Ten lessons connecting systems, spaces and matrix geometry, with two case-switching explorations per lesson.
Introductory conceptual bridge, not a full semester. SVD, abstract proofs and large-scale numerical algorithms are outside scope. No jurisdiction-specific certification standard is claimed.
Algebra, equations, vectors and matrix notation. Calculus is not required here.
Course outline
Columns are the ingredients
Interpret Ax as a combination of columns after checking dimensions.
Elimination keeps the promise
Preserve the solution set when reducing an augmented matrix.
Free does not mean chaos
Parameterize the nullspace and apply rank-nullity.
No spare passengers in a basis
Distinguish spanning from independence and find basis coordinates.
Composition has an order
Build linear maps from basis images and compose in the right order.
The determinant is not a length
Interpret signed area scale and invertibility.
Inspect the leftover
Compute projection and check orthogonality.
Best fit is not exact fit
Solve least squares and verify the normal equation.
Eigenvectors keep their line
Recognize eigenvectors and the condition for diagonalization.
Positive entries are not positive energy
Test a symmetric quadratic form for positive definiteness.
Sources and curriculum note
Sources fetched October 4, 2026. These are archived courses, not a current universal syllabus. All teaching scenarios are explicitly hypothetical.
Read every lesson below. The interactive reader above contains the same explanations, with visual tools and quizzes.
1. Columns are the ingredients
Learning goal: Interpret Ax as a combination of columns after checking dimensions.
A matrix with m rows and n columns accepts n input coordinates and produces m output coordinates. Its jth column is the image of the jth coordinate unit vector. Thus Ax is x₁ times column 1 plus x₂ times column 2, and so on. Rows offer another view: each output coordinate is a row dot product with x.
The column view explains which outputs are reachable. The row view explains which equations constrain the input. Both give the same product, but neither permits entrywise multiplication or ignoring unmatched coordinates. A 2 by 3 matrix needs three input coordinates, not two.
In the original hypothetical model A=[[1,2],[3,1]], x=(2,1), the column contributions are (2,6) and (2,1), giving (4,7). Nothing here is measured data. Finding an input for a target b means finding column weights that make b; a target outside the column span cannot be made.
Official source grounding: LA0. Learning pain points and all hypothetical examples are editorial, not reported error statistics.
Worked example
Hypothetical A=[[1,2],[3,1]], x=(2,1). Compute Ax two ways.
Check that A has two columns and x has two entries.
Columns: 2(1,3)+1(2,1)=(2,6)+(2,1)=(4,7).
Rows: 1·2+2·1=4 and 3·2+1·1=7.
Both methods return (4,7); input coordinates are weights, not outputs.
Practice problem and solution
New hypothetical A=[[1,2],[0,1],[3,0]] has three rows and two columns. For x=(2,-1), form Ax as a combination of columns and enter the sum of the three output entries.
Column 1 is (1,0,3) and column 2 is (2,1,0). 2(1,0,3)-1(2,1,0)=(0,-1,6). The entries sum to 0-1+6=5. Rows give the same output.
Mental model: Ax is a weighted sum of columns.
Common trap: Choosing input size from the number of rows.
2. Elimination keeps the promise
Learning goal: Preserve the solution set when reducing an augmented matrix.
Row operations transform the equations without changing their simultaneous solutions. You may swap rows, scale a row by a nonzero number, or add a multiple of one row to another. Every operation must include the augmented right-hand side. Changing only coefficients changes the original problem.
A pivot is a leading nonzero entry used to eliminate another coefficient. If the intended pivot is zero, search for a row to swap before dividing. A row [0,0 | c] with c≠0 is a contradiction; [0,0 | 0] is redundant. A missing pivot may leave a free variable, not automatically make the system inconsistent.
In the hypothetical system x+y=5 and 2x+3y=12, subtracting twice the first equation leaves y=2, then x=3. Keep the original equations for verification. Substitution checks both arithmetic and the meaning of the reduced rows; the appearance of a triangular matrix alone is not enough.
Official source grounding: LA1. Learning pain points and all hypothetical examples are editorial, not reported error statistics.
Worked example
Solve hypothetical x+y=5, 2x+3y=12.
Augmented rows are [1,1 | 5] and [2,3 | 12].
R₂ <- R₂-2R₁ gives [0,1 | 2].
Read y=2; x=5-2=3.
Check 3+2=5 and 2·3+3·2=12.
Practice problem and solution
New hypothetical system x+y+z=6, x+2y+3z=14, 2x+3y+4z=k. Eliminate to find the one value of k that keeps the system consistent. Enter k.
Row 1 plus row 2 gives 2x+3y+4z=20, which matches the left side of row 3. Row 3 minus rows 1 and 2 becomes 0=k-20. Consistency needs k=20; any other k gives a row [0 0 0 | nonzero].
Mental model: Row operations preserve solutions.
Common trap: Dropping the augmented column.
3. Free does not mean chaos
Learning goal: Parameterize the nullspace and apply rank-nullity.
The nullspace N(A) contains inputs mapped to zero. It is closed under addition and scalar multiplication because A is linear. Eliminate in Ax=0, assign free variables, and express pivot variables in terms of them. Each independent parameter produces a basis direction, not a completely independent choice of every coordinate.
Rank is the number of pivots. For an m by n matrix, rank plus nullity equals n, the input dimension. The number of rows is not the input dimension. More columns than pivots means some nonzero input changes can leave the output unchanged.
For hypothetical A=[[1,2,0],[0,0,1]], Ax=0 says x₁=-2x₂ and x₃=0. Choosing x₂=s gives s(-2,1,0). If Ax=b has a particular solution xₚ, all its solutions are xₚ+z for z in N(A). That translated set need not be a subspace, whereas N(A) always is.
Official source grounding: LA0. Learning pain points and all hypothetical examples are editorial, not reported error statistics.
Worked example
Find N(A) for hypothetical A=[[1,2,0],[0,0,1]].
Write x₁+2x₂=0 and x₃=0.
Choose x₂=s, then x₁=-2s.
All solutions are s(-2,1,0); a basis is {(-2,1,0)}.
Check A(-2,1,0)=(0,0), and rank 2 plus nullity 1 equals 3.
Practice problem and solution
New hypothetical 4 by 6 matrix already reduced: rows (1,0,2,0,0,3), (0,1,-1,0,0,1), (0,0,0,1,0,2), (0,0,0,0,0,0). Pivots are in columns 1, 2 and 4. Set x₃=0, x₅=0, x₆=1 and solve Ax=0. Enter the sum of the six entries of that special solution.
Free columns are 3, 5 and 6. Row 1: x₁+3=0 so x₁=-3. Row 2: x₂+1=0 so x₂=-1. Row 3: x₄+2=0 so x₄=-2. The vector is (-3,-1,0,-2,0,1), with sum -5. Rank 3 and six columns give nullity 3.
Learning goal: Distinguish spanning from independence and find basis coordinates.
A spanning set reaches every vector in a named target space. An independent set has no nontrivial zero combination. A basis does both. A redundant spanning set is not a basis; an independent set may still be too small to span. Two independent vectors can form a basis of a plane in R³, but not of all R³.
Coordinates are relative to an ordered basis. If B=(b₁,b₂), then [v]B=(c₁,c₂) means v=c₁b₁+c₂b₂. Put the basis vectors in the columns of P to obtain v=P[v]B. Solve that system to convert standard coordinates to basis coordinates. The list of coefficients is not the standard vector unless the basis is standard.
In a hypothetical exercise B=((1,1),(1,-1)), v=(4,2). Solving c₁+c₂=4 and c₁-c₂=2 gives (3,1). The same vector has standard coordinates (4,2) and B-coordinates (3,1). Always reconstruct the vector with the stated basis to check your answer.
Official source grounding: LA0. Learning pain points and all hypothetical examples are editorial, not reported error statistics.
Worked example
Find B-coordinates of hypothetical v=(4,2) in B=((1,1),(1,-1)).
New hypothetical basis B=((1,2),(3,1)) of R², and v=(9,7). Check that B is a basis, then enter the first B-coordinate c₁ in v=c₁(1,2)+c₂(3,1).
The determinant of [[1,3],[2,1]] is 1-6=-5, nonzero, so the columns are independent and span R². Solve c₁+3c₂=9 and 2c₁+c₂=7: c₂=7-2c₁, then c₁+21-6c₁=9, so c₁=12/5=2.4 and c₂=2.2. Check: 2.4(1,2)+2.2(3,1)=(9,7).
Mental model: A basis supplies unique weights in a named space.
Common trap: Checking only independence.
5. Composition has an order
Learning goal: Build linear maps from basis images and compose in the right order.
A linear map obeys T(au+bv)=aT(u)+bT(v). The columns of its standard matrix are the images of the standard basis vectors. A fixed nonzero translation is affine rather than linear: it fails T(0)=0. Passing the zero test alone does not prove linearity; it is a quick way to disprove it.
If A acts first and B second, the final result is B(Ax)=BAx. The rightmost matrix acts first. AB may give a different result or may not be defined. Square matrices can commute in special cases, but no general commutativity rule exists. Undoing an invertible composite reverses the order of the inverse actions.
In the hypothetical model A=[[1,1],[0,1]] and B=[[2,0],[0,1]], BA=[[2,2],[0,1]] while AB=[[2,1],[0,1]]. Acting on (0,1) gives (2,1) versus (1,1). This toy shear and scale makes the order visible without claiming a real physical experiment.
Official source grounding: LA0. Learning pain points and all hypothetical examples are editorial, not reported error statistics.
Worked example
For hypothetical A=[[1,1],[0,1]], B=[[2,0],[0,1]], find BA(0,1).
Apply A first: (0+1,1)=(1,1).
Apply B next: (2,1).
For the full product, B acts on A columns (1,0),(1,1), giving (2,0),(2,1).
Thus BA=[[2,2],[0,1]] and BA(0,1)=(2,1).
Practice problem and solution
New hypothetical A=[[1,1],[0,2]] and B=[[0,1],[3,0]], x=(2,1). Apply A first, then B, then A again. Enter the first coordinate of the final vector.
Ax=(3,2). B(3,2)=(2,9). A(2,9)=(11,18). The first coordinate is 11. The combined matrix is ABA, with the first-applied matrix written last on the right.
Mental model: The rightmost factor acts first.
Common trap: Assuming all matrix products commute.
6. The determinant is not a length
Learning goal: Interpret signed area scale and invertibility.
For A=[[a,b],[c,d]], det A=ad-bc. Its absolute value is the area multiplier for a real plane map, while its sign records orientation. A negative determinant does not mean negative geometric area. Zero determinant means dependent columns and no inverse for the square map, not that all entries vanish.
Determinants multiply under composition, but do not generally add under addition. Swapping two rows flips the sign, scaling one row scales the determinant, and adding a multiple of one row to another leaves it unchanged. If elimination is used, keep a record of swaps and scalings before reading the triangular diagonal product.
In the hypothetical geometric model A=[[2,1],[0,-3]], determinant is -6. A region of area four becomes area twenty-four with reversed orientation. Contrast [[1,2],[2,4]], whose columns lie on one line. It collapses a plane direction, so its determinant is zero despite nonzero entries.
Official source grounding: LA0. Learning pain points and all hypothetical examples are editorial, not reported error statistics.
Worked example
Hypothetical A=[[2,1],[0,-3]]. Find determinant and image area for a square of area four.
det=2(-3)-1(0)=-6.
Area scale is |-6|=6.
Image area is 4·6=24 square units.
Nonzero determinant means invertible; negative sign means orientation reversal.
Practice problem and solution
New hypothetical A=[[a,1],[3,2]] sends a region of area 4 to a region of area 20 and reverses orientation. Enter a.
Scale is 20/4=5, so |det A|=5. Reversal means det A=-5. det A=2a-3, so 2a-3=-5 and a=-1. Orientation preserved would give 2a-3=5 and a=4.
Mental model: Zero determinant means lost dimension.
Common trap: Assuming determinants distribute over addition.
7. Inspect the leftover
Learning goal: Compute projection and check orthogonality.
For nonzero u, projection onto its line is p=u(u·b)/(u·u). The coefficient divides by squared norm, not norm. The residual e=b-p is perpendicular to u. Projection separates the component that the line can represent from the part it cannot.
The orthogonal residual makes p the nearest point on the line. Scaling the spanning vector changes its coefficient but not the projected vector. A zero spanning vector makes the formula undefined; projection onto the zero subspace is instead zero. State the Euclidean dot product before using this formula.
In a hypothetical exercise u=(1,2), b=(3,1), both dot products u·b and u·u equal five. Thus p=(1,2), e=(2,-1), and u·e=0. A residual can be perpendicular and still large. It is the residual, not the original b, that must be perpendicular to the line.
Official source grounding: LA2. Learning pain points and all hypothetical examples are editorial, not reported error statistics.
Worked example
Project hypothetical b=(3,1) onto u=(1,2).
u·b=3+2=5; u·u=1+4=5.
Coefficient is one; p=(1,2).
Residual is (3,1)-(1,2)=(2,-1).
Dot with u gives 2-2=0; p lies in the line and is nearest.
Practice problem and solution
New hypothetical b=(2,6,3) and u=(1,2,2). Project b onto u, then enter the squared length of the leftover e=b-cu, to 4 decimals.
u·b=2+12+6=20 and u·u=9, so c=20/9. e=b-cu=(2-20/9, 6-40/9, 3-40/9)=(-2/9,14/9,-13/9). e·e=(4+196+169)/81=369/81=41/9≈4.5556. Check u·e=(-2+28-26)/9=0.
Mental model: Projection has orthogonal leftover.
Common trap: Using norm instead of squared norm.
8. Best fit is not exact fit
Learning goal: Solve least squares and verify the normal equation.
When Ax=b is inconsistent, least squares minimizes ||Ax-b||². Its residual e=b-Ax must be perpendicular to every column of A. Thus Aᵀe=0, which rearranges to AᵀAx=Aᵀb. This is not a claim that every residual entry vanishes.
Independent columns make AᵀA invertible and give a unique coefficient vector. Dependent columns may give nonunique coefficients while the fitted projection stays unique. These tiny exercises explain the geometry; they are not a recommendation to explicitly invert normal equations for a large numerical system.
For a hypothetical constant fit to toy values (1,2,6), A is a column of three ones. The normal equation 3c=9 gives c=3. Residual (-2,-1,3) sums to zero, so it is orthogonal to that column. These are invented values, not collected data or a statistical inference about a population.
Official source grounding: LA2. Learning pain points and all hypothetical examples are editorial, not reported error statistics.
Worked example
Fit a constant to hypothetical (1,2,6).
A=(1,1,1)ᵀ and Ac=(c,c,c).
AᵀA=3, Aᵀb=9; solve 3c=9 to obtain c=3.
p=(3,3,3); e=(-2,-1,3).
Aᵀe=0 and squared residual 4+1+9=14. The fit is best but not exact.
Practice problem and solution
New hypothetical data points (0,1), (1,3), (2,4). Fit y=c+mx by least squares. Enter the predicted value at x=3, to 4 decimals.
A has rows (1,0),(1,1),(1,2); b=(1,3,4). AᵀA=[[3,3],[3,5]], Aᵀb=(8,11). Solving gives c=7/6≈1.1667 and m=1.5. Prediction at 3 is 7/6+4.5=5.6667. Residual (-1/6,1/3,-1/6) is perpendicular to both columns.
Mental model: Best fit means orthogonal leftover.
Common trap: Calling a best fit an exact solution.
9. Eigenvectors keep their line
Learning goal: Recognize eigenvectors and the condition for diagonalization.
An eigenvector is nonzero and satisfies Av=λv. Its line is preserved, but a negative λ reverses direction and λ=0 sends it to zero. Saying same direction is too narrow. Zero is excluded because it would satisfy the equation for every λ and identify no direction.
Eigenvalues solve det(A-λI)=0. For each value, find nonzero vectors in N(A-λI). An n by n matrix with n independent eigenvectors has A=SΛS⁻¹. Columns of S and entries of Λ must be paired in the same order. A repeated eigenvalue does not automatically give enough independent eigenvectors.
In hypothetical A=[[2,1],[0,3]], v=(1,1) obeys Av=3v, while (1,0) obeys Av=2v. These directions are independent. Powers scale each mode by its eigenvalue raised to the power; a general input combines mode contributions rather than being scaled by one universal number.
Official source grounding: LA3. Learning pain points and all hypothetical examples are editorial, not reported error statistics.
Worked example
For hypothetical A=[[2,1],[0,3]], compute A²(1,1).
Av=(3,3)=3v, with v nonzero.
Apply A again: A²v=A(3v)=3Av.
Therefore A²v=9v=(9,9).
Directly A(3,3)=(9,9), confirming the mode computation.
Practice problem and solution
New hypothetical A=[[5,1],[0,2]]. Eigenvectors (1,0) for eigenvalue 5 and (1,-3) for eigenvalue 2. Starting from x=(4,-3), enter the first coordinate of A⁴x.
x=3(1,0)+1(1,-3). A⁴x=3·5⁴(1,0)+1·2⁴(1,-3)=1875(1,0)+16(1,-3)=(1891,-48). The first coordinate is 1891. Direct multiplication four times gives the same vector.
Mental model: An eigenvector basis enables diagonalization.
Common trap: Counting repeated roots as independent vectors.
10. Positive entries are not positive energy
Learning goal: Test a symmetric quadratic form for positive definiteness.
A real symmetric matrix is positive definite if xᵀAx>0 for every nonzero real x. Positive semidefinite allows zero at nonzero inputs. Indefinite means the form takes both positive and negative values. None of these conditions is simply entrywise positivity.
For symmetric [[a,b],[b,c]], positive definiteness is equivalent to a>0 and ac-b²>0. Completing the square gives a(x+(b/a)y)²+(c-b²/a)y² when a≠0. Both coefficients are positive under those conditions. Positive determinant alone also occurs for a negative definite matrix, so it is insufficient.
In a hypothetical energy-like model A=[[2,1],[1,2]], q=2x²+2xy+2y²=2(x+y/2)²+1.5y². This proves positivity for every nonzero input. The toy expression is not a measured energy law. In contrast, positive-entry matrix [[1,2],[2,1]] gives q(1,-1)=-2 and q(1,1)=6.
Official source grounding: LA4. Learning pain points and all hypothetical examples are editorial, not reported error statistics.
Worked example
For hypothetical A=[[2,1],[1,2]], prove positive definite and evaluate q(1,-1).
Symmetry holds; first minor 2>0 and det=4-1=3>0.
q=2x²+2xy+2y².
Complete square: 2(x+y/2)²+1.5y²>0 for all nonzero (x,y).
At (1,-1), q=2-2+2=2; this one value alone would not prove definiteness.
Practice problem and solution
New hypothetical A=[[2,k],[k,3]] with k real. Enter the positive boundary value of k where the matrix stops being positive definite, to 4 decimals.
The first leading minor is 2>0. The determinant is 6-k². Positive definite needs 6-k²>0, so |k|<√6≈2.4495. At k=√6 the determinant is zero and the form is only semidefinite.
Mental model: Positive definite tests every nonzero input.