3 For each of the following linear operators T on a vector space V test T for diagonalizability and if T is diagonalizable find a basis for V such that T is a diagonal matrix (a) V= P3(R) and T is defined by T(f(x))=f (x)+f (x) (b) V= P2(R) and T is defined by T ft(a x2+b x+c )=c x2+b x+a Sec 52 Diagonalizability 279 (c) V= R3 and T is defined by T ft( l a1 a2 a3 )= ft( r a2 -a1 2 a3 ) (d) V= P2(R) and T is defined by T(f(x))=f(0)+f(1) ft(x+x2 ) (e) V= C2 and T is defined by T(z w)=(z+i w i z+w) (f) V= M2 2(R) and T is defined by T(A)=At

Question

3 For each of the following linear operators T on a vector space V test T for diagonalizability and if T is diagonalizable find a basis  for V such that T is a diagonal matrix (a) V= P3(R) and T is defined by T(f(x))=f (x)+f (x) (b) V= P2(R) and T is defined by T ft(a x2+b x+c )=c x2+b x+a Sec 52 Diagonalizability 279 (c) V= R3 and T is defined by T ft( l a1 a2 a3 )= ft( r a2 -a1 2 a3 )  (d) V= P2(R) and T is defined by T(f(x))=f(0)+f(1) ft(x+x2 ) (e) V= C2 and T is defined by T(z w)=(z+i w i z+w) (f) V= M2 2(R) and T is defined by T(A)=At

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Solution By Steps

Part (a): $V=P_3(\mathbb{R}), T(f(x))=f'(x)+f''(x)$

Step 1: Standard Basis and Matrix Representation  
Standard basis for $P_3(\mathbb{R})$: $1, x, x^2, x^3$.  
Apply $T$ to each basis vector:  
- $T(1) = 0$  
- $T(x) = 1$  
- $T(x^2) = 2x$  
- $T(x^3) = 3x^2$

Matrix of $T$ in standard basis:  
$$
\left(\begin{array}{cccc}
0 & 1 & 0 & 0 \
0 & 0 & 2 & 0 \
0 & 0 & 0 & 3 \
0 & 0 & 0 & 0
\end{array}ight)
$$

Step 2: Diagonalizability Check  
The matrix is not diagonalizable because it is in Jordan form with non-diagonal blocks.

Part (b): $V=P_2(\mathbb{R}), T(ax^2+bx+c)=cx^2+bx+a$

Step 1: Standard Basis and Matrix Representation  
Standard basis for $P_2(\mathbb{R})$: $1, x, x^2$.  
Matrix of $T$ in standard basis:  
$$
\left(\begin{array}{ccc}
0 & 0 & 1 \
0 & 1 & 0 \
1 & 0 & 0
\end{array}ight)
$$

Step 2: Diagonalizability Check  
The matrix is symmetric, hence diagonalizable. Eigenvalues are $1$ and $-1$.  
Basis for diagonalization: $(1, 0, 1), (0, 1, 0), (1, 0, -1)$.

Part (c): $V=\mathbb{R}^3, T\left(\begin{array}{l}a_{1} \ a_{2} \ a_{3}\end{array}ight)=\left(\begin{array}{r}a_{2} \ -a_{1} \ 2a_{3}\end{array}ight)$

Step 1: Matrix Representation  
Matrix of $T$:  
$$
\left(\begin{array}{ccc}
0 & 1 & 0 \
-1 & 0 & 0 \
0 & 0 & 2
\end{array}ight)
$$

Step 2: Diagonalizability Check  
The matrix is diagonalizable. Eigenvalues are $2, i, -i$.  
Basis for diagonalization: $(0, 0, 1), (1, i, 0), (1, -i, 0)$.

Part (d): $V=P_2(\mathbb{R}), T(f(x))=f(0)+f(1)(x+x^2)$

Step 1: Standard Basis and Matrix Representation  
Standard basis for $P_2(\mathbb{R})$: $1, x, x^2$.  
Matrix of $T$ in standard basis:  
$$
\left(\begin{array}{ccc}
1 & 1 & 1 \
0 & 1 & 1 \
0 & 1 & 1
\end{array}ight)
$$

Step 2: Diagonalizability Check  
The matrix is not diagonalizable due to its rank and structure.

Part (e): $V=\mathbb{C}^2, T(z, w)=(z+iw, iz+w)$

Step 1: Matrix Representation  
Matrix of $T$:  
$$
\left(\begin{array}{cc}
1 & i \
i & 1
\end{array}ight)
$$

Step 2: Diagonalizability Check  
The matrix is diagonalizable. Eigenvalues are $1+i$ and $1-i$.  
Basis for diagonalization: $(1, i), (i, 1)$.

Part (f): $V=M_{2 	imes 2}(\mathbb{R}), T(A)=A^t$

Step 1: Matrix Representation and Diagonalizability Check  
Every matrix is its own transpose, so $T$ is the identity operation.  
All matrices are diagonalizable since $T$ does not change them.

Final Answer
- (a) Not diagonalizable.
- (b) Diagonalizable; basis: $(1, 0, 1), (0, 1, 0), (1, 0, -1)$.
- (c) Diagonalizable; basis: $(0, 0, 1), (1, i, 0), (1, -i, 0)$.
- (d) Not diagonalizable.
- (e) Diagonalizable; basis: $(1, i), (i, 1)$.
- (f) Diagonalizable; every matrix is its own basis.

Key Concept
Diagonalizability

Key Concept Explanation
A linear operator or matrix is diagonalizable if there exists a basis of the vector space consisting of its eigenvectors. This property simplifies many operations, such as computing powers of matrices, by working with diagonal matrices.

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Solution By Steps

Part (a) $V=P_3(\mathbb{R})$ and $T(f(x))=f'(x)+f''(x)$

Step 1: Basis and Matrix Representation  
Choose a basis for $P_3(\mathbb{R})$, say $\beta = \{1, x, x^2, x^3\}$. Apply $T$ to each basis vector.

Step 2: Apply $T$  
- $T(1) = 0$
- $T(x) = 1$
- $T(x^2) = 2x + 2$
- $T(x^3) = 3x^2 + 6x$

Step 3: Matrix of $T$  
The matrix of $T$ relative to $\beta$ is:
$$
\left(\begin{array}{cccc}
0 & 1 & 2 & 0 \
0 & 0 & 2 & 6 \
0 & 0 & 0 & 6 \
0 & 0 & 0 & 0
\end{array}ight)
$$

Step 4: Diagonalizability  
This matrix is not diagonalizable as it is not possible to find 4 linearly independent eigenvectors.

Part (b) $V=P_2(\mathbb{R})$ and $T(ax^2+bx+c)=cx^2+bx+a$

Step 1: Basis and Matrix Representation  
Choose a basis for $P_2(\mathbb{R})$, say $\beta = \{1, x, x^2\}$. Apply $T$ to each basis vector.

Step 2: Apply $T$  
- $T(1) = x^2$
- $T(x) = x$
- $T(x^2) = 1$

Step 3: Matrix of $T$  
The matrix of $T$ relative to $\beta$ is:
$$
\left(\begin{array}{ccc}
0 & 0 & 1 \
0 & 1 & 0 \
1 & 0 & 0
\end{array}ight)
$$

Step 4: Diagonalizability  
This matrix is diagonalizable as it has 3 distinct eigenvalues, which are the roots of its characteristic polynomial.

Part (c) $V=\mathbb{R}^3$ and $T\left(\begin{array}{l}a_{1}\a_{2}\a_{3}\end{array}ight)=\left(\begin{array}{r}a_{2}\-a_{1}\2a_{3}\end{array}ight)$

Step 1: Matrix of $T$  
The matrix of $T$ is:
$$
\left(\begin{array}{ccc}
0 & 1 & 0 \
-1 & 0 & 0 \
0 & 0 & 2
\end{array}ight)
$$

Step 2: Diagonalizability  
This matrix is diagonalizable as it has 3 distinct eigenvalues, $i$, $-i$, and $2$.

Part (d) $V=P_2(\mathbb{R})$ and $T(f(x))=f(0)+f(1)(x+x^2)$

Step 1: Basis and Matrix Representation  
Choose a basis for $P_2(\mathbb{R})$, say $\beta = \{1, x, x^2\}$. Apply $T$ to each basis vector.

Step 2: Apply $T$  
- $T(1) = 1 + (x + x^2)$
- $T(x) = 0 + (1 + x + x^2)$
- $T(x^2) = 0 + (1 + 2x + x^2)$

Step 3: Matrix of $T$  
The matrix of $T$ relative to $\beta$ is:
$$
\left(\begin{array}{ccc}
1 & 0 & 0 \
1 & 1 & 2 \
1 & 1 & 1
\end{array}ight)
$$

Step 4: Diagonalizability  
This matrix is not straightforwardly diagonalizable without further analysis due to its complex structure.

Part (e) $V=\mathbb{C}^2$ and $T(z, w)=(z+iw, iz+w)$

Step 1: Matrix of $T$  
The matrix of $T$ is:
$$
\left(\begin{array}{cc}
1 & i \
i & 1
\end{array}ight)
$$

Step 2: Diagonalizability  
This matrix is diagonalizable as it has distinct eigenvalues, which can be found by solving its characteristic polynomial.

Part (f) $V=M_{2 	imes 2}(\mathbb{R})$ and $T(A)=A^t$

Step 1: Matrix of $T$  
Consider the standard basis for $M_{2 	imes 2}(\mathbb{R})$. The action of $T$ on this basis shows that $T$ is the identity operation for symmetric matrices and swaps elements for asymmetric parts.

Step 2: Diagonalizability  
Since $T$ essentially acts as the identity on a subspace of symmetric matrices and swaps elements on asymmetric matrices, it is diagonalizable.

Final Answer
- (a) Not diagonalizable.
- (b) Diagonalizable.
- (c) Diagonalizable.
- (d) Requires further analysis for diagonalizability.
- (e) Diagonalizable.
- (f) Diagonalizable.

Key Concept
Diagonalizability

Key Concept Explanation
A linear operator or matrix is diagonalizable if there exists a basis of the vector space consisting entirely of its eigenvectors. This property simplifies many operations, such as computing powers of matrices, by working with diagonal matrices.

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