Lecture 2e Orthogonal Complement (pages )

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1 Lecture 2e Orthogonl Complement (pges -) We hve now seen tht n orthonorml sis is nice wy to descrie suspce, ut knowing tht we wnt n orthonorml sis doesn t mke one fll into our lp. In theory, the process for finding n orthonorml sis is esy. Strt with one vector, dd vector tht in the suspce tht is orthogonl to your first vector, then dd vector in the suspce tht is orthogonl to the first two vectors. The prolem comes when you strt to del with very lrge spces, nd you re trying to find vector tht is orthogonl to the 7 vectors tht cme efore it! (Especilly when you consider tht these vectors must hve t lest 8 entries.) On the fce of it, we lredy hve ll the techniques we need, since the vector eqution x v = is simply homogeneous liner eqution whose vriles re the entries of x. So, if we needed to mke x orthogonl to 7 vectors, we would e looking t solving system of 7 liner equtions. But tht would e relly ig mtrix to row reduce (ctully, it wouldn t most people who use these techniques del with wy more vriles nd equtions), so if we cn find n esier wy, we should! Our first step on this journey will e to dd some more definitions to our voculry. For, while we hve lredy defined wht it mens for one vector to e orthogonl to nother, we re now looking t hving our vector e orthogonl to SET of vectors. Definition: Let S e suspce of R n. We shll sy tht vector x is orthogonl to S if Exmple: Let S = Spn ecuse given ny element we see tht x s = for ll s S +. Then = =. We lso see tht is orthogonl to S, of Spn is orthogonl to S, since

2 It is lso esy to notice tht of S, ut =. But = = () = is orthogonl to S, since we lso hve tht is not orthogonl to S, since is n element =. (As usul, the est wy to show tht x is not orthogonl to suspce S is to find specific element s of S such tht x s.) Note tht in order to show tht x is orthogonl to suspce S, we only need to show tht x is orthogonl to the sis vectors for S. In fct, we cn show tht x is orthogonl to ny spnning set for S we don t need to worry out the independence of the vectors for this property. For, if A = {,..., k } is spnning for S, then every element s of S cn e written s s + + s k k for some s,..., s k R. And if x is orthogonl every i for i k, then we hve the following: x s = x (s + + s k k ) = ( x s ) + + ( x s k k ) = s ( x ) + + s k ( x k ) = s () + + s k () = We lso see tht, if x is orthogonl to S, then t x is orthogonl to S for ll sclrs t R, since (t x) s = t( x s) = t() =. Also, if oth x nd y re orthogonl to S, then so is x + y, since ( x + y) s = ( x s) + ( y s) = + =. If we dd to these fcts the fct tht is orthogonl to S, since v = for ny vector v, nd thus definitely for the ones in S, then we hve shown tht the set of ll vectors orthogonl to S is never the empty set. And this mens tht we hve shown tht the set of ll vectors orthogonl to S is itself suspce of R n, since it is non-empty suset of R n tht is closed under ddition nd sclr multipliction. We give this suset the following nme: 2

3 Definition: We cll the set of ll vectors orthogonl to S the orthogonl complement of S nd denote it S. Tht is S = { x R n x s = for ll s S} Exmple: Find sis for S, where S = Spn 2 6. Agin, vector is orthogonl to S if itis orthogonl to the vectors in its spnning x x set, so we re looking for vectors x = x 2 x such tht x 2 2 x = nd x x x x 2 x x system: 6 =. This is the sme s looking for solutions to the following x +2x 2 +x = x +6x 2 +x +x = To solve this homogeneous system, we row reduce its coefficient mtrix: [ ] [ ] From our RREF mtrix, we see tht our system is equivlent to x +2x 2 +x = x +x = Replcing the vrile x 2 with the prmeter s nd the vrile x with the prmeter t, we get tht the generl solution to this system is x x 2 x x = 2s t s t t = s 2 + t The generl solution toour system is list of ll the vectors x tht re orthogonl 2 to S, so we see tht, is spnning set for S. Moreover,

4 these vectors re not sclr multiple of ech other, nd thus re linerly 2 independent, so we hve tht, is sis for S. In our exmple, we were esily le to show tht our spnning set ws linerly independent ecuse it only hd two vectors in it. But wht if our spnning set ws igger? Would we need to row reduce mtrix to prove tht our spnning set is linerly independent? No! Becuse our sis for S cn lwys e thought of s the nullspce for mtrix, nd ck in Mth 6 we showed tht the technique for finding the spnning set for the nullspce in fct results in sis for the nullspce. And which mtrix re we finding the nullspce of? It is the mtrix whose ROWS re the vectors in the spnning set for our suset. NOT THE COLUMNS!!! I recommend going hed nd setting up the system of equtions using the defining dot products ( x s, x s 2, etc.) to void confusion on this mtter. But we will use this fct when we prove the following theorem. Theorem 7.2.: Let S e k-dimensionl suspce of R n. Then () S S = { } (2) dim(s ) = n k () If { v,..., v k } is n orthonorml sis for S nd { v k+,..., v n } is n orthonorml sis for S, then { v,..., v k, v k+,..., v n } is n orthonorml sis for R n. Proof of Theorem 7.2.: To see tht S S = { }, let x S S. Then x is n element of S, so x is orthogonl to every element of S. But we lso hve tht x is n element of S, so this mens tht x is orthogonl to itself. Tht is, x x =, which mens tht x =. (See Theorem...) Next, to see tht dim(s ) = n k, let A e the mtrix whose ROWS re the sis vectors of S. Then A is k n mtrix, nd S is the rowspce of A. This mens tht the rnk of A is the sme s the dimension of S, so rnk(a) = k. But we lso hve tht S is the nullspce of A, nd thus the dimension of S is the nullity of A. By the Rnk Theorem, we know tht rnk(a)+nullity(a) = n, so the dim(s ) = nullity(a) = n rnk(a) = n k. Finlly, to see tht { v,..., v k, v k+,..., v n } is n orthonorml sis for R n, rememer tht we in fct only need to show tht { v,..., v k, v k+,..., v n } is n orthonorml set (s it will then utomticlly e sis). Tht mens we need to show tht v i v j = whenever i j. We will rek this into four different scenrios: () i, j k. Then oth v i nd v j re in { v,..., v k }, which is n orthonorml set, so we know tht v i v j =.

5 () i k nd k + j n. Then v i S nd v j S, so y the definition of S we know tht v i v j =. (c) j k nd k + i n. Then v j S nd v i S, so y the definition of S we know tht v i v j =. (d) k + i, j n. Then oth v i nd v j re in { v k+,..., v n }, which is n orthonorml set, so we know tht v i v j =. 5