Chapter 2 · Linear Algebra¶
End-of-chapter exercises. Each has a definitions box, explanation, worked steps, and the answer.
2.1 · Custom operation \(a \star b = ab + a + b\) on \(\mathbb{R}\setminus\{-1\}\)¶
Topics & Definitions
- Group — a set with an operation obeying 4 axioms: closure, associativity, identity, inverses.
- Abelian group — a group whose operation is also commutative (\(a\star b = b\star a\)).
- Identity \(e\) — the element with \(a\star e = a\).
- Inverse — the element that undoes \(a\), returning \(e\).
The key move is factoring: \(a \star b = (a+1)(b+1) - 1\). So the operation behaves like ordinary multiplication shifted by \(1\), which makes every axiom easy. It is closed (never gives \(-1\)), associative, has identity \(0\), every element has an inverse, and it is commutative.
For part b, solve \(3 \star x \star x = 4(x+1)^2 - 1 = 15\).
Answer
\((\mathbb{R}\setminus\{-1\},\ \star)\) is an Abelian group. Identity \(e = 0\), inverse \(a^{-1} = \dfrac{-a}{a+1}\).
2.1b: \(x = 1\) or \(x = -3\).
The four group axioms, and what makes a group Abelian
A group is a set \(G\) with one operation \(\star\) that obeys four rules. Checking a group means checking these one by one; here every check rides on the factoring \(a\star b = (a+1)(b+1) - 1\), which turns \(\star\) into ordinary multiplication shifted by \(1\).
- Closure — combining any two members lands you back inside the set. For \(\star\): \((a+1)(b+1)-1\) equals \(-1\) only if \((a+1)(b+1)=0\), which cannot happen when \(a,b\neq -1\). So the result is never the forbidden \(-1\), and you stay in \(\mathbb{R}\setminus\{-1\}\).
- Associativity — regrouping does not change the result: \((a\star b)\star c = a\star(b\star c)\). This is inherited from ordinary multiplication of the shifted factors, which is associative.
- Identity — one special element \(e\) leaves everything unchanged: \(a\star e = a\) for all \(a\). Here \(e = 0\), since \((a+1)(0+1)-1 = a\).
- Inverses — every \(a\) has a partner \(a^{-1}\) that undoes it, returning the identity: \(a\star a^{-1} = e\). Here \(a^{-1} = \dfrac{-a}{a+1}\), and it is never \(-1\), so it lives in the set too.
Those four make \(G\) a group. It is also Abelian (commutative) if one bonus rule holds: \(a\star b = b\star a\) for all \(a,b\). Since \(\star\) is built from ordinary multiplication, and multiplication does not care about order, \(\star\) commutes, so this is an Abelian group. (A group can pass the four axioms yet fail this test, like the matrix group in 2.3, that is a non-Abelian group.)
2.2 · Integers mod \(n\) under \(\oplus\)¶
Topics & Definitions
- Congruence class \([a]\) — all integers with the same remainder as \(a\) mod \(n\).
- Modular addition — \([a] \oplus [b] = [a+b]\), then reduce mod \(n\).
- Prime condition — \((\mathbb{Z}_n\setminus\{0\}, \otimes)\) is a group only when \(n\) is prime.
\(\mathbb{Z}_n = \{[0],[1],\dots,[n-1]\}\) with \([a]\oplus[b] = [a+b]\) reduced mod \(n\). It is a group because ordinary integer addition already is, and the classes inherit that.
Answer
\((\mathbb{Z}_n, \oplus)\) is an Abelian group. Identity \([0]\), inverse of \([a]\) is \([n-a]\).
\((\mathbb{Z}_n\setminus\{0\}, \otimes)\) is a group iff \(n\) is prime.
2.3 · Upper-unitriangular \(3\times 3\) matrices under multiplication¶
Topics & Definitions
- Matrix group — a set of matrices closed under multiplication satisfying the 4 axioms.
- Upper unitriangular — \(1\)s on the diagonal, \(0\)s below, free entries above.
- Non-commutative — \(AB \neq BA\) in general, so a group can be non-Abelian.
\(G\) is all matrices \(\begin{bmatrix}1&x&z\\0&1&y\\0&0&1\end{bmatrix}\). Multiplying two keeps the shape (closed), multiplication is associative, the identity is in \(G\), and inverses stay in \(G\). But the \((1,3)\) entry of a product depends on order, so it is not Abelian.
Answer
\((G, \cdot)\) is a group, but not Abelian.
2.4 · Matrix products (check dimensions first!)¶
Topics & Definitions
- Dimension rule — \(AB\) is defined only if \(A\)'s columns \(=\) \(B\)'s rows: \((m\times n)(n\times p)\to(m\times p)\).
- Entry rule — entry \((i,j)\) = row \(i\) of \(A\) dotted with column \(j\) of \(B\).
- Non-commutative — compare b and c: same matrices swapped, different results.
Part a is a \(3\times 2\) times a \(3\times 3\): inner dimensions \(2\) and \(3\) do not match, so it is undefined.
Answer
a) undefined (dimension mismatch)
b) \(\begin{bmatrix}4&3&5\\10&9&11\\16&15&17\end{bmatrix}\) c) \(\begin{bmatrix}5&7&9\\11&13&15\\8&10&12\end{bmatrix}\)
d) \(\begin{bmatrix}14&6\\-21&2\end{bmatrix}\) e) \(\begin{bmatrix}12&3&-3&-12\\-3&1&2&6\\6&5&1&0\\13&12&3&2\end{bmatrix}\)
2.5 · Find all solutions of \(A\mathbf{x} = \mathbf{b}\)¶
Topics & Definitions
- Gaussian elimination — row operations to reach echelon form.
- Contradiction row — \(0 = \text{nonzero}\) means inconsistent, no solution.
- Free variable — a column with no pivot, giving infinitely many solutions.
Answer
2.5a: no solution, \(S = \varnothing\) (a contradiction row appears).
2.5b: infinitely many solutions, two free parameters.
2.6 · Solving \(A\mathbf{x} = \mathbf{b}\) with Gaussian elimination¶
Topics & Definitions
- RREF — pivots are \(1\), alone in their column, stepping down-right.
- Pivot vs free column — pivot columns give basic variables; no-pivot columns give free variables.
- General solution — one particular solution plus the free-variable directions.
Reducing \([A\mid \mathbf{b}]\) to RREF puts pivots in columns \(2, 4, 5\); columns \(1, 3, 6\) are free.
RREF of \([A \mid \mathbf{b}]\)
Pivots in columns \(2, 4, 5 \Rightarrow x_2, x_4, x_5\) basic; \(x_1, x_3, x_6\) free.
Answer
2.7 · Solving \(A\mathbf{x} = 12\mathbf{x}\)¶
Topics & Definitions
- Eigenvalue / eigenvector — \(A\mathbf{x} = \lambda\mathbf{x}\): a vector \(A\) only scales, never rotates; \(\lambda\) is the scale factor.
- Rearrangement — \(A\mathbf{x} = \lambda\mathbf{x} \Rightarrow (A - \lambda I)\mathbf{x} = 0\), a homogeneous system.
- Eigenspace — all eigenvectors for a given \(\lambda\) (a line/plane through the origin).
- Constraint — an extra equation like \(\sum x_i = 1\) picks one point on that line.
\((A - 12I)\mathbf{x} = 0\) elimination
Zero row \(\Rightarrow x_3\) free. Then \(x_2 = \tfrac32 x_3,\ x_1 = \tfrac32 x_3 \Rightarrow \mathbf{x} = t(3,3,2)\). The constraint \(3t + 3t + 2t = 1\) gives \(t = \tfrac18\).
Answer
2.8 · Inverses¶
Topics & Definitions
- Inverse \(A^{-1}\) — the matrix with \(A\,A^{-1} = I\).
- Determinant — a number measuring how much \(A\) scales space.
- Invertible iff \(\det \neq 0\) — \(\det = 0\) means \(A\) squashes a dimension, so no inverse.
- Gauss–Jordan — reduce \([A\mid I]\) until the left is \(I\); the right becomes \(A^{-1}\).
Part a. \(\det = 0\) (row 1 + row 3 = \(2\times\) row 2), so no inverse.
\([A \mid I] \to [I \mid A^{-1}]\)
Answer
a) not invertible (\(\det = 0\)).
b) \(A^{-1} = \begin{bmatrix}0&-1&0&1\\-1&0&0&1\\1&1&0&-1\\1&1&1&-2\end{bmatrix}\)
2.9 · Which sets are subspaces of \(\mathbb{R}^3\)?¶
Topics & Definitions
- Subspace — a subset containing \(\mathbf{0}\), closed under addition and scaling.
- The 3 tests — contains \(\mathbf{0}\); closed under \(+\); closed under scalar \(\times\).
- Span — all linear combinations of given vectors; always a subspace.
- The 3 killers — a square term, an integer-only condition, or \(=\) a nonzero constant.
- a. A span of two vectors \(\Rightarrow\) subspace.
- b. Has \(\lambda^2\) (never negative); scaling \((1,-1,0)\) by \(-1\) escapes \(\Rightarrow\) not.
- c. A plane; through the origin only when \(\gamma = 0\).
- d. Integer middle coordinate; scaling \((0,1,0)\) by \(0.5\) escapes \(\Rightarrow\) not.
Answer
a) subspace · b) not · c) subspace iff \(\gamma = 0\) · d) not
2.10 · Linear independence¶
Topics & Definitions
- Linearly independent — the only combination giving \(\mathbf{0}\) is all-zero coefficients.
- Linearly dependent — some vector is a combination of the others.
- Method — vectors as columns, row-reduce, count pivots.
- Rule — pivot in every column \(\Rightarrow\) independent; a missing column pivot \(\Rightarrow\) dependent.
Vectors as columns, then reduce
a) only 2 pivots, a zero row \(\Rightarrow\) dependent. b) 3 pivots (rank 3) \(\Rightarrow\) independent.
Answer
a) linearly dependent, with \(\mathbf{x}_3 = 2\mathbf{x}_1 - \mathbf{x}_2\).
b) linearly independent (rank \(3\)).
2.11 · Write \(\mathbf{y}\) as a linear combination¶
Topics & Definitions
- Linear combination — \(\lambda_1\mathbf{v}_1 + \lambda_2\mathbf{v}_2 + \lambda_3\mathbf{v}_3\): scale each vector, then add.
- Set-up — build \([\mathbf{x}_1\ \mathbf{x}_2\ \mathbf{x}_3 \mid \mathbf{y}]\), same as solving a system.
- Method — reduce to RREF; the coefficients appear in the final column.
\([\mathbf{x}_1\ \mathbf{x}_2\ \mathbf{x}_3 \mid \mathbf{y}] \to\) RREF
Check: \(-6(1) + 3(1) + 2(2) = 1\), matching \(\mathbf{y}\).
Answer
2.12 · Basis of \(U_1 \cap U_2\)¶
Topics & Definitions
- Intersection \(U_1 \cap U_2\) — all vectors that lie in both subspaces at once.
- Key idea — a vector is in the intersection when it is a combination of the \(\mathbf{u}\)'s and of the \(\mathbf{w}\)'s: \(a_1\mathbf{u}_1 + a_2\mathbf{u}_2 + a_3\mathbf{u}_3 = b_1\mathbf{w}_1 + b_2\mathbf{w}_2 + b_3\mathbf{w}_3\).
- Set-up — move everything to one side: \(a_1\mathbf{u}_1+a_2\mathbf{u}_2+a_3\mathbf{u}_3 - b_1\mathbf{w}_1 - b_2\mathbf{w}_2 - b_3\mathbf{w}_3 = \mathbf{0}\), a homogeneous system with columns \([\mathbf{u}_1\ \mathbf{u}_2\ \mathbf{u}_3 \mid -\mathbf{w}_1\ -\mathbf{w}_2\ -\mathbf{w}_3]\).
- Free variable — a column with no pivot; here \(b_3\), which parameterises the (one-dimensional) intersection.
Stack the six vectors as columns of a \(4\times 6\) matrix and row-reduce. Two \(\mathbf{u}\) columns and two \(\mathbf{w}\) columns take pivots, leaving \(b_3\) free. Back-substituting recovers the \(b\) coefficients, and feeding them into \(b_1\mathbf{w}_1+b_2\mathbf{w}_2+b_3\mathbf{w}_3\) rebuilds the shared vector.
Reduce, then rebuild the intersection vector
Row echelon form of \([\mathbf{u}_1\ \mathbf{u}_2\ \mathbf{u}_3 \mid -\mathbf{w}_1\ -\mathbf{w}_2\ -\mathbf{w}_3]\):
Column \(b_3\) has no pivot, so set \(b_3 = t\). Back-substitution gives
Choosing \(t = 2\) to clear fractions: \(b_1 = 0,\ b_2 = 7,\ b_3 = 2\).
Rebuild the intersection vector from the \(\mathbf{w}\) side:
Dividing by \(2\) gives the simplest representative \((4,-1,-2,-1)\).
Answer
\(U_1 \cap U_2\) is one-dimensional, with basis
2.13 · Subspaces as solution spaces \(U_1 = \ker A_1,\ U_2 = \ker A_2\)¶
Topics & Definitions
- Solution space (null space) — every \(\mathbf{x}\) with \(A\mathbf{x} = \mathbf{0}\); a subspace of the domain \(\mathbb{R}^3\).
- Rank–nullity — \(\dim(\ker A) = n - \operatorname{rank}(A)\), where \(n\) is the number of columns.
- Basis from free variables — each column without a pivot gives one basis vector.
- Mind the definition — here the \(U\)'s are solution spaces, so the dimension counts free variables, not pivots (contrast with 2.14).
Both \(A_1\) and \(A_2\) are \(4\times 3\), so \(\mathbf{x}\in\mathbb{R}^3\). Row-reducing each one gives rank \(2\), leaving a single free variable, so both solution spaces are one-dimensional. Strikingly, the two matrices reduce to the same echelon form, so \(U_1\) and \(U_2\) turn out to be the identical line.
Reduce each \(A_i\) and read the null space
Both give \(x_1 = -x_3\) and \(x_2 = -x_3\). Setting \(x_3 = t\):
Same for both, so \(U_1 = U_2\) and the intersection is that same line.
Answer
a) \(\dim U_1 = \dim U_2 = 1\).
b) Basis of \(U_1\) = basis of \(U_2\) = \(\left\{\,(-1,\,-1,\,1)\,\right\}\).
c) Since \(U_1 = U_2\), the intersection is the same line: \(\left\{\,(-1,\,-1,\,1)\,\right\}\).
2.14 · Subspaces as column spaces \(U_1 = \operatorname{col} A_1,\ U_2 = \operatorname{col} A_2\)¶
Topics & Definitions
- Column space — the span of a matrix's columns; here a subspace of \(\mathbb{R}^4\).
- Dimension \(=\) rank — the number of pivot columns.
- Basis \(=\) original pivot columns — reduce to find which columns are pivots, then take those columns from the original matrix (row ops change the columns, so never read the basis off the RREF).
- Intersection — solve \(c_1\mathbf{v}_1 + c_2\mathbf{v}_2 = d_1\mathbf{w}_1 + d_2\mathbf{w}_2\), a homogeneous system in \((c_1,c_2,d_1,d_2)\).
Same two matrices as 2.13, but now the \(U\)'s are spanned by the columns. Each has rank \(2\) with pivots in columns \(1\) and \(2\), so the first two columns of each original matrix form a basis. For the intersection, stack the two bases (negating \(U_2\)'s) and solve.
Intersection system \([\mathbf{v}_1\ \mathbf{v}_2 \mid -\mathbf{w}_1\ -\mathbf{w}_2]\)
Pivots in \(c_1, c_2, d_2\); the \(d_1\) column is free. Back-substitution gives \(d_2 = 0\), \(c_2 = d_1\), \(c_1 = 3d_1\). Taking \(d_1 = 1\):
which is exactly \(\mathbf{w}_1\), confirming it lies in both spaces. By Grassmann, \(\dim(U_1\cap U_2) = 2 + 2 - 3 = 1\).
Answer
a) \(\dim U_1 = \dim U_2 = 2\).
b) Basis of \(U_1 = \left\{\,(1,1,2,1),\ (0,-2,1,0)\,\right\}\); basis of \(U_2 = \left\{\,(3,1,7,3),\ (-3,2,-5,-1)\,\right\}\).
c) Basis of \(U_1 \cap U_2 = \left\{\,(3,\,1,\,7,\,3)\,\right\}\).
Why 2.13 and 2.14 differ (the thing to actually remember)
Same matrices \(A_1, A_2\), but the two problems ask about opposite subspaces, so both the dimension formula and the way you read off a basis flip:
| 2.13 · solution space | 2.14 · column space | |
|---|---|---|
| What \(U\) is | all \(\mathbf{x}\) with \(A\mathbf{x}=\mathbf{0}\) (\(\ker A\)) | the span of \(A\)'s columns (\(\operatorname{col} A\)) |
| Dimension | \(\dim = n - \operatorname{rank}(A)\) (free variables) | \(\dim = \operatorname{rank}(A)\) (pivots) |
| Basis from | solving the system, one vector per free variable | the original columns sitting in pivot positions |
| Here | \(3 - 2 = 1\) | \(2\) |
So the \(\dim = n - \operatorname{rank}\) formula is a null-space fact. It is easy to reach for it in 2.14, but a column space is measured by the rank directly. Rank–nullity ties them together: \(\operatorname{rank}(A) + \dim(\ker A) = n\), here \(2 + 1 = 3\).
The rank–nullity theorem, in plain words
Think of a matrix as a machine with one slot for each column. When you row-reduce it, every column ends up in one of two piles:
- Pivot columns are the ones pulling their own weight, each adds a genuinely new direction the matrix can reach. How many there are is the rank.
- Non-pivot columns are the leftovers, each one is just a mix of earlier columns and shows up as a free choice when you solve \(A\mathbf{x}=\mathbf{0}\). How many there are is the size of the solution space.
The theorem is simply the observation that every column lands in exactly one pile, so the two counts always add back up to the total number of columns:
For these \(4\times 3\) matrices: \(2\) pulling their weight \(+\ 1\) leftover \(=\ 3\) columns. Handy check: once you know the rank, the other number is free.
What \"basis = the pivot columns\" really means
Row operations change the columns, so you can never copy the basis straight out of the RREF. What row reduction does preserve is the dependency pattern, telling you which columns are independent. So: reduce to spot the pivot positions (here columns \(1\) and \(2\)), then go back and take those columns from the original matrix. For \(A_1\) that is \((1,1,2,1)\) and \((0,-2,1,0)\), the untouched first two columns.
2.15 · Subspaces \(F, G \subseteq \mathbb{R}^3\) and their intersection¶
Topics & Definitions
- Subspace — a subset that contains \(\mathbf{0}\) and is closed under addition and scalar multiplication.
- Intersection \(F \cap G\) — all vectors lying in both subspaces at once.
- Basis of a plane — solve the defining equation for one variable, then split the free variables into separate vectors.
- Two routes to \(F \cap G\) — substitute \(G\)'s general form into \(F\)'s equation (part b), or set a combination of \(F\)'s basis equal to a combination of \(G\)'s basis and solve (part c). Both give the same line.
\(F = \{(x,y,z)\in\mathbb{R}^3 \mid x+y-z = 0\}\) is a plane through the origin. \(G = \{(a-b,\ a+b,\ a-3b)\mid a,b\in\mathbb{R}\}\) is the span of two vectors. Both pass the three subspace tests: they contain the zero vector, and they are closed under addition and scaling (each condition is inherited from the linear form of the definition).
Part a (subspaces). For \(F\): \(0+0-0 = 0\) so \(\mathbf{0}\in F\); adding two solutions of \(x+y-z=0\) gives another solution; scaling by \(c\) gives \(c(x+y-z) = 0\). For \(G\): choosing \(a=b=0\) gives \(\mathbf{0}\in G\); sums and scalar multiples keep the same \((a-b,\ a+b,\ a-3b)\) shape. So both are subspaces of \(\mathbb{R}^3\).
Part b, intersection without a basis
A general vector in \(G\) is \((a-b,\ a+b,\ a-3b)\). For it to lie in \(F\) it must satisfy \(x+y-z=0\):
Substituting \(a=-3b\) back into \(G\)'s coordinates:
So \(F\cap G = \{(-4b,-2b,-6b)\mid b\in\mathbb{R}\} = \{k(2,1,3)\mid k\in\mathbb{R}\}\).
Part c, intersection using bases
Basis for \(F\): rewrite \(x+y-z=0\) as \(z=x+y\), so \((x,y,x+y) = x(1,0,1)+y(0,1,1)\), giving \(B_F = \{(1,0,1),(0,1,1)\}\). Basis for \(G\): separate the parameters, \(a(1,1,1)+b(-1,1,-3)\), giving \(B_G = \{(1,1,1),(-1,1,-3)\}\). Set a combination of \(B_F\) equal to a combination of \(B_G\):
Matching coordinates gives \(c_1 = d_1-d_2\), \(c_2 = d_1+d_2\), \(c_1+c_2 = d_1-3d_2\). Substituting the first two into the third: \(2d_1 = d_1 - 3d_2 \Rightarrow d_1 = -3d_2\). Let \(d_2 = t\), so \(d_1 = -3t\):
This matches part b exactly.
Answer
Both \(F\) and \(G\) are subspaces of \(\mathbb{R}^3\).
(a one-dimensional line, obtained identically with or without bases).
2.16 · Are the following mappings linear?¶
Topics & Definitions
- Linear mapping — a map \(\Phi\) that passes two tests for all inputs \(f,g\) and scalars \(c\).
- Additivity (plus test) — \(\Phi(f+g) = \Phi(f) + \Phi(g)\).
- Homogeneity (scalar test) — \(\Phi(c\cdot f) = c\cdot\Phi(f)\).
- To disprove linearity — a single counterexample that breaks either test is enough.
A map is linear exactly when it respects addition and scaling. Integration and differentiation both do (they split across plus signs and let constants slide out), so parts a and b are linear. Cosine does not (it bends its input), so part c fails. Any fixed-matrix map is automatically linear, so parts d and e pass, including the rotation matrix in e where the sines and cosines are just constants.
Parts a & b, integration and differentiation
a) Integration \(\Phi(f) = \int_a^b f(x)\,dx\). Additivity: \(\int_a^b (f+g)\,dx = \int_a^b f\,dx + \int_a^b g\,dx = \Phi(f)+\Phi(g)\). Homogeneity: \(\int_a^b c\,f\,dx = c\int_a^b f\,dx = c\,\Phi(f)\). Linear.
b) Differentiation \(\Phi(f) = f'\). Additivity: \((f+g)' = f' + g' = \Phi(f)+\Phi(g)\). Homogeneity: \((c\,f)' = c\,f' = c\,\Phi(f)\). Linear.
Part c, cosine (counterexample)
\(\Phi(x) = \cos(x)\). Test additivity: \(\Phi(x+y) = \cos(x+y)\), but \(\Phi(x)+\Phi(y) = \cos(x)+\cos(y)\), and \(\cos(x+y)\neq\cos(x)+\cos(y)\) in general. Concrete counterexample with \(x=y=0\): \(\Phi(0+0) = \cos 0 = 1\), while \(\Phi(0)+\Phi(0) = 1+1 = 2\). Since \(1\neq 2\), it is not linear.
Parts d & e, matrix maps
d) \(\Phi(\mathbf{x}) = A\mathbf{x}\) with \(A = \begin{bmatrix}1&2&3\\1&4&3\end{bmatrix}\). Additivity: \(A(\mathbf{x}+\mathbf{y}) = A\mathbf{x}+A\mathbf{y}\). Homogeneity: \(A(c\mathbf{x}) = c(A\mathbf{x})\). Linear.
e) \(\Phi(\mathbf{x}) = A\mathbf{x}\) with \(A = \begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}\) for fixed \(\theta\). The entries are constants, so this is still just matrix multiplication, and the same two properties hold. Linear (this is a rotation matrix).
Answer
- a) linear (integration)
- b) linear (differentiation)
- c) not linear (\(\cos(x+y)\neq\cos x+\cos y\); e.g. \(x=y=0\) gives \(1\neq 2\))
- d) linear (matrix map)
- e) linear (rotation matrix)
Linearity: two checkpoints worth memorizing
Turn the formal definition into two everyday tests:
- The Plus Test — does splitting the input across a \(+\) survive? \(\Phi(f+g) = \Phi(f)+\Phi(g)\).
- The Multiplier Test — can a constant slide out to the front? \(\Phi(cf) = c\,\Phi(f)\).
The real trick is where the variable sits. If it is trapped inside a curvy function like \(\cos\), the map bends the input and both tests fail. If the variable only rides on the outside (a derivative, an integral, or multiplication by a fixed matrix) the tests pass. A rotation matrix still passes, because its sines and cosines are baked-in constants, not the input.
Two things that trip people up here: the notation \(L^1, C^1, C^0\) is just naming the space of functions in play (integrable, once-differentiable, continuous), so "\(\Phi:C^1\to C^0,\ \Phi(f)=f'\)" is simply differentiation. And integration and differentiation are themselves linear operators: a constant multiple slides straight through, which is a separate idea from the mechanical power-rule arithmetic you use to actually evaluate an integral.
2.17 · Transformation matrix, rank, kernel and image of \(\Phi:\mathbb{R}^3\to\mathbb{R}^4\)¶
Topics & Definitions
- Transformation matrix \(A_\Phi\) — read the coefficients of \(x_1,x_2,x_3\) row by row; column \(j\) holds the coefficients of \(x_j\).
- Rank — number of pivots after row reduction.
- Kernel \(\ker(\Phi)\) — all inputs sent to \(\mathbf{0}\); \(\dim(\ker) = (\text{columns}) - \operatorname{rank}\).
- Image \(\operatorname{Im}(\Phi)\) — all reachable outputs; \(\dim(\operatorname{Im}) = \operatorname{rank}\), spanned by the original pivot columns.
Reading the four output rows gives the coefficient matrix \(A_\Phi\). Row-reducing it produces a pivot in all three columns, so the rank is \(3\). By rank–nullity the kernel is trivial (\(3-3 = 0\)), meaning only the zero vector maps to zero, and the image is three-dimensional, spanned by the three original columns.
Build \(A_\Phi\) and row-reduce
Pull the coefficients out of \(\Phi(x_1,x_2,x_3) = (3x_1+2x_2+x_3,\ x_1+x_2+x_3,\ x_1-3x_2,\ 2x_1+3x_2+x_3)\):
Swap to get a leading \(1\), then clear below:
Three pivots, so \(\operatorname{rk}(A_\Phi) = 3\).
Answer
\(\ker(\Phi) = \{\mathbf{0}\}\), so \(\dim(\ker(\Phi)) = 0\).
\(\dim(\operatorname{Im}(\Phi)) = 3\), with
Transformation matrix, kernel, image: what they actually are
- Transformation matrix — the map's action machine: feed in \(\mathbf{x}\), get \(\Phi(\mathbf{x})\). It is not an inverse or "undo" matrix. You build it by pulling each variable's coefficients into a column, one output row at a time.
- Kernel — the "drops to zero" club: the inputs the map sends to \(\mathbf{0}\). Full rank means only \(\mathbf{0}\) qualifies, so \(\dim(\ker)=0\). Rank–nullity hands it to you for free: \((\text{columns}) - \operatorname{rank}\).
- Image — the full list of outputs the machine can actually produce. Its dimension always equals the rank, and it is spanned by the original pivot columns.
2.18 · Automorphisms: kernels and images of \(f\) and \(g\)¶
Topics & Definitions
- Homomorphism — a structure-preserving map between vector spaces; equivalently, a linear map (it preserves addition and scalar multiplication).
- Endomorphism — a homomorphism from a space to itself, \(\Phi: V \to V\).
- Automorphism — a bijective endomorphism: an invertible linear map \(V \to V\) whose inverse is also linear.
- Kernel — \(\ker(f) = \{v : f(v) = \mathbf{0}\}\). Image — \(\operatorname{Im}(g) = \{g(v) : v \in V\}\).
- Key fact used here — \(f\) and \(g\) are automorphisms with \(f \circ g = \mathrm{id}\), so both are bijections; a bijection sends only \(\mathbf{0}\) to \(\mathbf{0}\) and is onto.
Because \(f \circ g = \mathrm{id}_E\), both \(f\) and \(g\) are invertible (automorphisms). Bijectivity is what makes all three claims fall out: an injective map has trivial kernel, and a surjective map has full image.
The three claims
1. \(\ker(f) = \ker(g \circ f)\). If \(f(v) = \mathbf{0}\) then \((g\circ f)(v) = g(\mathbf{0}) = \mathbf{0}\), so \(\ker(f) \subseteq \ker(g\circ f)\). Conversely, if \((g\circ f)(v) = \mathbf{0}\), apply \(g^{-1}\): \(f(v) = g^{-1}(\mathbf{0}) = \mathbf{0}\), so \(\ker(g\circ f) \subseteq \ker(f)\). Hence equal.
2. \(\operatorname{Im}(g) = \operatorname{Im}(g \circ f)\). Clearly \(\operatorname{Im}(g\circ f) \subseteq \operatorname{Im}(g)\). For the reverse, take any \(g(w) \in \operatorname{Im}(g)\). Since \(f\) is surjective, \(w = f(u)\) for some \(u\), so \(g(w) = (g\circ f)(u) \in \operatorname{Im}(g\circ f)\). Hence equal.
3. \(\ker(f) \cap \operatorname{Im}(g) = \{\mathbf{0}\}\). \(f\) is an automorphism, so \(\ker(f) = \{\mathbf{0}\}\). The intersection of \(\{\mathbf{0}\}\) with any set is \(\{\mathbf{0}\}\).
Answer
All three hold: \(\ker(f) = \ker(g\circ f)\), \(\operatorname{Im}(g) = \operatorname{Im}(g\circ f)\), and \(\ker(f)\cap\operatorname{Im}(g) = \{\mathbf{0}\}\), each following from \(f\) and \(g\) being automorphisms (bijective linear maps).
2.19 · Endomorphism on \(\mathbb{R}^3\): kernel, image, and a basis change¶
Topics & Definitions
- Endomorphism — a linear map from a space to itself, here \(\Phi:\mathbb{R}^3\to\mathbb{R}^3\).
- Kernel / Image — inputs sent to \(\mathbf{0}\) / all reachable outputs.
- Invertibility test — \(\det A_\Phi \neq 0\) means full rank, trivial kernel, and image equal to all of \(\mathbb{R}^3\).
- Basis change of a map — \(\tilde{A}_\Phi = B^{-1} A_\Phi B\), where \(B\) has the new basis vectors as its columns.
The transformation matrix is \(A_\Phi = \begin{bmatrix}1&1&0\\1&-1&0\\1&1&1\end{bmatrix}\). Its determinant is \(-2\), which is nonzero, so \(\Phi\) is invertible: the kernel is just the zero vector and the image is all of \(\mathbb{R}^3\). Part b re-expresses the same map in the new basis \(B\) using the similarity formula \(B^{-1}A_\Phi B\).
Part a, kernel and image
\(\det A_\Phi = -2 \neq 0\), so \(\operatorname{rk}(A_\Phi) = 3\).
Part b, basis change to \(B\)
Put the new basis vectors as columns: \(B = \begin{bmatrix}1&1&1\\1&2&0\\1&1&0\end{bmatrix}\). Then
Answer
a) \(\ker(\Phi) = \{\mathbf{0}\}\) (dim 0), \(\operatorname{Im}(\Phi) = \mathbb{R}^3\) (dim 3).
b) \(\displaystyle \tilde{A}_\Phi = \begin{bmatrix}6&9&1\\-3&-5&0\\-1&-1&0\end{bmatrix}.\)
2.20 · Basis changes and a homomorphism \(\mathbb{R}^2 \to \mathbb{R}^3\)¶
Topics & Definitions
- Ordered basis — a basis with a fixed order, so coordinates are well defined.
- Change-of-basis matrix — converts a coordinate vector from one basis to another; its columns are the old basis vectors written in the new basis.
- Transformation matrix w.r.t. bases \(B, C\) — its \(j\)-th column is \(\Phi(b_j)\) written in coordinates of \(C\).
- Similarity for different in/out bases — if input coords change by \(P_1\) and output coords by \(P_2\), then \(A' = P_2\, A_\Phi\, P_1\).
- Basis test — a set of \(n\) vectors in \(\mathbb{R}^n\) is a basis iff the matrix with them as columns has nonzero determinant.
The vectors \(b_1,b_2\) and \(b_1',b_2'\) each form a basis of \(\mathbb{R}^2\) (nonzero determinants). \(P_1\) rewrites \(B'\)-coordinates as \(B\)-coordinates. The three \(c\) vectors form a basis of \(\mathbb{R}^3\), and \(P_2\) converts \(C\)-coordinates to the standard basis. The homomorphism \(\Phi\) is pinned down by its action on \(b_1+b_2\) and \(b_1-b_2\); solving recovers \(\Phi(b_1)\) and \(\Phi(b_2)\), which build \(A_\Phi\). Combining the change-of-basis matrices gives \(A'\), and part f chases one vector through both routes as a check.
Parts a & b, bases and \(P_1\)
\(\det[b_1\ b_2] = \det\begin{bmatrix}2&-1\\1&-1\end{bmatrix} = -1 \neq 0\), and \(\det[b_1'\ b_2'] = \det\begin{bmatrix}2&1\\-2&1\end{bmatrix} = 4 \neq 0\). Both are bases. \(P_1\) changes coordinates from \(B'\) to \(B\): \([x]_B = P_1 [x]_{B'}\), so \(P_1 = [b_1\ b_2]^{-1}[b_1'\ b_2']\):
Part c, the basis \(C\) and \(P_2\)
\(C = [c_1\ c_2\ c_3] = \begin{bmatrix}1&0&1\\2&-1&0\\-1&2&-1\end{bmatrix}\). (i) \(\det C = 4 \neq 0\), so \(C\) is a basis of \(\mathbb{R}^3\). (ii) \(C'\) is the standard basis. Since \(v = C[v]_C = [v]_{C'}\), the change from \(C\) to \(C'\) is just \(P_2 = C\).
Part d, transformation matrix \(A_\Phi\) w.r.t. \(B\) and \(C\)
Given \(\Phi(b_1+b_2) = c_2 + c_3\) and \(\Phi(b_1-b_2) = 2c_1 - c_2 + 3c_3\). In \(C\)-coordinates these are \((0,1,1)\) and \((2,-1,3)\). Solving:
The columns of \(A_\Phi\) are these coordinate vectors:
Part e, \(A'\) w.r.t. \(B'\) and \(C'\)
Input coords change by \(P_1\) (\(B' \to B\)) and output coords by \(P_2\) (\(C \to C'\)), so
Part f, chasing \(x\) with coords \([2,3]^\top\) in \(B'\)
(i) \([x]_B = P_1[x]_{B'} = \begin{bmatrix}4&0\\6&-1\end{bmatrix}\begin{bmatrix}2\\3\end{bmatrix} = (8,9)\). (ii) \([\Phi(x)]_C = A_\Phi[x]_B = (-1,9,7)\). (iii) In the standard basis, \(\Phi(x) = C[\Phi(x)]_C = (6,-11,12)\). (iv) Directly, \(A'[x]_{B'} = (6,-11,12)\), which matches (iii). Check confirmed.
Answer
a) Both \(B\) and \(B'\) are bases (\(\det = -1\) and \(4\)).
b) \(P_1 = \begin{bmatrix}4&0\\6&-1\end{bmatrix}\).
c) \(C\) is a basis (\(\det C = 4\)); \(P_2 = C = \begin{bmatrix}1&0&1\\2&-1&0\\-1&2&-1\end{bmatrix}\).
d) \(A_\Phi = \begin{bmatrix}1&-1\\0&1\\2&-1\end{bmatrix}\).
e) \(A' = \begin{bmatrix}0&2\\-10&3\\12&-4\end{bmatrix}\).
f) \([x]_B = (8,9)\); \([\Phi(x)]_C = (-1,9,7)\); \(\Phi(x) = (6,-11,12)\) in the standard basis, confirmed both ways.
The standard basis is secretly the identity
When a question asks for coordinates or a matrix "relative to the standard basis \(C'\)" without ever defining \(C'\), that is because the standard basis change-of-basis matrix is just the identity \(I = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\). So any \(I^{-1}\) hiding in the chain collapses to \(I\), and there is no matrix to invert at that step. Ordinary coordinates are the standard-basis coordinates.
Reading a change-of-basis chain (right to left)
Compose from the inside out: \(A' = P_2\, A_\Phi\, P_1\) reads as "convert the input basis with \(P_1\), apply the map with \(A_\Phi\), then convert the output basis with \(P_2\)." The inner dimensions dovetail, \((3\times 3)(3\times 2)(2\times 2)\), which is the fast sanity check that the pieces are in the right order.
- The rule \((\text{destination})^{-1}(\text{start})\) still holds; it just vanishes when the destination is the standard basis, since that inverse is \(I\).
- Always verify at the end. Computing \(A'\,[x]_{B'}\) directly should land on the exact same vector as the long route (change basis, apply the map, change basis back). If they disagree, re-check the arithmetic one entry at a time, because a single sign slip in a dot product, for example \((-4)(2)+(3)(3) = +1\), not \(-2\), shifts the whole output vector even when the method is flawless.
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