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4.1 RANK of a MATRIX Rank List Given Matrix M, the Following Are Equal

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4.1 RANK of a MATRIX Rank List Given Matrix M, the Following Are Equal page 1 of Section 4.1 CHAPTER 4 MATRICES CONTINUED SECTION 4.1 RANK OF A MATRIX rank list Given matrix M, the following are equal: (1) maximal number of ind cols (i.e., dim of the col space of M) (2) maximal number of ind rows (i.e., dim of the row space of M) (3) number of cols with pivots in the echelon form of M (4) number of nonzero rows in the echelon form of M You know that (1) = (3) and (2) = (4) from Section 3.1. To see that (3) = (4) just stare at some echelon forms. This one number (that all four things equal) is called the rank of M. As a special case, a zero matrix is said to have rank 0. how row ops affect rank Row ops don't change the rank because they don't change the max number of ind cols or rows. example 1 12-10 24-20 has rank 1 (maximally one ind col, by inspection) 48-40 000 []000 has rank 0 example 2 2540 0001 LetM= -2 1 -1 0 21170 To find the rank of M, use row ops R3=R1+R3 R4=-R1+R4 R2ØR3 R4=-R2+R4 2540 0630 to get the unreduced echelon form 0001 0000 Cols 1,2,4 have pivots. So the rank of M is 3. how the rank is limited by the size of the matrix IfAis7≈4then its rank is either 0 (if it's the zero matrix), 1, 2, 3 or 4. The rank can't be 5 or larger because there can't be 5 ind cols when there are only 4 cols to begin with. example 3 IfMis7≈6andhasrank 4 then the 6 cols are dep, any set of 5 cols is dep, some foursome of cols is ind. Similarly, the 7 rows are dep, any 6 rows are dep, any 5 rows are dep, some set of 4 rows is ind. page 2 of Section 4.1 example 4 IfAis6≈7andthe6rows are dependent then the rank is 0,1,2,3,4 or 5. invertible rule LetMbeann≈nmatrix. The following are equivalent; i.e., either all are true or all are false. (1) M is invertible (nonsingular). (2) |M| ≠ 0. (3) Echelon form of M is I. (4) Rows of M are independent. (5) Cols of M are independent. (6) Rank of M is n. proof that (6) belongs on the list Item (6) is equivalent to (5) because for ann≈nmatrix, ind cols means maximally n ind cols which means the rank is n. order of a determinant If a matrix isn≈nthen its determinant is said to have order n. subdeterminants The fat dots in Fig 1 illustrate the idea of3≈3submatrix. The determinant of the fat dots is a subdeterminant of order 3 as opposed to the original det which is of order 5. A5≈5matrix has lots of subdets of orders 4,3,2,1 and has one subdet (the whole thing) of order 5. …ø…øø …………… …………… …ø…øø …ø…øø FIG 1 rank list Given a matrix M. The following are all equal and are called the rank of M: (1) maximal number of ind cols (i.e., dim of the col space of M) (2) maximal number of ind rows (i.e., dim of the row space of M) (3) number of cols with pivots in the echelon form of M (4) number of nonzero rows in the echelon form of M (5) largest order of all the nonzero subdeterminants of M The proof that (5) belongs on the list is omitted because it gets messy. example 5 Ifa5≈7matrix has rank 3 then every5≈5subdeterminant is 0, every4≈4 subdeterminant is 0 and at least one (but not necessarily all) of the3≈3 subdeterminants is not 0. page 3 of Section 4.1 PROBLEMS FOR SECTION 4.1 1. Find the rank 115 2 1000 3 23456 013 0100 (a) (b) (c) 2519 (d) 4 []00000 0001 5 -1 1 1 0036 3418 2. Suppose M has 9 cols and its rank is 5. Decide, if possible, whether the following are ind or dep. (a) first six cols of M (b) first five cols of M (c) first four cols of M 3. Suppose M is7≈9 (7rows, 9 cols). What can you conclude about the rank of M if (a) that's all the information you have (b)row3=2row7+8row4 (c) first five rows are ind (d) first 5 rows are ind and every set of 6 rows is dep. (e) first 7 cols are ind (f) rows 1 and 2 are ind, rows 1,2,3 are dep 4. Suppose the echelon form of M is 120 001 000 What can you conclude about |M| and about subdeterminants of M. 5. Suppose rankM=5.Find rank MT. 6. Suppose rankA=6andAisinvertible. Find the rank of A-1. 7.IfrankB=2find rank(3B). 8. Let t √20 M= √ 2t√ 2 0 √ 2t Find the rank of M (it will depend on t). 9.Mis5≈5.What can you conclude about the rank of M if (a)oneofits3≈3subdeterminants is 6 (b)the3≈3subdeterminant in the NE corner is 0 and the3≈3subdet in the SW corner is 8 (c)oneofits3≈3subdets is 0, another of its3≈3subdets is 8 and all the4≈4 subdets are 0 (d)allthe3≈3subdets are 0 (e)oneofits3≈3subdets is 0 page 1 of Section 4.2 SECTION 4.2 ORTHOGONAL MATRICES orthogonal matrix rule Let M be a (real)n≈nmatrix. The following are equivalent (i.e., if any one of the three holds then they all hold, if any one fails then they all fail). (1) M-1 =MT, i.e., MMT = I; i.e. MTM=I. n (2) Cols of M are orthonormal vectors in R (orthogonal and unit length). (3) Rows of M are orthonormal vectors in Rn. A matrix with these properties is called orthogonal. Note that, for a square matrix, this rule says all of the following (and then some): If the rows are orthonormal then the cols are also orthonormal. If the cols are orthonormal then the rows are also orthonormal. If the rows are orthonormal then the inverse of the matrix is easy to find; it's the transpose. If the cols are orthonormal then the inverse of the matrix is easy to find; it's the transpose. warning Items (2) and (3) in the orthogonal matrix rule involve orthonormal rows and cols, not just orthogonal rows and cols. proof that (1) is equivalent to (3) Suppose M is3≈3.Letu,v,w be the rows of M so that u,v,w are the cols of MT. then u…u u…v u…w MMT = v…u v…v v…w w…u w…v w…w So MMT = I iff u…u = v…v = w…w = 1 and u…v = u…w = v…w = 0 iff u,v,w are orthog and «u« = «v« = «w« = 1 iff u,v,w are orthonormal proof that (1) is equivalent to (2) This can be done similarly by letting p,q,r be the cols of M and considering MTM instead of MMT. example 1 Let 1/√61/√62/√6 M= -2/√50 1/√5 1/√30 -5√30 2/√30 By inspection, the rows are unit vectors (i.e., «row 1« = «row 2« = «row 3« = 1. And the rows are orthogonal (row 1…row 2 = row 1…row 3 = row 2…row3=0).SoMisan orthogonal matrix. Its cols are orthonormal and M-1 =MT. the basis changing matrix If a new basis for Rn is orthonormal then the basis changing matrix P has orthonormal cols so it is an orthogonal matrix and easy to invert; namely P-1 =PT. For example, if the axes in R2 are rotated by 32o then the new orthonormal basis vectors are u = ( cos 32o, sin 32o) v = (-sin 32o, cos 32o). page 2 of Section 4.2 And cos 32o -sin 32o P= sin 32o cos 32o cos 32o sin 32o P-1 =PT = -sin 32o cos 32o determinant of an orthogonal matrix (1) If M is orthogonal then |M|=–1. This also means that if|M| ≠ –1 then M can't be orthogonal. (2) But orthogonal matrixes are not the only matrices whose dets are –1; in other words, if |M| = –1 then M is not necessarily an orthogonal matrix. proof of (1) Suppose A is orthogonal. Then AAT =I |AAT|=1 |A||AT|=1 |A|2 =1 |A|=–1 proof of (2) 41 Here's a counterexample. Let M = []31 . Then detM=1butMisnotorthogonal. warning repeated A matrix with merely orthogonal cols is a nothing. An orthogonal matrix has to have orthonormal cols. how to show that a matrix is orthogonal method 1 Look at its rows (or cols) to see if they are orthonormal. method 1 Multiply the matrix by its transpose to see if the product is I. how to show that a matrix is not orthogonal method 1 Look at its rows (or cols) to see that they are not orthonormal. method 2 Or show that the matrix times its transpose is not I. method 3 Find the det. If the det is not –1 then the matrix is not orthog. (But if the det is – 1 then you have no conclusion.) example 2 Show that if M is orthogonal then -M is also orthogonal. solution method 1 If M is orthogonal then its col vectors are orthonormal. Intuitively, multiplying by -1 reverses the direction of an arrow but doesn't change it length, so the cols of -M are still orthogonal and still have unit length.
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