SME
Linear Algebra
Linear transformations, kernel and range, change of basis
Use this for “T has matrix A = [[4, 1], [2, 3]]. Find its matrix in the basis v₁ = (1, −2), v₂ = (1, 1).”
Diagram
P⁻¹ is 2 × 2 and AP is 2 × 2, so B is 2 × 2.
Row 1 of P⁻¹ times column 1 of AP gives entry (1, 1): b₁₁ = 1/3 × 2 + (−1/3) × (−4) = 2.
Tap an entry of B to see its row and column.
This is an example. Type your own values to start a new problem.
A
=
,
P
=
B
=
P⁻¹AP
=
det P
Determinant of P
worked out (not typed)
c₁₁
Product AP, row 1, column 1
worked out (not typed)
c₁₂
Product AP, row 1, column 2
worked out (not typed)
c₂₁
Product AP, row 2, column 1
worked out (not typed)
c₂₂
Product AP, row 2, column 2
worked out (not typed)
q₁₁
Inverse P⁻¹, row 1, column 1
worked out (not typed)
q₁₂
Inverse P⁻¹, row 1, column 2
worked out (not typed)
q₂₁
Inverse P⁻¹, row 2, column 1
worked out (not typed)
q₂₂
Inverse P⁻¹, row 2, column 2
worked out (not typed)
Formulas
det
P
=
p₁₁
×
p₂₂
−
p₁₂
×
p₂₁
3 = 1 × 1 − 1 × (−2)
c₁₁
=
a₁₁
×
p₁₁
+
a₁₂
×
p₂₁
2 = 4 × 1 + 1 × (−2)
c₁₂
=
a₁₁
×
p₁₂
+
a₁₂
×
p₂₂
5 = 4 × 1 + 1 × 1
c₂₁
=
a₂₁
×
p₁₁
+
a₂₂
×
p₂₁
−4 = 2 × 1 + 3 × (−2)
c₂₂
=
a₂₁
×
p₁₂
+
a₂₂
×
p₂₂
5 = 2 × 1 + 3 × 1
q₁₁
=
p₂₂
det P
1
3
=
1
3
q₁₂
=
−
p₁₂
det P
−
1
3
=
−
1
3
q₂₁
=
−
p₂₁
det P
2
3
=
−
−2
3
q₂₂
=
p₁₁
det P
1
3
=
1
3
b₁₁
=
q₁₁
×
c₁₁
+
q₁₂
×
c₂₁
2
=
1
3
×
2
+
(−
1
3
)
×
(−4)
b₁₂
=
q₁₁
×
c₁₂
+
q₁₂
×
c₂₂
0
=
1
3
×
5
+
(−
1
3
)
×
5
b₂₁
=
q₂₁
×
c₁₁
+
q₂₂
×
c₂₁
0
=
2
3
×
2
+
1
3
×
(−4)
b₂₂
=
q₂₁
×
c₁₂
+
q₂₂
×
c₂₂
5
=
2
3
×
5
+
1
3
×
5
Assumptions
•
P has the new basis vectors as its columns: P turns new coordinates into standard ones, and P⁻¹ turns them back.
•
B = P⁻¹AP: change to standard coordinates, apply A, change back. Column j of B is T(vⱼ) in the new basis.
•
B and A are similar: the same trace, det and eigenvalues. When both columns of P are eigenvectors, B is diagonal.
Step-by-step
WE KNOW
a₁₁ = 4, a₁₂ = 1, a₂₁ = 2, a₂₂ = 3, p₁₁ = 1, p₁₂ = 1, p₂₁ = −2, p₂₂ = 1
FIND
determinant of P (det P), product AP, row 1, column 1 (c₁₁), product AP, row 1, column 2 (c₁₂), product AP, row 2, column 1 (c₂₁), product AP, row 2, column 2 (c₂₂), inverse P⁻¹, row 1, column 1 (q₁₁), inverse P⁻¹, row 1, column 2 (q₁₂), inverse P⁻¹, row 2, column 1 (q₂₁), inverse P⁻¹, row 2, column 2 (q₂₂), new matrix B of T, row 1, column 1 (b₁₁), new matrix B of T, row 1, column 2 (b₁₂), new matrix B of T, row 2, column 1 (b₂₁), new matrix B of T, row 2, column 2 (b₂₂)
Step 1 · Find determinant of P (det P)
Use det P = p₁₁ × p₂₂ − p₁₂ × p₂₁
Down the main diagonal, take away the other diagonal. det P ≠ 0, so the columns are a basis.
det P = 1 × 1 − 1 × (−2)
det P = 3
Step 2 · Find product AP, row 1, column 1 (c₁₁)
Use c₁₁ = a₁₁ × p₁₁ + a₁₂ × p₂₁
Apply A to each column of P first: entry (1, 1) of AP is row 1 of A times column 1 of P.
c₁₁
=
4
×
1
+
1
×
(−2)
[[4, 1], [2, 3]] × [[1, 1], [−2, 1]] = [[4 × 1 + 1 × (−2), 4 × 1 + 1 × 1], [2 × 1 + 3 × (−2), 2 × 1 + 3 × 1]] = [[2, 5], [−4, 5]]
c₁₁
=
2
Step 3 · Find product AP, row 1, column 2 (c₁₂)
Use c₁₂ = a₁₁ × p₁₂ + a₁₂ × p₂₂
Row 1 of A times column 2 of P.
c₁₂
=
4
×
1
+
1
×
1
c₁₂
=
5
Step 4 · Find product AP, row 2, column 1 (c₂₁)
Use c₂₁ = a₂₁ × p₁₁ + a₂₂ × p₂₁
Row 2 of A times column 1 of P.
c₂₁
=
2
×
1
+
3
×
(−2)
c₂₁
=
2
+
(−6)
c₂₁
=
−4
Step 5 · Find product AP, row 2, column 2 (c₂₂)
Use c₂₂ = a₂₁ × p₁₂ + a₂₂ × p₂₂
Row 2 of A times column 2 of P.
c₂₂
=
2
×
1
+
3
×
1
c₂₂
=
5
Step 6 · Find inverse P⁻¹, row 1, column 1 (q₁₁)
Use q₁₁ = p₂₂ ÷ (det P)
The 2 × 2 rule: swap the two diagonal entries, change the sign of the other two, and divide each by det P.
q₁₁
=
1
3
q₁₁
=
1
3
Step 7 · Find inverse P⁻¹, row 1, column 2 (q₁₂)
Use q₁₂ = −p₁₂ ÷ (det P)
An entry off the diagonal keeps its place, changes sign, and is divided by det P.
q₁₂
=
−
1
3
q₁₂
=
−
1
3
Step 8 · Find inverse P⁻¹, row 2, column 1 (q₂₁)
Use q₂₁ = −p₂₁ ÷ (det P)
An entry off the diagonal keeps its place, changes sign, and is divided by det P.
q₂₁
=
−
−2
3
q₂₁
=
2
3
Step 9 · Find inverse P⁻¹, row 2, column 2 (q₂₂)
Use q₂₂ = p₁₁ ÷ (det P)
The other diagonal entry, moved to the opposite corner, over det P.
q₂₂
=
1
3
q₂₂
=
1
3
Step 10 · Find new matrix B of T, row 1, column 1 (b₁₁)
Use b₁₁ = q₁₁ × c₁₁ + q₁₂ × c₂₁
B
=
P
-1
(AP):
row
1
of
P
-1
times
column
1
of
AP.
Column
1
of
B
is
T(v₁)
in
the
new
coordinates.
b₁₁
=
1
3
×
2
+
(−
1
3
)
×
(−4)
[[
1
3
,
−
1
3
],
[
2
3
,
1
3
]]
×
[[2,
5],
[−4,
5]]
=
[[
1
3
×
2
+
(−
1
3
)
×
(−4),
1
3
×
5
+
(−
1
3
)
×
5],
[
2
3
×
2
+
1
3
×
(−4),
2
3
×
5
+
1
3
×
5]]
=
[[2,
0],
[0,
5]]
b₁₁
=
2
Step 11 · Find new matrix B of T, row 1, column 2 (b₁₂)
Use b₁₂ = q₁₁ × c₁₂ + q₁₂ × c₂₂
Row
1
of
P
-1
times
column
2
of
AP.
b₁₂
=
1
3
×
5
+
(−
1
3
)
×
5
b₁₂
=
1
2
3
+
(−
1
2
3
)
b₁₂
=
0
Step 12 · Find new matrix B of T, row 2, column 1 (b₂₁)
Use b₂₁ = q₂₁ × c₁₁ + q₂₂ × c₂₁
Row
2
of
P
-1
times
column
1
of
AP.
b₂₁
=
2
3
×
2
+
1
3
×
(−4)
b₂₁
=
1
1
3
+
(−
1
1
3
)
b₂₁
=
0
Step 13 · Find new matrix B of T, row 2, column 2 (b₂₂)
Use b₂₂ = q₂₁ × c₁₂ + q₂₂ × c₂₂
Row
2
of
P
-1
times
column
2
of
AP.
B
is
diagonal:
each
column
of
P
is
an
eigenvector
of
A,
its
eigenvalue
on
the
diagonal.
b₂₂
=
2
3
×
5
+
1
3
×
5
tr B = 2 + 5 = 7, tr A = 4 + 3 = 7
det B = 2 × 5 − 0 × 0 = 10, det A = 4 × 3 − 1 × 2 = 10
b₂₂
=
5
Check
Put the numbers back in. Both sides should match.
3 = 1 × 1 − 1 × (−2) ✓
2 = 4 × 1 + 1 × (−2) ✓
5 = 4 × 1 + 1 × 1 ✓
−4 = 2 × 1 + 3 × (−2) ✓
5 = 2 × 1 + 3 × 1 ✓
1
3
=
1
3
✓
−
1
3
=
−
1
3
✓
2
3
=
−
−2
3
✓
1
3
=
1
3
✓
2
=
1
3
×
2
+
(−
1
3
)
×
(−4)
✓
0
=
1
3
×
5
+
(−
1
3
)
×
5
✓
0
=
2
3
×
2
+
1
3
×
(−4)
✓
5
=
2
3
×
5
+
1
3
×
5
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
Kernel and range of a transformation
Use this for “T(x) = Ax with A = [[1, 2, −1], [2, 4, 1], [3, 6, 0]]. Find the kernel and the range of T. Is T one-to-one? Onto?”
A rotation then a reflection
Use this for “Rotate (3, 1) by 90° and then reflect it in the x-axis. Find the image and the one matrix that does both.”