366 CHAPTER 13. MATRICES AND THE INNER PRODUCT

Now this means(AkBl (A−B)x,(A−B)x

)= 0 for all k+ l = m− 1 since the sum of

such terms is 0 and each of them is nonnegative. Now this implies(√AkBl (A−B)x,

√AkBl (A−B)x

)= 0

and so√

AkBl (A−B)x = 0⇒ AkBl (A−B)x = 0,k+ l = m−1. Then, using the binomialtheorem,

0 =

=(A−B)m−1

m−1

∑j=0

(m−1

j

)Am−1− jB j (−1) j (A−B)x = (A−B)m x

This clearly implies A = B. To see this, consider m = 7.If m = 7,(A−B)7 x = 0 so (A−B)8 x = 0 so

((A−B)4 x,(A−B)4 x

)= 0 so it follows

that (A−B)4 x = 0 so ((A−B)2 x,(A−B)2 x

)= 0

so((A−B)2 x,x

)= 0 so ((A−B)x,(A−B)x) = 0 so (A−B) = 0. ■

13.13 Spectral Theory of Self Adjoint OperatorsFirst is some notation which may be useful since it will be used in the following presenta-tion.

Definition 13.13.1 Let X ,Y be inner product space and let u∈Y,v∈X . Then define u⊗v∈L (X ,Y ) as follows.

u⊗ v(w)≡ (w,v)u

where (w,v) is the inner product in X. Then this is clearly linear. That it is continuousfollows right away from

|(w,v)u| ≤ |u|Y |w|X |v|Xand so

sup|w|X≤1

|u⊗ v(w)|Y ≤ |u|Y |v|X

Sometimes this is called the tensor product, although much more can be said about thetensor product.

Note how this is similar to the rank one transformations used to consider the dimensionof the space L (V,W ) in Theorem 5.1.4. This is also a rank one transformation but herethere is no restriction on the dimension of the vector spaces although, as usual, the interestis in finite dimensional spaces. In case you have {v1, · · · ,vn} an orthonormal basis for V and{u1, · · · ,um} an orthonormal basis for Y, (or even just a basis.) the linear transformationsui⊗ v j are the same as those rank one transformations used before in the above theoremand are a basis for L (V,W ). Thus for A = ∑i, j ai jui⊗ v j, the matrix of A with respect tothe two bases has its i jth entry equal to ai j. This is stated as the following proposition.

Proposition 13.13.2 Suppose {v1, · · · ,vn} is an orthonormal basis for V and {u1, · · · ,um}is a basis for W. Then if A ∈L (V,W ) is given by A = ∑i, j ai jui⊗ v j, then the matrix of Awith respect to these two bases is an m×n matrix whose i jth entry is ai j.

366 CHAPTER 13. MATRICES AND THE INNER PRODUCTNow this means (A‘B! (A — B).x,(A—B)x) =0 for all k+/ =m-—1 since the sum ofsuch terms is 0 and each of them is nonnegative. Now this implies(vat! (A—B)x, VAKBI (A — B)x) =0and so VA‘B! (A — B)x = 0 = A‘B! (A —B)x =0,k +1 =m-—1. Then, using the binomialtheorem,=(A—B)""!mi m—1 foe .0 F(a ial —ays= 4-2)"jo \ JThis clearly implies A = B. To see this, consider m = 7.If m=7,(A—B)'x=0s0 (A—B)*x =0 so ((A —B)*x,(A ~B)'x) = 0 so it followsthat (A —B)*x=0 so((4 —B)?x,(A — Bx) =0$0 ((A-B)*x,x) =0 so ((A—B).x,(A—B)x) =0 so (A—B) =0.13.13 Spectral Theory of Self Adjoint OperatorsFirst is some notation which may be useful since it will be used in the following presenta-tion.Definition 13.13.1 Let X,Y be inner product space and letu € Y,v € X. Then define u®v €L (X,Y) as follows.u®v(w) =(w,v)uwhere (w,v) is the inner product in X. Then this is clearly linear. That it is continuousfollows right away from\(w,v) ul < July [wy |vlyand sosup |u@v(w)|y < |uly |vlx|wly SlSometimes this is called the tensor product, although much more can be said about thetensor product.Note how this is similar to the rank one transformations used to consider the dimensionof the space £(V,W) in Theorem 5.1.4. This is also a rank one transformation but herethere is no restriction on the dimension of the vector spaces although, as usual, the interestis in finite dimensional spaces. In case you have {v1,--- ,¥,} an orthonormal basis for V and{u1,+++ Um} an orthonormal basis for Y, (or even just a basis.) the linear transformationsu; ® vj; are the same as those rank one transformations used before in the above theoremand are a basis for “(V,W). Thus for A = ¥); ; a;ju;@vj, the matrix of A with respect tothe two bases has its ij” entry equal to a; j- This is stated as the following proposition.Proposition 13.13.2 Suppose {v1,--- ,v,} is an orthonormal basis for V and {uy,--+ ,Um}is a basis for W. Then if A € £2 (V,W) is given by A = J; ;ajjuj @ vj, then the matrix of Awith respect to these two bases is an m Xn matrix whose ij” entry is aij.