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Thread: Some problem on eigenvalues,eigenvectors and diagonalizability

  1. #1
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    Some problem on eigenvalues,eigenvectors and diagonalizability

    I'm preparing final exam on linear algebra and come across with these problem:

    #1
    Let A be an n*n matrix.Assume that for some $\displaystyle k>=1$, $\displaystyle A^k=/=0$ and $\displaystyle A^kA=0$.
    a)Find all the eigenvalues of A.
    b)Prove that A is not diagonalizable.
    c)Prove that k<n
    d)If k=n-1,find the dimension of null A

    #2
    Let A be an n*n matrix and $\displaystyle a=/=b$ be two real numbers.Assume that
    $\displaystyle (A-aI)(A-bI)=0$
    Prove that A is diagonalizable.

    #3
    Let $\displaystyle k>=2$ and $\displaystyle X_1,X_2,...,X_n$ be eigenvectors correspond to distinct eigenvalues $\displaystyle n_1,n_2,...,n_k$ of an n*n matrix A.
    Prove that $\displaystyle X_1+X_2+...+X_k$ is not an eigenvector of A.

    For #3,
    Suppose on the contrary, $\displaystyle X_1+X_2+...+X_k$ is an eigenvector of A
    Then there exists a number n such that
    $\displaystyle A(X_1+X_2+...+X_k)=n(X_1+X_2+...+X_k)$
    $\displaystyle => AX_1+AX_2+...+AX_n=n(X_1+X_2+...+X_k)$
    $\displaystyle => n_1X_1+n_2X_2+...+n_kX_k=n(X_1+X_2+...+X_k)$

    $\displaystyle =>(n_1-n)X1+(n_2-n)X2+...+(n_k-n)X_k=0$

    Since eigenvectors are lin. ind.
    therefore, $\displaystyle (n_1-n)=(n_2-n)=...=(n_k-n)=0 => n_1=n_2=...=n_k=n$
    Contradict to $\displaystyle n_1,n_2,...,n_k$are distinct.

    Is it right ?
    Last edited by maoro; Nov 30th 2011 at 09:05 AM.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Some problem on eigenvalues,eigenvectors and diagonalizability

    Quote Originally Posted by maoro View Post
    Let A be an n*n matrix.Assume that for some $\displaystyle k>=1$, $\displaystyle A^k=/=0$ and $\displaystyle A^kA=0$.
    a)Find all the eigenvalues of A.
    If $\displaystyle \lanbda$ is an eigenvalue of $\displaystyle A$ there exists $\displaystyle 0\neq x\in\mathbb{R}^n$ such that $\displaystyle Ax=\lambda x$ . Then, $\displaystyle 0=A^{k+1}x=\ldots=\lambda^{k+1}x\;(x\neq 0)$ which implies $\displaystyle \lambda=0$

    b)Prove that A is not diagonalizable.
    $\displaystyle A^{k+1}=0\Rightarrow (\det A)^{k+1}=0\Rightarrow \det A=0\Rightarrow \textrm{rank}A<n$ etc ...

    Try the rest of problem #1

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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Re: Some problem on eigenvalues,eigenvectors and diagonalizability

    Quote Originally Posted by maoro View Post
    #2Let A be an n*n matrix and $\displaystyle a=/=b$ be two real numbers.Assume that $\displaystyle (A-aI)(A-bI)=0$ Prove that A is diagonalizable.
    One way: $\displaystyle p(x)=(x-a)(x-b)$ is an annihilator polynomial of $\displaystyle A$ so the possible minimal polynomials of $\displaystyle A$ are $\displaystyle \mu(x)=x-a$ , $\displaystyle \mu(x)=x-b$ or $\displaystyle \mu(x)=(x-a)(x-b)$ . In all cases, the blocks of the canonical form of $\displaystyle A$ have dimension $\displaystyle 1$, so $\displaystyle A$ is diagonalizable.

    P.S. Comment us if you have not covered Jordan canonical forms.
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  4. #4
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    Re: Some problem on eigenvalues,eigenvectors and diagonalizability

    Sorry for posting too many questions on a single thread.

    btw, I have not covered Jordan canonical forms and so is there any alternative approach for #2 ?
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    Re: Some problem on eigenvalues,eigenvectors and diagonalizability

    Quote Originally Posted by maoro View Post
    is there any alternative approach for #2 ?
    Yes, there is. Hint : Consider $\displaystyle F_1=\textrm{Im}(A-aI),\;F_2=\textrm{Im}(A-bI)$ . Prove that $\displaystyle F_1\subset \ker (A-bI)$ , $\displaystyle F_2\subset \ker (A-aI)$ and $\displaystyle \mathbb{R}^n=F_1\oplus F_2$ .
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