13.7. EXERCISES 353

and you note that in Jk, the diagonal entries are of the form λk for λ an eigenvalueof A. Also J = D +N where N is nilpotent and commutes with D. Argue then that

∞∑k=0

tkJk

k!

is an upper triangular matrix which has on the diagonal the expressions eλt whereλ ∈ σ (A) . Thus conclude

σ (Ψ (t)) ⊆ exp (tσ (A))

Next take etλ ∈ exp (tσ (A)) and argue it must be in σ (Ψ (t)) . You can do this asfollows:

Ψ (t)− etλI =

∞∑k=0

tkAk

k!−

∞∑k=0

tkλk

k!I =

∞∑k=0

tk

k!

(Ak − λkI

)

=

 ∞∑k=0

tk

k!

k−1∑j=1

Ak−jλj

 (A− λI)

Now you need to argue∞∑k=0

tk

k!

k−1∑j=1

Ak−jλj

converges to something in L (Rn,Rn). To do this, use the ratio test and Lemma 13.4.2after first using the triangle inequality. Since λ ∈ σ (A) , Ψ(t)− etλI is not one to oneand so this establishes the other inclusion. You fill in the details. This theorem is aspecial case of theorems which go by the name “spectral mapping theorem”.

19. Suppose Ψ (t) ∈ L (V,W ) where V,W are finite dimensional inner product spaces andt→ Ψ(t) is continuous for t ∈ [a, b]: For every ε > 0 there there exists δ > 0 such thatif |s− t| < δ then ||Ψ(t)−Ψ(s)|| < ε. Show t → (Ψ (t) v, w) is continuous. Here it isthe inner product in W. Also define what it means for t → Ψ(t) v to be continuousand show this is continuous. Do it all for differentiable in place of continuous. Nextshow t→ ||Ψ(t)|| is continuous.

20. If z (t) ∈W, a finite dimensional inner product space, what does it mean for t→ z (t)to be continuous or differentiable? If z is continuous, define∫ b

a

z (t) dt ∈W

as follows. (w,

∫ b

a

z (t) dt

)≡∫ b

a

(w, z (t)) dt.

Show that this definition is well defined and furthermore the triangle inequality,∣∣∣∣∣∫ b

a

z (t) dt

∣∣∣∣∣ ≤∫ b

a

|z (t)| dt,

and fundamental theorem of calculus,

d

dt

(∫ t

a

z (s) ds

)= z (t)

hold along with any other interesting properties of integrals which are true.

13.7. EXERCISES 35319.20.and you note that in J*, the diagonal entries are of the form \* for \ an eigenvalueof A. Also J= D+ N where N is nilpotent and commutes with D. Argue then thatoS tk y*k)k=0is an upper triangular matrix which has on the diagonal the expressions e** where € 0 (A). Thus concludeo (W (t)) exp (ta (A))Next take e € exp(ta(A)) and argue it must be in o(W(t)). You can do this asfollows:= ea es1=S5 (4" — X*1)k=0 k=0 k=0= - A AI) | (A— XI)U(t)—e*l =Now you need to argueoo pk ke 1SEL ayk= of “ j=1converges to something in £ (R", R”). To do this, use the ratio test and Lemma 13.4.2after first using the triangle inequality. Since \ € a (A), V(t) —e™T is not one to oneand so this establishes the other inclusion. You fill in the details. This theorem is aspecial case of theorems which go by the name “spectral mapping theorem”.Suppose W(t) € £(V,W) where V, W are finite dimensional inner product spaces andt + W(t) is continuous for t € [a,b]: For every ¢ > 0 there there exists 5 > 0 such thatif |s —t| < 6 then ||W (¢) — U(s)|| < ©. Show t > (W(t) v, w) is continuous. Here it isthe inner product in W. Also define what it means for t > W(t) v to be continuousand show this is continuous. Do it all for differentiable in place of continuous. Nextshow t > ||W (¢)|| is continuous.If z(t) € W, a finite dimensional inner product space, what does it mean for t > z (t)to be continuous or differentiable? If z is continuous, define[swaew(« [ 0a) =[ (w, z (t)) dt.Show that this definition is well defined and furthermore the triangle inequality,[wa < [\e(olaeand fundamental theorem of calculus,£([ tas) = z(t)hold along with any other interesting properties of integrals which are true.as follows.