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 $x=1$.

## Charlie Fan

9 Jun 2020Problem 1 should be false. The problem requires that the basis consists of eigenvectors, while in the solution given above, e2 is NOT an eigenvector. Actually if you take e1, e1+e2 and e3 as the basis, then this operator is diagonalizable and thus self-adjoint.

## Linearity

10 Jun 2020A basis consisting of eigenvectors is different from an orthonomral basis consisting of eigenvectors. Operator which is diagonalizable may not be self-adjoint.

## Nick

6 Oct 2019Problem #1, I think it is false based on the Real Spectral Theorem (7.29) which says:

Suppose F = R and T L(V) Then the following are equivalent:

(a) T is self-adjoint.

(b) V has an orthonormal basis consisting of eigenvectors of T.

(c) T has a diagonal matrix with respect to some orthonormal basis

of V.

By definition, if there is a orthonormal basis consisting of eigenvectors of T, then T is self-adjoint. I think you want to use a basis that is not orthonormal, then the statement will be true..

## Linearity

6 Oct 2019It seems you know that in order to be self-adjoint, a basis of $T$ has to be orthonormal. Then I do not know why you think it is false...

## Aditya

29 Apr 2020I agree with Nick. We can convert the basis of eigenvectors into an orthonormal basis of eigenvectors. Thus we will obtain a diagonal matrix wrt basis of orthonormal eigenvectors, implying T is self-adjoint.

## Linearity

29 Apr 2020There is a basis of $\mathbb R^3$ consisting of eigenvectors of $T$. But not orthonormal one.

You are saying:

We can convert the basis of eigenvectors into an orthonormal basis of eigenvectors. If you think it is true, you have to show it. But this is definitely wrong. A linear combination of eigenvectors may bot be an eigenvector.

Your statement means every diagonalizable operator is self-adjoint, which is certainly wrong.

## Chi Yuan Lau

6 Aug 2019I'm sorry , how to solve ex13, well, I try my best to modify the proof of the real spectral theorem , but when I deduced the conclusion , I found I had'nt used the condition $T$ is normal ..........

## Linearity

7 Aug 2019You have to show the following. Suppose $T$ is invariant under $T$, then $U^\perp$ is invariant under $T^*$. In the proof of the real spectral theorem, $U$ is invariant under both $T$ and $T^*$, see 7.21 (the fact that $T$ is normal is used here). Hence $U^\perp$ is also invariant under both $T$ and $T^*$. Then the restriction of $T$ on $U^\perp$ is also normal. So you can use induction hypothesis.

I may make a mistake. Please comment if you find it.

## Marcel Ackermann

30 Jul 20177B1) True. Example: Let (1,0,0), (1,1,0), (1,1,1) be a basis of R³. Let T be defined by T(1,0,0)=(1,0,0), T(1,1,0)=(2,2,0), T(1,1,1)=(3,3,3). By construction there is a basis consisting of eigenvectors of T. And T is not self-adjoint: $ < T(1,0,0), (1,1,0)>= 1$ while $ < (1,0,0), T(1,1,0)> = 2$.

## Mohammad Rashidi

30 Jul 2017Thank you

## Wu Jinyang

23 Aug 2017Still question 1, second line, it should be Te1 = e1

so that T(e1+e2) = Te1 + Te2 = e1+e1+2e2 = 2(e1 + e2)