Matrix Calculations: Determinants and Basis Transformation A. - PowerPoint PPT Presentation

Determinants Change of basis Radboud University Nijmegen Matrices and basis transformations Matrix Calculations: Determinants and Basis Transformation A. Kissinger Institute for Computing and Information Sciences Radboud University Nijmegen Version: autumn 2017 A. Kissinger Version: autumn 2017 Matrix Calculations 1 / 32

Determinants Change of basis Radboud University Nijmegen Matrices and basis transformations Outline Determinants Change of basis Matrices and basis transformations A. Kissinger Version: autumn 2017 Matrix Calculations 2 / 32

Determinants Change of basis Radboud University Nijmegen Matrices and basis transformations Last time • Any linear map can be represented as a matrix: f ( v ) = A · v g ( v ) = B · v • Last time, we saw that composing linear maps could be done by multiplying their matrices: f ( g ( v )) = A · B · v • Matrix multiplication is pretty easy: � 1 2 � � 1 − 1 � � 1 · 1 + 2 · 0 1 · ( − 1) + 2 · 4 � � 1 7 � = = · 3 4 0 4 3 · 1 + 4 · 0 3 · ( − 1) + 4 · 4 3 13 ...so if we can solve other stuff by matrix multiplication, we are pretty happy. A. Kissinger Version: autumn 2017 Matrix Calculations 3 / 32

Determinants Change of basis Radboud University Nijmegen Matrices and basis transformations Last time • For example, we can solve systems of linear equations: A · x = b ...by finding the inverse of a matrix: x = A − 1 · b • There is an easy shortcut formula for 2 × 2 matrices: � d � a b � � 1 − b A − 1 = A = = ⇒ − c c d ad − bc a ...as long as ad − bc � = 0. • We’ll see today that “ ad − bc ” is an example of a special number we can compute for any square matrix (not just 2 × 2) called the determinant. A. Kissinger Version: autumn 2017 Matrix Calculations 4 / 32

Determinants Change of basis Radboud University Nijmegen Matrices and basis transformations Determinants What a determinant does For an n × n matrix A , the determinant det( A ) is a number (in R ) It satisfies: det( A ) = 0 ⇐ ⇒ A is not invertible ⇒ A − 1 does not exist ⇐ ⇐ ⇒ A has < n pivots in its echolon form Determinants have useful properties, but calculating determinants involves some work. A. Kissinger Version: autumn 2017 Matrix Calculations 6 / 32

Determinants Change of basis Radboud University Nijmegen Matrices and basis transformations Determinant of a 2 × 2 matrix � a b � • Assume A = c d • Recall that the inverse A − 1 exists if and only if ad − bc � = 0, and in that case is: � d � − b A − 1 = 1 ad − bc − c a • In this 2 × 2-case we define: � � � a b � a b � � det = � = ad − bc � � c d c d � ⇒ A − 1 does not exist. • Thus, indeed: det( A ) = 0 ⇐ A. Kissinger Version: autumn 2017 Matrix Calculations 7 / 32

Determinants Change of basis Radboud University Nijmegen Matrices and basis transformations Determinant of a 2 × 2 matrix: example • Example: � 0 . 8 0 . 1 � � 8 1 � = 1 P = 0 . 2 0 . 9 10 2 9 • Then: 10 · 9 8 10 − 1 10 · 2 det( P ) = 10 72 2 = 100 − 100 70 7 = 100 = 10 • We have already seen that P − 1 exists, so the determinant must be non-zero. A. Kissinger Version: autumn 2017 Matrix Calculations 8 / 32

Determinants Change of basis Radboud University Nijmegen Matrices and basis transformations Determinant of a 3 × 3 matrix   a 11 a 12 a 13 • Assume A = a 21 a 22 a 23   a 31 a 32 a 33 • Then one defines: � � a 11 a 12 a 13 � � � � det A = a 21 a 22 a 23 � � � � a 31 a 32 a 33 � � � � � � � � a 22 a 23 a 12 a 13 a 12 a 13 � � � � � � = + a 11 · � − a 21 · � + a 31 · � � � � � � a 32 a 33 a 32 a 33 a 22 a 23 � � � � • Methodology: • take entries a i 1 from first column, with alternating signs (+, -) • take determinant from square submatrix obtained by deleting the first column and the i -th row A. Kissinger Version: autumn 2017 Matrix Calculations 9 / 32

Determinants Change of basis Radboud University Nijmegen Matrices and basis transformations Determinant of a 3 × 3 matrix, example � � 1 2 − 1 � � � � � � � � 3 4 2 − 1 2 − 1 � � � � � � � � 5 3 4 = 1 � − 5 � + − 2 � � � � � � � � 0 1 0 1 3 4 � � � � � � − 2 0 1 � � � � � � � � = 3 − 0 − 5 2 − 0 − 2 8 + 3 = 3 − 10 − 22 = − 29 A. Kissinger Version: autumn 2017 Matrix Calculations 10 / 32

Determinants Change of basis Radboud University Nijmegen Matrices and basis transformations The general, n × n case � � a 12 · · · a 1 n � � � � � � a 11 · · · a 1 n a 22 · · · a 2 n � � � � � � a 32 · · · a 3 n � � � � � � . . . . � � � � � � . . . . = + a 11 · − a 21 · . . . . . . � � � � � � . . . . � � � � � � a n 1 . . . a nn a n 2 . . . a nn � � � � � � a n 2 . . . a nn � � � � � � · · · a 12 a 1 n � � · · · � � � � � � . . � � . . + a 31 · · · · · · ± a n 1 � � . . � � � � � � · · · � � a ( n − 1)2 . . . a ( n − 1) n � � (where the last sign ± is + if n is odd and - if n is even) Then, each of the smaller determinants is computed recursively. A. Kissinger Version: autumn 2017 Matrix Calculations 11 / 32

Determinants Change of basis Radboud University Nijmegen Matrices and basis transformations Applications • Determinants detect when a matrix is invertible • Though we showed an inefficient way to compute determinants, there is an efficient algorithm using, you guessed it...Gaussian elimination! • Solutions to non-homogeneous systems can be expressed directly in terms of determinants using Cramer’s rule (wiki it!) • Most importantly: determinants will be used to calculate eigenvalues in the next lecture A. Kissinger Version: autumn 2017 Matrix Calculations 12 / 32

Determinants Change of basis Radboud University Nijmegen Matrices and basis transformations Vectors in a basis Recall: a basis for a vector space V is a set of vectors B = { v 1 , . . . , v n } in V such that: 1 They uniquely span V , i.e. for all v ∈ V , there exist unique a i such that: v = a 1 v 1 + . . . + a n v n Because of this, we use a special notation for this linear combination:   a 1 .  .  := a 1 v 1 + . . . + a n v n .   a n B A. Kissinger Version: autumn 2017 Matrix Calculations 14 / 32

Determinants Change of basis Radboud University Nijmegen Matrices and basis transformations Same vector, different outfits The same vector can look different, depending on the choice of basis. Consider the standard basis: S = { (1 , 0) , (0 , 1) } vs. another basis: �� 100 � � 100 �� B = , 0 1 Is this a basis? Yes... � 100 100 � • It’s independent because: has 2 pivots. 0 1 • It’s spanning because... we can make every vector in S using linear combinations of vectors in B : � 1 � � 100 � � 0 � � 100 � � 100 � 1 = = − 0 100 0 1 1 0 ...so we can also make any vector in R 2 . A. Kissinger Version: autumn 2017 Matrix Calculations 15 / 32

Determinants Change of basis Radboud University Nijmegen Matrices and basis transformations Same vector, different outfits �� 1 � � 0 �� �� 100 � � 100 �� S = B = , , 0 1 0 1 Examples: � 100 � � 1 � � 300 � � 2 � = = 0 0 1 1 S B S B � 1 � 1 � � � 0 � � − 1 � 100 = = 0 0 1 1 S B S B A. Kissinger Version: autumn 2017 Matrix Calculations 16 / 32

Determinants Change of basis Radboud University Nijmegen Matrices and basis transformations Why??? • Many find the idea of multiple bases confusing the first time around. • S = { (1 , 0) , (0 , 1) } is a perfectly good basis for R 2 . Why bother with others? 1 Some vector spaces don’t have one “obvious” choice of basis. Example: subspaces S ⊆ R n . 2 Sometimes it is way more efficient to write a vector with respect to a different basis, e.g.:     93718234 1 − 438203 1         110224 0     =     − 5423204980 0         . . . . . . S B 3 The choice of basis for vectors affects how we write matrices as well. Often this can be done cleverly. Example: JPEGs, MP3s, search engine rankings, ... A. Kissinger Version: autumn 2017 Matrix Calculations 17 / 32

Determinants Change of basis Radboud University Nijmegen Matrices and basis transformations Transforming bases, part I • Problem: given a vector written in B = { (100 , 0) , (100 , 1) } , how can we write it in the standard basis? Just use the definition: � x � � 100 � � 100 � � 100 x + 100 y � = x · + y · = 0 1 y y B S • Or, as matrix multiplication: � 100 100 � � x � � 100 x + 100 y � · = 0 1 y y � �� � ���� � �� � T B⇒S in basis B in basis S • Let T B⇒S be the matrix whose columns are the basis vectors B . Then T B⇒S transforms a vector written in B into a vector written in S . A. Kissinger Version: autumn 2017 Matrix Calculations 19 / 32