292 CHAPTER 11. INNER PRODUCT SPACES

4. Find the equation of the plane through the three points (1, 2, 3) , (2,−3, 1) , (1, 1, 7) .

5. Let T map a vector space V to itself. Explain why T is one to one if and only if T isonto. It is in the text, but do it again in your own words.

6. ↑Let all matrices be complex with complex field of scalars and let A be an n×n matrixand B a m×m matrix while X will be an n×m matrix. The problem is to considersolutions to Sylvester’s equation. Solve the following equation for X

AX −XB = C

where C is an arbitrary n×m matrix. Show there exists a unique solution if and onlyif σ (A)∩ σ (B) = ∅. Hint: If q (λ) is a polynomial, show first that if AX −XB = 0,then q (A)X − Xq (B) = 0. Next define the linear map T which maps the n × mmatrices to the n×m matrices as follows.

TX ≡ AX −XB

Show that the only solution to TX = 0 is X = 0 so that T is one to one if and only ifσ (A)∩σ (B) = ∅. Do this by using the first part for q (λ) the characteristic polynomial

for B and then use the Cayley Hamilton theorem. Explain why q (A)−1

exists if andonly if the condition σ (A) ∩ σ (B) = ∅.

7. Recall the Binet Cauchy theorem, Theorem 3.3.14. What is the geometric meaning ofthe Binet Cauchy theorem?

8. For W a subspace of V, W is said to have a complementary subspace [15] W ′ ifW ⊕W ′ = V. Suppose that both W,W ′ are invariant with respect to A ∈ L (V, V ).Show that for any polynomial f (λ) , if f (A)x ∈ W, then there exists w ∈ W suchthat f (A)x = f (A)w. A subspace W is called A admissible if it is A invariant andthe condition of this problem holds.

9. ↑ Return to Theorem 9.3.5 about the existence of a basis β ={βx1

, · · · , βxp

}for V

where A ∈ L (V, V ) . Adapt the statement and proof to show that ifW is A admissible,then it has a complementary subspace which is also A invariant. Hint:

The modified version of the theorem is: Suppose A ∈ L (V, V ) and the minimal poly-nomial of A is ϕ (λ)

mwhere ϕ (λ) is a monic irreducible polynomial. Also suppose

that W is an A admissible subspace. Then there exists a basis for V which is of

the form β ={βx1

, · · · , βxp, v1, · · · , vm

}where {v1, · · · , vm} is a basis of W . Thus

span(βx1

, · · · , βxp

)is the A invariant complementary subspace forW . You may want

to use the fact that ϕ (A) (V ) ∩W = ϕ (A) (W ) which follows easily because W is Aadmissible. Then use this fact to show that ϕ (A) (W ) is also A admissible.

10. Let U,H be finite dimensional inner product spaces. (More generally, complete innerproduct spaces.) Let A be a linear map from U to H. Thus AU is a subspace ofH. For g ∈ AU, define A−1g to be the unique element of {x : Ax = g} which isclosest to 0. Then define (h,g)AU ≡

(A−1g, A−1h

)U. Show that this is a well defined

inner product. Let U,H be finite dimensional inner product spaces. (More generally,complete inner product spaces.) Let A be a linear map from U to H. Thus AU is asubspace of H. For g ∈ AU, define A−1g to be the unique element of {x : Ax = g}which is closest to 0. Then define (h,g)AU ≡

(A−1g, A−1h

)U. Show that this is a

well defined inner product and that if A is one to one, then ∥h∥AU =∥∥A−1h

∥∥U

and∥Ax∥AU = ∥x∥U .