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1. Solution: It is true. Consider the standard orthonormal basis $e_1,e_2,e_3$ of $\mb R^3$.

Define $T\in \ca L(\mb R^3)$ by the rule:\[Te_1=e_1,\quad Te_2=2e_2+e_1,\quad Te_3=3e_3.\]Since we have\[\langle Te_1,e_2\rangle =\langle e_1,e_2\rangle =0,\]\[\langle e_1,Te_2\rangle =\langle e_1,e_1+2e_2\rangle =1,\]it follows that $T$ is not self-adjoint.

On the other hand, it is easy to see that $e_1$, $e_2+e_1$, $e_3$ is a basis consisting eigenvectors of $T$.

Actually, $T$ is also not normal. Indeed, one can get $T^*e_1=e_1+e_2$, then\[\|Te_1\|=1,\quad \|T^*e_1\|=\|e_1+e_2\|=2.\]Hence by 7.20, $T$ is not normal.

2. Solution: Since $T$ is self-adjoint, it follows from 7.23 that there exists an orthonormal basis $e_1,\dots,e_n$ of $V$. Note that $2$ and $3$ are the ony eigenvalues of $T$, hence for any given $i\in \{1,\cdots,n\}$ we have $$(T-2I)e_i=0\quad \text{or}\quad (T-3I)e_i=0.$$Note that $$T^2-5T+6I=(T-2I)(T-3I)=(T-3I)(T-2I),$$one has\[(T^2-5T+6I)e_i=(T-2I)(T-3I)e_i=(T-3I)(T-2I)e_i.\]Because $(T-2I)e_i=0$ or $(T-3I)e_i=0$ for any given $i\in \{1,\cdots,n\}$, we have\[(T^2-5T+6I)e_i=0\] for any given $i\in \{1,\cdots,n\}$. Therefore $T^2-5T+6I=0$.

3. Solution: Let $e_1$, $e_2$, $e_3$ be the standard basis of $\C^3$, define $T\in\ca L(\C^3)$ by the rule:\[Te_1=2e_1,\quad Te_2=2e_2+e_1,\quad Te_3=3e_3.\]The matrix of $T$ with respect to the basis $e_1$, $e_2$, $e_3$ is\[\begin{bmatrix}
2 & 1 & 0\\
0 & 2 & 0\\
0 & 0 & 3
\end{bmatrix}.\]This is an upper-triangular matrix. By 5.32, the eigenvalues of $T$ are $2$ and $3$.

On the other hand,\[(T^2-5T+6I)e_2=(T-3I)(T-2I)e_2=(T-3I)e_1=-e_1\ne 0.\]Hence $T^2-5T+6I\ne 0$.

4. Solution: Suppose $T$ is normal, by 7.22 we have all pairs of eigenvectors corresponding to distinct eigenvalues of $T$ are orthogonal. By 7.24 and 5.41, we have\[V=E(\lambda_1,T)\oplus\cdots\oplus E(\lambda_m,T),\] where $\lambda_1$, $\cdots$, $\lambda_m$ denote the distinct eigenvalues of $T$.

Suppose all pairs of eigenvectors corresponding to distinct eigenvalues of $T$ are orthogonal and \[V=E(\lambda_1,T)\oplus\cdots\oplus E(\lambda_m,T),\] where $\lambda_1$, $\cdots$, $\lambda_m$ denote the distinct eigenvalues of $T$, then $V$ has an orthonormal basis consisting of eigenvectors of $T$. In fact, this basis can be chosen as the union of orthonormal basis of $E(\lambda_i,T)$, $i=1,\cdots,m$. Again by 7.24, we have $T$ is normal.

5. Solution: Suppose $T$ is self-adjoint, $v$ and $u$ are eigenvectors of $T$ corresponding to eigenvalues $\lambda$ and $\xi$ respectively, where $\lambda\ne \xi$. Then $T$ is self-adjoint \[ \langle Tv,u\rangle=\langle v,Tu\rangle, \]hence \[ \lambda \langle v,u\rangle=\xi\langle v,u\rangle. \]This implies $\langle v,u\rangle=0$ since $\lambda\ne \xi$. By 7.29 and 5.41, we have\[V=E(\lambda_1,T)\oplus\cdots\oplus E(\lambda_m,T),\] where $\lambda_1$, $\cdots$, $\lambda_m$ denote the distinct eigenvalues of $T$.

Suppose all pairs of eigenvectors corresponding to distinct eigenvalues of $T$ are orthogonal and \[V=E(\lambda_1,T)\oplus\cdots\oplus E(\lambda_m,T),\] where $\lambda_1$, $\cdots$, $\lambda_m$ denote the distinct eigenvalues of $T$, then $V$ has an orthonormal basis consisting of eigenvectors of $T$. In fact, this basis can be chosen as the union of orthonormal basis of $E(\lambda_i,T)$, $i=1,\cdots,m$. Again by 7.29, we have $T$ is self-adjoint.

6. Solution: See Linear Algebra Done Right Solution Manual Chapter 7 Problem 9.

7. Solution: See Linear Algebra Done Right Solution Manual Chapter 7 Problem 10.

8. Solution: Consider $T\in\ca L(\C^3)$. Let $e_1=(1,0,0)$, $e_2=(0,1,0)$ and $e_3=(0,0,1)$. Define $T$ such that\[Te_1=0,\quad Te_2=e_1,\quad Te_3=e_2.\]Then we have $T^3e_1=0$, $T^3e_2=T^2e_1=0$ and $T^3e_3=T^2e_2=Te_1=0$, i.e. $T^3=0$. It follows that $T^9=T^8$. On the other hand, we have $T^2e_2=Te_1=0$ and $Te_2=e_1$, it follows that $T^2\ne T$.

9. Solution: See Linear Algebra Done Right Solution Manual Chapter 7 Problem 11.

10. Solution: See Linear Algebra Done Right Solution Manual Chapter 7 Problem 12.

11. Solution: See Linear Algebra Done Right Solution Manual Chapter 7 Problem 13.

12. Solution: See Linear Algebra Done Right Solution Manual Chapter 7 Problem 14.

14. Solution: See Linear Algebra Done Right Solution Manual Chapter 7 Problem 15.

15. Solution: Let the entry be $x$. Then by definition we have\[\begin{pmatrix}
1 & 1 & 0\\
0 & 1 & 1\\
1 & 0 & x
\end{pmatrix}\begin{pmatrix}
1 & 0 & 1\\
1 & 1 & 0\\
0 & 1 & x
\end{pmatrix}=\begin{pmatrix}
1 & 0 & 1\\
1 & 1 & 0\\
0 & 1 & x
\end{pmatrix}\begin{pmatrix}
1 & 1 & 0\\
0 & 1 & 1\\
1 & 0 & x
\end{pmatrix}.\]Consider the (1,3) entry of the two products, we get $x=1$.