 & 1 \end{pmatrix}.

Then

A = C^{-1}(BC) = \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 2 & 3 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix}.

Check by columns: $T(1,0) = (2,1) = 1\cdot(1,0) + 1\cdot(1,1)$ and $T(1,1) = (3,1) = 2\cdot(1,0) + 1\cdot(1,1)$ .

Problem 3 · The Inverse in Two Bases

Given: the basis $u_1 = (1, 1)$ , $u_2 = (0, 1)$ of $\mathbf{F}^2$ and an invertible $T \in \mathcal{L}(\mathbf{F}^2)$ with $\mathcal{M}\big(T, (u)\big) = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$ . Find the matrix of $T^{-1}$ in the basis $(u)$ and in the standard basis $(v)$ .

What is $\mathcal{M}\big(T^{-1}, (u)\big)$ ?

What is $\mathcal{M}\big(T^{-1}, (v)\big)$ ?

✅ Correct!

\mathcal{M}(T^{-1}) = \big(\mathcal{M}(T)\big)^{-1}

in each basis, and the two answers are related by

C\,(\cdot)\,C^{-1}

❌ Check the product.

\mathcal{M}\big(T^{-1}, (u)\big)

is the inverse of

\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}

, so multiplying the two in either order must give

I

❌ Not quite. Inverting a matrix is not taking reciprocals of entries:

\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & -1/2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 3/2 \\ 0 & 1 \end{pmatrix} \ne I

❌ Check the second part. Rearranging

A = C^{-1}BC

gives

B = CAC^{-1}

; apply it to

T^{-1}

❌ That is $\mathcal{M}(T, (v))$ . You changed basis correctly but used

T

instead of

T^{-1}

❌ The basis was not changed. That is

\mathcal{M}\big(T^{-1}, (u)\big)

again; the standard basis gives a different matrix.

❌ The formula is used backwards. This is

C^{-1}\,\mathcal{M}(T^{-1}, (u))\,C

. Since

\mathcal{M}(\cdot, (u)) = C^{-1}\,\mathcal{M}(\cdot, (v))\,C

, the standard-basis matrix is

C\,\mathcal{M}(\cdot, (u))\,C^{-1}

Show solution

Step 1. By $\mathcal{M}(T^{-1}) = \big(\mathcal{M}(T)\big)^{-1}$ in the basis $(u)$ :

\mathcal{M}\big(T^{-1}, (u)\big) = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}^{-1} = \begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}, \quad\text{since}\quad \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix} = I.

Step 2. The columns of $C = \mathcal{M}\big(I, (u), (v)\big)$ are $u_1$ , $u_2$ :

C = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}, \qquad C^{-1} = \begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix}.

Step 3. The change-of-basis formula for the operator $T^{-1}$ reads $\mathcal{M}(T^{-1}, (u)) = C^{-1}\,\mathcal{M}(T^{-1}, (v))\,C$ . Multiply by $C$ on the left and $C^{-1}$ on the right:

\mathcal{M}\big(T^{-1}, (v)\big) = C\begin{pmatrix} 1 & -2 \\ 0 & 1 \end{pmatrix}C^{-1} = \begin{pmatrix} 1 & -2 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} 3 & -2 \\ 2 & -1 \end{pmatrix}.

Check. In the standard basis, $\mathcal{M}(T) = C\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}C^{-1} = \begin{pmatrix} -1 & 2 \\ -2 & 3 \end{pmatrix}$ , and

\begin{pmatrix} -1 & 2 \\ -2 & 3 \end{pmatrix}\begin{pmatrix} 3 & -2 \\ 2 & -1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.

Problem 4 · Inverting the Formula

Given: an invertible $T \in \mathcal{L}(V)$ , bases $(u)$ and $(v)$ of $V$ , and $A = \mathcal{M}(T, (u))$ , $B = \mathcal{M}(T, (v))$ , $C = \mathcal{M}(I, (u), (v))$ , so that $A = C^{-1}BC$ . Which expression equals $A^{-1}$ ?

✅ Correct! Reverse the order and invert each factor:

(C^{-1}BC)^{-1} = C^{-1}B^{-1}(C^{-1})^{-1} = C^{-1}B^{-1}C

. It is the change-of-basis formula for

T^{-1}

❌ The order was not reversed. Inverting each factor in place gives

(C^{-1})^{-1}B^{-1}C^{-1} = CB^{-1}C^{-1}

, but

(XY)^{-1} = Y^{-1}X^{-1}

reverses the order.

❌ Not quite. The last factor is the inverse of

C^{-1}

, and

(C^{-1})^{-1} = C

, not

C^{-1}

❌ The $C$ 's do not cancel. Matrix multiplication is not commutative, so

C^{-1}

and

C

cannot meet. In Problem 2,

A^{-1} = \begin{pmatrix} -1 & 2 \\ 1 & -1 \end{pmatrix}

while

B^{-1} = \begin{pmatrix} 0 & 1 \\ 1 & -2 \end{pmatrix}

Show solution

Route 1: the product rule. $B = \mathcal{M}(T, (v))$ is invertible, with inverse $\mathcal{M}(T^{-1}, (v))$ . Applying $(XY)^{-1} = Y^{-1}X^{-1}$ twice to the three invertible factors gives

A^{-1} = (C^{-1}BC)^{-1} = C^{-1}B^{-1}(C^{-1})^{-1} = C^{-1}B^{-1}C.

Check: $(C^{-1}BC)(C^{-1}B^{-1}C) = C^{-1}B(CC^{-1})B^{-1}C = C^{-1}(BB^{-1})C = C^{-1}C = I$ , and the product in the other order also equals $I$ by the same regrouping.

Route 2: the formula for $T^{-1}$ . $A^{-1} = \mathcal{M}(T^{-1}, (u))$ and $B^{-1} = \mathcal{M}(T^{-1}, (v))$ . The change-of-basis formula for the operator $T^{-1}$ , with the same $C$ , gives $\mathcal{M}(T^{-1}, (u)) = C^{-1}\,\mathcal{M}(T^{-1}, (v))\,C$ , that is, $A^{-1} = C^{-1}B^{-1}C$ .

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