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Com 40 Linear Algebra II Spectral Theory and Abstract Vector Spaces Spectral Theory Next let i = k = 3 and obtain that T looks like a diagonal matrix in so far as the first 3 rows and columns are concerned. Continuing in this way, it follows T is a diagonal matrix. 11 Let A be a normal matrix. Then there exists a unitary matrix U such that U ∗ AU is a diagonal matrix. 4 there exists a unitary matrix U such that U ∗ AU equals an upper triangular matrix. The theorem is now proved if it is shown that the property of being normal is preserved under unitary similarity transformations.

Let z′ = Az and choose c such that z (t0 ) = Φ (t0 ) c Then both z (t) , Φ (t) c solve x′ = Ax, x (t0 ) = z (t0 ) Apply uniqueness to conclude z = Φ (t) c. Finally, consider that Φ (t) c for c ∈ Fn −1 either is the general solution or it is not the general solution. If it is, then Φ (t) −1 exists for all t. If it is not, then Φ (t) cannot exist for any t from what was just shown. 41. Let Φ′ (t) = AΦ (t) . Then Φ (t) is called a fundamental matrix if Φ (t) t. 25) and it is given by the formula x (t) = Φ (t) Φ (t0 ) −1 x0 + Φ (t) ∫ t Φ (s) −1 f (s) ds t0 Now these few problems have done virtually everything of significance in an entire undergraduate differential equations course, illustrating the superiority of linear algebra.

Therefore, K= f (n+1) (ξ) f (n+1) (ξ) = (n + 1) n! (n + 1)! and the formula is true for n. The following is a special case and is what will be used. 4 Let h : (−δ, 1 + δ) → R have m+1 derivatives. Then there exists t ∈ [0, 1] such that m ∑ h(k) (0) h(m+1) (t) h (1) = h (0) + + . k! (m + 1)! k=1 Now let f : U → R where U ⊆ R and suppose f ∈ C m (U ) . Let x ∈ U and let r > 0 be such that B (x,r) ⊆ U. n Then for ||v|| < r, consider f (x+tv) − f (x) ≡ h (t) for t ∈ [0, 1] . Then by the chain rule, h′ (t) = n n ∑ n ∑ ∑ ∂f ∂2f (x + tv) vk , h′′ (t) = (x + tv) vk vj ∂xk ∂xj ∂xk j=1 k=1 k=1 Then from the Taylor formula stopping at the second derivative, the following theorem can be obtained.

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