91Ó°ÊÓ

For each of the following linear operators \(T\) on the vector space \(V\), determine whether the given subspace \(W\) is a \(T\)-invariant subspace of V. (a) \(\mathrm{V}=\mathrm{P}_{3}(R), \mathrm{T}(f(x))=f^{\prime}(x)\), and \(\mathrm{W}=\mathrm{P}_{2}(R)\) (b) \(\mathrm{V}=\mathrm{P}(R), \mathrm{T}(f(x))=x f(x)\), and \(\mathrm{W}=\mathrm{P}_{2}(R)\) (c) \(\mathrm{V}=\mathrm{R}^{3}, \mathrm{~T}(a, b, c)=(a+b+c, a+b+c, a+b+c)\), and \(\mathrm{W}=\\{(t, t, t): t \in R\\}\) (d) $\mathrm{V}=\mathrm{C}([0,1]), \mathrm{T}(f(t))=\left[\int_{0}^{1} f(x) d x\right] t$, and \(\mathrm{W}=\\{f \in \mathrm{V}: f(t)=a t+b\) for some \(a\) and \(b\\}\) (e) $\mathrm{V}=\mathrm{M}_{2 \times 2}(R), \mathrm{T}(A)=\left(\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right) A$, and \(\mathrm{W}=\left\\{A \in \mathrm{V}: A^{t}=A\right\\}\)

Step by step solution

01

Problem (a)

We have V = P3(R), T(f(x)) = f'(x), and W = P2(R). Let's take an arbitrary polynomial w(x) ∈ W, then w(x) has the form: \(w(x) = ax^2 + bx + c\), where a, b, c are constants. Now apply T to w(x): \(T(w(x)) = w'(x) = 2ax + b\) Since T(w(x)) results in a polynomial of degree at most 1, T(w(x)) still belongs to W. Therefore, W is a T-invariant subspace of V.

02

Problem (b)

We have V = P(R), T(f(x)) = x * f(x), and W = P2(R). Let's take an arbitrary polynomial w(x) ∈ W, then w(x) has the form: \(w(x) = ax^2 + bx + c\), where a, b, c are constants. Now apply T to w(x): \(T(w(x)) = x * w(x) = ax^3 + bx^2 + cx\) Since T(w(x)) results in a polynomial of degree 3, T(w(x)) does not belong to W. Therefore, W is not a T-invariant subspace of V.

03

Problem (c)

We have V = R^3, T(a, b, c) = (a+b+c, a+b+c, a+b+c), and W = {(t, t, t): t ∈ R}. Let's take an arbitrary vector w ∈ W, then w has the form: \(w = (t, t, t)\), where t is a constant. Now apply T to w: \(T(w) = (t+t+t, t+t+t, t+t+t) = (3t, 3t, 3t)\) Since T(w) is still in the form (k, k, k) with k = 3t, T(w) belongs to W. Therefore, W is a T-invariant subspace of V.

04

Problem (d)

We have V = C([0,1]), T(f(t)) = \(\left[\int_{0}^{1} f(x) d x\right] t\), and W = {f ∈ V: f(t) = at + b for some a and b}. Let's take an arbitrary function w ∈ W, then w has the form: \(w(t) = at + b\), where a and b are constants. Now apply T to w(t): \(T(w(t)) = \left[\int_{0}^{1} (at + b) dx\right] t = \left[\frac{a}{2} + b\right] t\) Since T(w(t)) is still in the form (kt) with k = a/2 + b, T(w(t)) belongs to W. Therefore, W is a T-invariant subspace of V.

05

Problem (e)

We have V = M2x2(R), T(A) = \(\left(\begin{array}{ll}0 & 1 \\\ 1 & 0\end{array}\right) A\), and W = {A ∈ V: A^t = A}. Let's take an arbitrary matrix w ∈ W, then w has the form: \(w = \left(\begin{array}{ll}a & b \\\ b & c\end{array}\right)\), where a, b, c are constants. Now apply T to w: \(T(w) = \left(\begin{array}{ll}0 & 1 \\\ 1 & 0\end{array}\right) \left(\begin{array}{ll}a & b \\\ b & c\end{array}\right) = \left(\begin{array}{ll}b & c \\\ a & b\end{array}\right)\) Now we need to check if T(w) is symmetric: \[\left(\begin{array}{ll}b & c \\\ a & b\end{array}\right)^t = \left(\begin{array}{ll}b & a \\\ c & b\end{array}\right)\] Since T(w)^t ≠ T(w), W is not a T-invariant subspace of V.

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