If you find any mistakes, please make a comment! Thank you.

Chapter 7 Exercise B


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 $x=1$.


Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.

This Post Has 11 Comments

  1. Problem 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.

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

  2. Problem #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..

    1. It 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...

      1. I 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.

        1. There 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.

  3. I'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 ..........

    1. You 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.

  4. 7B1) 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$.

    1. Thank you

      1. Still question 1, second line, it should be Te1 = e1
        so that T(e1+e2) = Te1 + Te2 = e1+e1+2e2 = 2(e1 + e2)

Leave a Reply

Close Menu