(II.D) There is exactly one rref matrix row-equivalent to any A

Transcription

1 IID There is exactly one rref matrix row-equivalent to any A Recall that A is row-equivalent to B means A = E M E B where E i are invertible elementary matrices which we think of as row-operations In ID to each matrix A we associated a welldefined row-equivalent matrix rre f A in reduced-row echelon form by means of an algorithmic sequence of row-operations E E N : writing E = E N E for their product we have R := rre f A := E A What if a different series of invertible row operations with product E yields an rre f matrix R = E A? Consider this mysterious example: A : I 3 }} R : E }} } } R : E The top row is our usual route while the vertical arrow deviates from it: E = E and yet R = R How can this be? If R = R then E A = E A or E EA = A and so E E = identity right? If A is invertible right; otherwise watch out! For example 0 a b 0 0 = c

2 2 IID THERE IS EXACTLY ONE RREF MATRIX ROW-EQUIVALENT TO ANY A In fact no matter what procedure you use you ll get the same R as we now prove DEFINITION If r j } m j= are the rows of an m n matrix A then the row space is V row A := span r r m } For P a k m matrix the rows of PA are m } k P ij r j j= i= which are linear combinations of the rows of A and thus in their span so If P is invertible then V row P A V row A V row PA V row P PA = V row A Since elementary row-operations are invertible row-equivalent matrices have the same row space Now take R = E A and R = E A any two rre f matrices row-equivalent to A Then V row R = V row A = V row R and the problem boils down to showing V row R = V row R = R = R for any two rre f matrices period DEFINITION 2 Given any rre f matrix R define KR := k N R has a leading in the k th column Clearly these are just the k < < k r indexing the pivot columns with r the rank of R Denote the rows of R by ρ R ρ m R only the first r are nonzero Then we have the following restatement of Theorem IIC4: }

3 IID THERE IS EXACTLY ONE RREF MATRIX ROW-EQUIVALENT TO ANY A 3 THEOREM 3 Let R be any m n rre f matrix and W = V row R Then KR = LW and ρ i R} r i= = Ψ i W} γ i= This says the rows of an rre f matrix are the standard basis of its row space and its rank is the dimension of its row space REMARK 4 An easy corollary of the last statement is that the nonzero rows of an rre f are always linearly independent this is not too hard to show by other means PROOF First observe that every row ρ of R is in W =row space of R If k KR then the k th column of R has a leading Let ρ be the row of R to which this leading belongs so that ρ k = is its first nonzero entry ρ = = ρ k = 0 Since also ρ W k must be in the leading set of W ie k LW We have just shown KR LW Write r for rankr = # of leading s = #KR} and recall that also r = # of nonzero rows of R Since r vectors span a space of dimension r dimv row R = dimspan ρ ρ r } r or dimw #KR} Now #LW} is the number of elements in a basis the standard basis! for W so dimw = #LW} and #KR} #LW} On the other hand as sets we showed KR LW above which = #KR} #LW} So #KR} = #LW} and the inclusion KR LW must be an equality of sets Therefore k < < k r } = l < < l γ } and r = γ or simply k i = l i for all i Now Ψ i W is the unique vector in the row space W = V row R of R with in the l i = k i th place and a 0 in all l j = k j th places for j = i But ρ i R gives such a vector since it has a leading in

4 4 IID THERE IS EXACTLY ONE RREF MATRIX ROW-EQUIVALENT TO ANY A the k th i place and 0 s in the places where the other leading s occur namely the k th j places j = i These are the same conditions By uniqueness of the vector satisfying them Ψ i = ρ i This concludes the proof of the Theorem So if V row R = W = V row R we apply the Theorem to each equality to get KR = LW = K R ρ i R} = Ψ i W} = ρ i R} But if the rows ρ i R and ρ i R of the two rre f matrices are the same they re the same matrix: R = R This concludes the proof of and of the title of this section! Now I give you a handful of vectors say 5 2 v = 3 v 6 2 = 7 v 3 = and ask you for the dimension of W = span v v 2 v 3 } and a basis to back it up fast! Then you perform the following magic trick: write A = rre f A = and voilá your basis is 0 0 ρ = ρ 2 = 2 2 3

5 IID THERE IS EXACTLY ONE RREF MATRIX ROW-EQUIVALENT TO ANY A 5 so that dimw = 2 Why does this work? Since A and rre f A are row-equivalent span v v 2 v 3 } = V row A = V row rre f A and by our theorem the nonzero entries of rre f A give the standard basis for V row rre f A Now recall from IIC that for W R n with standard basis Ψ i } and leading set LW any β W is of the form β = dimw i= β li Ψ i More generally according to our Theorem: if W = V row A and ρ i are the nonzero rows of rre f A then r 2 β = β ki ρ i i= If A = v v m this makes it very easy by taking rre f A and finding ρ i } to determine whether a given β is in span v v m } Does β satisfy 2? Then it is in the span If it is then it is easy to write β as a linear combination of the ρ i s: you know k i =columns of the leading s in rre f A and since you know β you know its k th i entries β ki On the other hand how do you write β as a linear combination of the original v i }? Either solve the inhomogeneous system α v + + α m v m = t A α = β ie v v m α α m = β together with the observation that Vrow rre f A = V row A

6 6 IID THERE IS EXACTLY ONE RREF MATRIX ROW-EQUIVALENT TO ANY A OR use ρ β = β ki ρ i ρ = β k β kr 0 0 m }} 0 m 0 = β k β kr 0 0 rre f A = β k β kr 0 0 EA A } v = β k β kr 0 0 EA v m which is a linear combination of the v i s Namely if we set α α m = β k β kr 0 0 EA then β = m i= α i v i Since this way of writing β as a linear combination of vi s as opposed to the first option the inhomogeneous system involves finding 2 EA it s useful primarily when you have more than one β and need a systematic way of dealing with them so that you don t rowreduce for each one Grassmanians We conclude with a geometric application in the same spirit as Pascal s hexagon Define G m R n = Grassmanian of m-planes in n-space to be the set of m -dimensional subspaces W R n thought of as a topological space That is each m -plane thru the origin is considered a point of G m R n ; alternately you can think of G m R n 2 it also depends on your choice of EA if A isn t invertible

7 IID THERE IS EXACTLY ONE RREF MATRIX ROW-EQUIVALENT TO ANY A 7 as a continuous manifold whose points are in -to- correspondence with m -planes We can use Theorem 3 to begin to see what these look like in particular to get a coordinate system on a big open set of them Since W is the row-space of a unique m n rre f matrix there is a - correspondence points of G m R n } rank m m n rre f matrices} W R If we write R R 2 R n where R = spanê } R 2 = spanê n ê n } R 3 = spanê n 2 ê n ê n } etc then the jumps for a given W in the sequence 0 dimw R dimw R 2 dimw R n = m correspond to the integers k i =columns in which leading s occur in R via dimw R n k i+ = + dimw R n k i So for the Grassmanian of 2 -planes in 4 -space the points correspond to row spaces of matrices of the form: where the s are the coordinates Since the first type of point has the most degrees of freedom it corresponds to a big open set of the same dimension as G 2 R 4 which in this case = 4 The remaining rre f s correspond to [points in] subsets of G 2 R 4 of lesser dimension More generally dimg m R n = n mm as we can see from the

8 8 IID THERE IS EXACTLY ONE RREF MATRIX ROW-EQUIVALENT TO ANY A m n rre f matrix 0 0 The most widely used of these is the case of projective n -space P n := G R n+ consisting of all lines considered as points in n + space Its importance is that it can be thought of as R n + directions at ; the points at are important for completing solution sets of polynomial equations to closed sets In any case the R n part is the rre f and the stuff at is all the other rre f s: In particular if n = 2 then the addition of a P at infinity to R 2 gives parallel lines a place to intersect! 3 More generally Bézout s Theorem says that if f x y and gx y are polynomials of degrees d and e respectively then the curves defined by f = 0 and g = 0 intersect in exactly de points provided one works in the complex projective plane CP 2 and counts an order m intersection point m times eg y = 0 and y = x 2 meet twice at 0 0 because they are tangent there Another way of thinking of P n : take the n -sphere solutions of x 2 + x x2 n+ = in Rn+ and identify all opposite points This makes RP just a circle wound up twice but RP 2 already hard to see 4 This is because it can t fit in R 3 without crossing itself! It 3 You may have noticed that a fix like this is needed for the Pascal construction to work when some pairs of lines are parallel 4 C looks like a sphere; if you remove the North pole you are left with the complex plane CP 2 has real dimension 4 so is harder to visualize

9 EXERCISES 9 can fit in R 4 More generally RP n for n a power of 2 can t fit inside R 2n!! Exercises Let V R 5 be the vector space spanned by the rows of the matrix A = a Find the standard basis Ψ i } of V b Given β V write out β = dimv i= β li Ψ i explicitly Express your answer in the form β = where some s are βli and the others are linear functions of the β li c Which of the following are in the row space: ? You should be able to do this without any additional work 2 For which y y 2 y 3 y 4 does the inhomogeneous linear system A x = y where A = have a solution? Determine this in two different ways which should give the same answer!: a by applying the rre f algorithm to the augmented matrix A y; and b by applying the rre f algorithm to t A then using criterion 2 for t y to be in its row space