Vectors, linear maps and matrices: the language of almost everything.
11 chapters written · 69 definitions · 97 theorems · builds on Unit 0
The story
Linear algebra studies one kind of function, the linear map, which respects addition and scaling. That sounds narrow, but it is the only kind of function we can understand completely, and calculus works by approximating every other function with one.
The unit builds the setting first (2.1 and 2.2: vector spaces, span, independence, basis, dimension), then the maps (2.3 to 2.5: null space, range, rank-nullity, inverses, matrices as maps). With that in place, elimination (2.6) and determinants (2.7) become tools with reasons behind them, not recipes. Eigenvalues (2.8) find the directions a map only stretches.
The last three chapters add length and angle (2.9: inner products and projection) and put everything to work: least squares and regression (2.10), and Markov chains and PageRank (2.11).
Every theorem in the unit, in the order it is proved. Each one is used by the ones after it. The label says whether it is on the 375/376 exams (T), an Apostol extra (A), or beyond both (B).
Theorem · TExtending to a basis In an n-dimensional space, any independent set can be extended to a basis, and any n independent vectors already form a basis.
Theorem · TOne-to-one in finite dimension For a linear T on an n-dimensional V, these are equivalent: T is one-to-one; independent sets stay independent; dimT(V)=n; bases go to bases.
Theorem · TDiagonal form of a linear map For T:V→W of rank r there are bases of V and W in which the matrix has 1s in the first r diagonal places and 0 everywhere else.
Theorem · TReading the solutions off the reduced form The system is inconsistent exactly when the last column is a pivot column. Otherwise each pivot variable equals its right side minus a combination of the free variables.
Theorem · TRank equals the number of pivots If an echelon form of A has r pivots, then rank A=r, nullity =n−r, and the pivot columns of the original A are a basis of the column space.
Theorem · TRow rank equals column rank The row space and the column space of a matrix have the same dimension, the number of pivots. So rank A = rank AT.
Theorem · BInvertible matrix theorem For an n×n matrix A, eleven conditions are equivalent: invertible, one-to-one, null space zero, independent columns, rank n, columns span Rn, onto, a unique solution for every b, n pivots, product of elementary matrices, invertible transpose.
Lemma · TWhat the first three axioms force on the rows A function with Axioms 1 to 3 is zero if a row is zero, changes sign when two rows are swapped, is unchanged when a multiple of one row is added to another, and is zero when the rows are dependent.
Theorem · TDeterminant by row reduction If row operations turn A into an upper triangular U using p swaps and scalings by c1,…,cq, then detA=(−1)p(c1⋯cq)−1u11⋯unn.
Theorem · TDiagonal form means an eigenbasis A map has a diagonal matrix in a basis exactly when every vector of that basis is an eigenvector. The diagonal entries are the eigenvalues.
Theorem · TDiagonalization as a factorization If A has n independent eigenvectors, then A=CDC−1, where the columns of C are those eigenvectors and D is diagonal with the matching eigenvalues.
Theorem · BQR from Gram–Schmidt A matrix A with independent columns factors as A=QR: Q has orthonormal columns and R is square, upper triangular and invertible. Then x^ solves Rx^=QTb and P=QQT.
Theorem · BConvergence for a diagonalizable chain with a simple eigenvalue 1 Let P be stochastic and diagonalizable. If 1 is an eigenvalue with a one-dimensional eigenspace and every other eigenvalue has size less than 1, then Pkx0 approaches the same stationary distribution π from every starting probability vector x0.
Theorem · BPerron–Frobenius theorem for regular chains If some power of the stochastic matrix P has all entries strictly positive (a regular chain), then 1 is a simple eigenvalue with a positive eigenvector, every other eigenvalue has size less than 1, and Pkx0→π for every probability vector x0.
Theorem · BThe Google matrix has a unique stationary distribution For 0≤d<1, G has exactly one stationary distribution π, all its entries are at least n1−d, and from any starting probability vector ∥Gkx0−π∥1≤2dk. Every eigenvalue other than 1 has size at most d.
A mixed set of 12 quick checks drawn at random from the 92 in this unit's chapters. Each one says which chapter it comes from, so a miss tells you what to reread. Press "New set" for another mix.
Quick checkTake V = the positive real numbers, with ordinary addition and ordinary multiplication by real numbers. Which axiom fails first?
Adding two positive numbers gives a positive number, so axiom 1 holds. But scaling by −1 turns 2 into −2, which is not positive. (Axiom 5 fails too: 0 is not positive.)
Quick checkS={(x,y):xy=0}, the two coordinate axes together. Is it a subspace of R2?
It does contain (0,0), and scaling a point on an axis keeps it on that axis. But (1,0)+(0,1)=(1,1) has xy=1. Passing two of the three checks is not enough.
Your answerLet A=[1324] and B=[05−12]. The matrix 2A−3B is computed entry by entry. What is its entry in row 2, column 1?
Only the (2,1) entries of A and B are needed: 3 and 5.
Row 2, column 1 of A is 3 and of B is 5. So the entry is 2⋅3−3⋅5=6−15=−9. Addition and scaling act on each position separately; no other entry matters.
Your answerIn the space of real functions, let f(t)=t2+1 and g(t)=3t. The vector 2f−g is itself a function, defined point by point. What is its value at t=2?
Find f(2) and g(2) first.
(2f−g)(2)=2f(2)−g(2)=2⋅5−6=4. Scaling and adding functions means scaling and adding their values at each t.
Quick checkIn the vector space of all real functions on [0,1], which of these is a vector?
A vector is an element of the space, and the elements of a function space are functions. f(0.3) is one number read off from one vector, and 0.3 is just an input.
The first misses O (its right side is 1, not 0). The third is closed under addition but not under scaling by −1. The second is a line through the origin: both conditions are linear and homogeneous, so sums and multiples keep x=0 and y=z.
Your answerIn this space ('sum' is the product, 'scaling by c' is the power xc), which positive number is the vector 3⋅4, meaning the vector 4 scaled by the scalar 3?
Scaling by c means raising to the power c.
Scaling x by c is defined as xc, so scaling 4 by 3 gives 43=64.
Quick checkWhat is the span of the single vector (1,2,3) in R3?
The span is every linear combination, and with one vector a combination is a multiple c(1,2,3) for any real c. It includes O (take c=0) and −(1,2,3), so it is more than the one vector. One vector cannot reach a whole 3-dimensional space.
Quick checkThree vectors satisfy x1−2x2+x3=O. Are they dependent or independent?
Dependent means SOME combination with coefficients not all zero equals O, and here we are handed one. The first option mixes up 'not a multiple of another' with independence: three vectors can be dependent with no two on a line. The third option is wrong because the all-zero combination gives O for any vectors at all, so it proves nothing.
Your answerFor which value of t are the vectors (1,2) and (3,t) dependent?
(3,t) must be 3 times (1,2).
Two vectors in the plane are dependent exactly when one is a multiple of the other. (3,t)=3(1,2)=(3,6) forces t=6. For any other t, the only solution of c1(1,2)+c2(3,t)=O is c1=c2=0.
Quick checkWhich statement does the k+1 theorem guarantee?
R3 is spanned by 3 vectors, so any 3+1=4 vectors in it are dependent. The theorem only forces dependence when there are MORE vectors than the spanning set. Three vectors in R3 may be dependent or independent (it depends on the vectors), and two vectors may be independent, as (1,0,0) and (0,1,0) are.
Quick checkThe set {(1,0),(0,1),(1,1)} spans R2. Is it a basis of R2?
A basis needs BOTH spanning and independence. This set spans, but the third vector is a combination of the first two, so one vector is wasted. Remove (1,1) and what is left is the standard basis.
Your answerWhat is the dimension of {(x1,…,x5)∈R5:x1+x2+x3=0 and x4=x5}?
Count the free coordinates after solving the conditions.
Two independent conditions, so 5−2=3. Concretely, x3=−x1−x2 and x5=x4, leaving x1,x2,x4 free, and the basis (1,0,−1,0,0), (0,1,−1,0,0), (0,0,0,1,1).
Your answerWhat is the dimension of the space of symmetric 3×3 matrices (those with Aij=Aji)?
Which entries can you choose freely? The rest are forced by symmetry.
A symmetric matrix is fixed by its 3 diagonal entries and the 3 entries above the diagonal (A12,A13,A23); the entries below are copies. That is 3+3=6 free numbers, and the six matrices with a single such entry (and its mirror) form a basis.
Quick checkYou have 3 independent vectors in R4. Which statement is guaranteed?
A basis of R4 has 4 vectors, so 3 are not enough to span. But part (a) says any independent set extends to a basis, and here it takes exactly one more vector, chosen outside their span. 'n independent vectors form a basis' needs n=4 of them.
Quick checkWhy can a vector have only ONE list of coordinates in a basis?
Spanning only guarantees that at least one list exists. Uniqueness comes from independence: if ∑ciei=∑diei, then ∑(ci−di)ei=O, and independence forces every ci=di. A dependent spanning set, such as (1,0),(0,1),(1,1), gives many lists for one vector: (1,1) is (1,1,0) and also (0,0,1).
Each output entry of the first map is a fixed combination of the inputs, so sums and multiples pass straight through. The second has a product x1x2 of two inputs, and doubling the input would quadruple that entry. The third moves the origin, since T(0,0)=(1,0).
Quick checkA function T satisfies T(O)=O. Must T be linear?
T(O)=O is necessary for linearity and not sufficient. The map x↦x2 sends 0 to 0, but T(1+1)=4=2=T(1)+T(1). The test is useful in one direction only: failing it proves a map is not linear.
Quick checkT:R3→R2 is linear. The null space N(T) is a subspace of which space?
The null space is the set of INPUTS sent to O, so it lives in the domain R3. The range is the set of outputs and lives in the codomain R2. Mixing these up is the most common slip with rank–nullity.
Your answerA linear map T:P5→R3 is onto. What is dimN(T)?
dimP5 is not 5. Count the coefficients.
Onto means the range is all of R3, so the rank is 3. The domain P5 has dimension 5+1=6 (coefficients c0,…,c5). Rank–nullity: dimN(T)=6−3=3. The common slip is using 5 for dimP5.
Your answerLet T:R3→R3 be T(x1,x2,x3)=(x1+x2+x3,2x1+2x2+2x3,3x1+3x2+3x3). What is its rank?
Write the output as s times a fixed vector.
Every output is (s,2s,3s) with s=x1+x2+x3, a multiple of (1,2,3). So the range is that line, with rank 1. (Nullity is then 3−1=2: the plane x1+x2+x3=0.)
Your answerLet T(f)=f′′ on P6. What is the rank of T?
Find the nullity first, and remember dimP6=7.
dimP6=7. The null space is {a+bt} (the polynomials with zero second derivative), nullity 2. Rank–nullity gives rank 7−2=5. Indeed the range is P4, which has dimension 5.
Quick checkCan a linear map T:R3→R2 be one-to-one?
The rank is at most 2, so by rank–nullity the nullity is at least 3−2=1. Then some x=O has T(x)=O=T(O), so two different inputs share an output. The third reason is wrong because R3 and R2 have the same number of points, in the sense that a one-to-one matching exists. That matching cannot be linear.
Your answerWith the same S(x,y,z)=(−y,x,z) and T(x,y,z)=(x,−z,y), compute TS(4,5,6). What is its FIRST entry?
The map written on the right acts first. Apply S, then T.
TS does S first: S(4,5,6)=(−5,4,6). Then T(−5,4,6)=(−5,−6,4). The first entry is −5. (The formula TS(x,y,z)=(−y,−z,x) gives the same.) Compare ST(4,5,6)=(6,4,5): a different vector.
Quick checkDoes f(x)=x2, as a function from R to R, have a left inverse?
A left inverse must return 1 from the output 1 and also return −1 from the same output. The square root only undoes the map on the non-negative numbers, so it is a left inverse of x2 restricted to x≥0, not of x2 on all of R.
Your answerA sprite point p is rotated by T(x,y)=(−y,x) and then shifted by S(x,y)=(x+5,y+2), and it ends on screen at (8,6). Undo the two steps in the right order. What was the x-coordinate of p?
First subtract the shift. Then undo the rotation, using T−1(x,y)=(y,−x).
The last step done is undone first. S−1(8,6)=(3,4). Then T−1(x,y)=(y,−x) gives T−1(3,4)=(4,−3). So p=(4,−3) and its x-coordinate is 4. Check: T(4,−3)=(3,4), and shifting gives (8,6).
Quick checkWhich pair of spaces is NOT isomorphic?
Isomorphic finite-dimensional spaces have the same dimension. R2×2 has dimension 4 and R3 has dimension 3. The first two pairs both match (4 and 4, 6 and 6).
Your answerThe space of 3×4 real matrices is isomorphic to Rn for which n?
Isomorphic means same dimension. Count the entries.
A 3×4 matrix has 12 entries, and the 12 matrices with a single 1 are a basis. So the dimension is 12, and by 'Same dimension exactly when isomorphic' the space is isomorphic to R12 (and to nothing of another dimension).
Quick checkCan you prescribe T(1,0)=(1,0), T(0,1)=(0,1) and T(1,1)=(5,5) for a linear map T?
Prescribing freely only works on an independent set, and a basis of the plane has two vectors, not three. Here the third value contradicts the first two.
Quick checkA linear map of the plane has T(e1)=(1,2) and T(e2)=(3,4). What is its matrix in the standard basis?
Each image is a COLUMN: column 1 is T(e1)=(1,2) and column 2 is T(e2)=(3,4). The first option writes them as rows, which is the transpose and a different map.
Quick checkFor a 3×2 matrix A, which statement is right?
The number of columns is the number of basis vectors of the source, so the source is R2. The number of rows is the target dimension, 3. The shape reads (target) by (source).
Quick checkA is 2×3 and B is 3×4. Which products exist?
AB pairs the 3 columns of A with the 3 rows of B. For BA you would pair the 4 columns of B with the 2 rows of A, and 4=2. Even when both products exist they are generally different.
Quick checkWhich of these matrices is NOT invertible?
Check Δ=ad−bc. The first has 4−6=−2. The third has 0−1=−1. The second has 2⋅2−4⋅1=0: its second column is twice its first, so it squashes the plane onto a line and cannot be undone.
Quick checkB=C−1AC where the columns of C are the new basis vectors in old coordinates. To find B's action on new coordinates y, which happens FIRST?
Product order is right to left. C acts first and converts new coordinates y into old coordinates. Then A acts in old coordinates, and C−1 converts the result back to new coordinates.
Clear column 1, then compare the two rows that are left.
R2→R2−2R1 gives (0,0,1) and R3→R3−3R1 gives (0,0,1). Then R3→R3−R2 gives a zero row. The echelon form has pivots in columns 1 and 3 only (column 2 is twice column 1), so there are 2 pivots and the rank is 2.
Your answerThe reduced row echelon form of a 3×5 matrix A has pivots in columns 1, 2 and 4. What is dimN(A)?
Nullity =n−r, where n is the number of columns.
The free variables are x3 and x5, the columns without a pivot. There are 5−3=2 of them, and the nullity is the number of free variables. This is rank–nullity read off the staircase.
Quick checkA 3×4 matrix has rank 2. Which statement is true?
The null space lives in R4 (the columns), so its dimension is n−r=4−2=2. The left null space lives in R3 (the rows), so its dimension is m−r=3−2=1. Mixing up m and n is the usual slip.
Your answerRow reduce [A∣I] for A=100110011. What is the entry in row 1, column 3 of A−1?
The left block is already upper triangular. Clear column 3 above its pivot first, then column 2.
Clear above the pivots. R2→R2−R3 turns row 2 of the right block from (0,1,0) into (0,1,−1). Then R1→R1−R2 turns row 1 from (1,0,0) into (1,−1,1). So A−1=100−1101−11, and the entry is 1. Check: row 1 of A times column 3 of A−1 is 1⋅1+1⋅(−1)+0=0, as it should be for an off-diagonal entry of I.
Your answerFor the system x+2y=3, 2x+ay=7, there is NOT exactly one solution for exactly one value of a. Which value?
Eliminate x from the second equation and look at the coefficient of y.
R2→R2−2R1 gives (a−4)y=1. If a=4 this gives one value of y, then one x. If a=4 the row reads 0=1, a pivot in the last column, so there is no solution at all. The value is a=4 (the columns (1,2) and (2,a) are parallel exactly then).
Look at where the zeros are. What does the triangular theorem say?
The matrix is upper triangular, so the determinant is the product of the diagonal: 2⋅(−3)⋅4=−24. The entries 5, 7 and 1 above the diagonal do not matter.
Quick checkYou divide row 2 of a matrix by 5 to make the numbers smaller, then row reduce the new matrix and find its determinant is 12. What is the determinant of the original matrix?
Dividing a row by 5 divides the determinant by 5 (Axiom 1: pulling 5 out of a row). So the original determinant is 5 times the new one, 60. The tempting answers forget the factor (12), undo it in the wrong direction (512), or confuse scaling with a swap (−12).
Expand along row 1: 1(5⋅10−6⋅8)−2(4⋅10−6⋅7)+3(4⋅8−5⋅7)=2−2(−2)+3(−3)=2+4−9=−3. (With 9 in place of 10 the rows would be dependent and the answer would be 0.)
Quick checkWhich is true of the zero vector O and the number 0 in the definition of an eigenvector?
T(O)=λO holds for every λ, which is exactly why O is excluded. But T(x)=0x=O for some non-zero x is perfectly possible: that is a non-zero vector in the null space, so 0 is an eigenvalue exactly when the map has a non-trivial null space.
Quick checkx is an eigenvector of A with eigenvalue 3. What does A+2I do to x?
(A+2I)x=Ax+2x=3x+2x=5x. The vector stays an eigenvector, and the eigenvalue shifts by 2. It does not stay at 3 (that forgets the 2I) and it is not multiplied by 2 (that confuses adding 2I with scaling A).
Quick checkOver R, which of these matrices is NOT diagonalizable?
The identity is already diagonal. The last matrix has two different eigenvalues, 3 and 2, so it is diagonalizable. The shear has the repeated eigenvalue 1 with algebraic multiplicity 2, but its eigenspace is only the line through (1,0), so g=1<m=2 and there is no eigenbasis.
Your answerUse Fk≈φk/5 with φ≈1.618034 to estimate F15, then give the exact whole number it rounds to.
The second term ψk/5 has size well below 1, so the nearest whole number is the answer.
φ15≈1364.0007 and dividing by 5≈2.236068 gives 610.0003. The correction ψ15/5 is tiny, so F15=610. You can confirm by adding up: 0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610.
Your answerLet A=200120002. Its only eigenvalue is 2. What is dimE(2), the number of independent eigenvectors?
Find the rank of A−2I and use rank plus nullity equals 3.
A−2I=000100000 has rank 1, so its null space has dimension 3−1=2. The solutions are y=0 with x and z free, spanned by (1,0,0) and (0,0,1). Here g=2<3=m, so A is not diagonalizable.
Quick checkWhen does equality ∣(x,y)∣=∥x∥∥y∥ hold in the Cauchy–Schwarz inequality?
Equality means the angle has cosine ±1: the vectors lie on one line, which is dependence. Perpendicular vectors give the opposite extreme, (x,y)=0. Equal lengths say nothing about the angle.
Quick checkTrue or false: every linearly independent set of non-zero vectors is orthogonal.
The theorem goes only one way. (1,0) and (1,1) are independent, but their dot product is 1, not 0. Independent means no waste; orthogonal means at right angles, which is much stronger.
Quick checkIn Gram–Schmidt, when we compute y3, which vectors do we subtract projections onto?
The formula only works when the vectors we project on are perpendicular to each other, so that the projections do not interfere. The yj are, but the xj usually are not. Projecting on x1 and x2 would remove some components twice.
Your answerq1,q2 are orthonormal vectors and x has ∥x∥2=30, (x,q1)=3, (x,q2)=4. What is the squared distance from x to the plane spanned by q1 and q2?
Squared error = ∥x∥2 minus the sum of the squared coefficients.
The projection onto the plane is s=3q1+4q2, with ∥s∥2=9+16=25. The squared distance is ∥x∥2−∥s∥2=30−25=5.
Quick checkIn the best approximation theorem, why is the projection s closer to x than any other point t of S?
The right angle at s is the whole reason: x−s is perpendicular to s−t, so Pythagoras gives ∥x−t∥2=∥x−s∥2+∥s−t∥2. The projection does not in general have the same length as x (it is shorter), and 'largest inner product' is not the criterion.
Quick checkWhat does the least squares line minimize?
The error vector lives in the space of heights, so each miss is measured straight up or down. The perpendicular distance to the line is a different problem (total least squares). The largest miss is a different problem again. And the plain sum of misses can be zero for a terrible line, because positive and negative misses cancel, which is why we square.
Your answerFit a line to the points (0,0),(1,1),(2,3). What is the slope m?
Build A with a column of ones and a column (0,1,2), then solve the 2×2 normal equations.
ATA=[3335] and ATb=(4,7). Subtracting the equations 3c+3m=4 and 3c+5m=7 gives 2m=3, so m=23 and c=−61. Check by the average rule: the line passes through (tˉ,yˉ)=(1,34), and −61+23=34.
Quick checkSolving (P−I)π=O for a 2×2 stochastic matrix gives the null-space vector (3,4). What is the stationary distribution?
A stationary distribution must have entries adding to 1, so divide the null-space vector by its sum, 7. Leaving it as (3,4) is the most common slip. The options (21,21) and (43,34) are not multiples of (3,4), so they are not even eigenvectors for this P.
Your answerA weather chain has P=[0.60.40.30.7]. What fraction of days are sunny in the long run (the first entry of π)?
Solve the one independent equation, then scale so the entries add to 1.
(P−I)π=O gives −0.4π1+0.3π2=0, so π1=0.75π2 and a null-space vector is (3,4). Dividing by the sum 7 gives π=(73,74). Check: 0.6⋅73+0.3⋅74=71.8+1.2=73.
Your answerA 2×2 stochastic matrix has P=[0.750.250.50.5]. What is its other eigenvalue (besides 1)?
Use trace = sum of eigenvalues, from 2.8.
The trace is 0.75+0.5=1.25 and the eigenvalues add to the trace, so the other eigenvalue is 1.25−1=0.25. The chances therefore approach π with error shrinking by a factor 0.25 each step.
Your answerIn the four-page web, a surfer has the chances x=(21,0,31,61). What is entry 3 of Px, the chance of being on page 3 one step later?
Row 3 of P says which pages send surfers to page 3, and what share of their chance they send.
Row 3 of P is (31,21,0,21): page 3 receives a third of page 1, half of page 2 and half of page 4. So (Px)3=31⋅21+21⋅0+0⋅31+21⋅61=61+121=41. This is the third entry of x3 in the step-through.