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Course Linear Algebra ★The Language of Mathematics

Week 1

Week 2

Week 3

Week 4

Week 5

Week 6 soon

L21Vector spaces
L22Subspaces, sums and intersections
L23Basis and dimension in general
L24Direct sums
R6Recitation 6

Week 7 soon

L25Coordinates
L26Row space and rank
L27The four fundamental subspaces
L28Change of basis
R7Recitation 7

Week 8 soon

L29Linear maps
L30The rank–nullity theorem
L31Isomorphisms and the space of maps
L32The matrix of a linear map
R8Recitation 8

Week 9 soon

L33Determinant as signed volume
L34The determinant defined
L35Computing determinants
L36Determinants and invertibility
R9Recitation 9

Week 10 soon

L37Eigenvalues and eigenvectors
L38The characteristic polynomial
L39Diagonalisation
L40Eigenvalues in action
R10Recitation 10

Week 11 soon

L41Complex eigenvalues
L42Invariant subspaces
L43Polynomials of matrices
L44The minimal polynomial
R11Recitation 11

Week 12 soon

L45Inner product spaces
L46Orthogonality and projections
L47Gram–Schmidt and QR
L48Least squares
R12Recitation 12

Week 13 soon

L49The spectral theorem
L50Quadratic forms
L51Singular value decomposition
L52The SVD in action and what comes next
R13Recitation 13

Lecture 18 · Week 5· 15:51

LU factorisation

Elimination as a factorisation, solving with LU.

Video: coming soon

The lecture (15:51) is not published yet. It will play right here as soon as it is on YouTube.

In this lecture

Today: elimination, recorded
Elimination as a product of matrices
From the steps to L
Why the multipliers land in L
Solving with L and U
How much work it saves
When a row swap is needed
Summary

Materials

Check yourself

Try it before you look

Find an LU factorisation of $A = [1328]$ .
Use it to solve $A x = (1, 7)$ by forward and back substitution, and check your answer.
Write down $L$ and $U$ for $120134015$ without multiplying any matrices.

Work it out on paper first. Then open the hint if you are stuck, and the answer to check.

1Hint

Everything you need is in Section 3.7 of the course book. Open it at page 46 of the chapter PDF; any theorem the question names is stated there, with an example worked the same way.

2Answer

The multiplier is $l_{21} = 3/1 = 3$ , and $R_{2} \to R_{2} - 3 R_{1}$ gives row $2$ $= (0, 2)$ . So $L = [1301]$ and $U = [1022]$ (Theorem 3.7.4). Check: $LU = [13 2 6 + 2] = A$ .
Forward substitution in $L y = (1, 7)$ : $y_{1} = 1$ , then $3 y_{1} + y_{2} = 7$ gives $y_{2} = 4$ . Back substitution in $U x = y$ : $2 x_{2} = 4$ gives $x_{2} = 2$ , then $x_{1} + 2 x_{2} = 1$ gives $x_{1} = - 3$ (Theorem 3.7.5). So $x = (- 3, 2)$ . Check: $A x = (- 3 + 4, - 9 + 16) = (1, 7)$ .
$R_{2} \to R_{2} - 2 R_{1}$ gives row $2$ $= (0, 1, 1)$ , and row $3$ already has a $0$ in column $1$ , so $l_{31} = 0$ . Then $l_{32} = 4/1 = 4$ , and $R_{3} \to R_{3} - 4 R_{2}$ gives row $3$ $= (0, 0, 1)$ . Writing each multiplier where it cleared an entry, $L = 120014001$ and $U = 100110011$ .