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Thread: diagonal

  1. #1
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    diagonal

    i just started on diagonalizable matrix and i am trying to prove that f: 2x2 matrix--> 2x2 matric where f(A)= A + A^t. find a basis for the 2x2 matrix which consist of eigenvectors for f.


    im wondering, the basis would be in the form of a 2x2 matrix yet eigenvectors are column matrices..so how do i go about doing it when the sizes are different?
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  2. #2
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    A basis is a set of vectors.These vectors must be linearly independent and must span the given space.

    So you are looking for 2 eigenvectors that form a basis - that is you are looking for 2 column vectors. You put these two vectors together to form the 2x2 matrix P.
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  3. #3
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    that makes sense.

    so to find the eigenvetors, we need to know A such that f(v)=Av where A has to be a 2x2 matrix right?

    since the funct is defined by f(A)= A+A^t, what i tried to do what such A to be 2x2 elementary matrices and obtain that the image is of the form

    (1 0 ),( 0 1), (0 0)
    (0 0),(1 0), (0 1)

    [ read it as 3 2x2 matrices]

    how do i find A then so that i can find its eigenvectors?
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  4. #4
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    Apologies - I misread the question. I didn't realize there was a linear transformation involved. The question actually doesn't quite make sense as written. You should check the wording and maybe repost it. Did you mean to prove that f is a linear transformation? If so, then this is straightforward. f is essentially a map from R^4 to R^4, so its eigenvectors have four components - not 2. Which means the P matrix would be 4x4. I think we can get that as follows. It's been a while since I've done a problem like this, so if someone could confirm that this is correct I'd appreciate it.


    First, identify each 2x2 matrix with a vector in R^4, and think of the linear transformation f as a mapping from R^4 to R^4 (in other words write the matrix with entries a, b, c, d as a column vector with entries a, b, c, d).

    Next, find the matrix of the transformation by computing f of each basis vector. The matrix of the transformation is the following:

    2 0 0 0
    0 1 1 0
    0 1 1 0
    0 0 0 2

    Now find the eigenvalues and eigenvectors of this matrix in the usual way. You should get eigenvalues of 0 and 2 (multiplicity 3)

    The corresponding eigenvectors are

    0 1 0 0
    -1 0 1 0
    1 0 1 0
    0 0 0 1

    Rewrite these eigenvectors as matrices. For example the eigenvector with eigenvalue 0 is

    0 -1
    1 0
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  5. #5
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    The exact qns is
    Let f : 2x2 matrix -> 2x2 matrix be the linear map defined by f(x)=A+A^t
    . Prove that f is diagonalizable by finding a basis for 2x2 matrix consisting of eigenvectors for f.
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  6. #6
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    How do you know when its from r^4 to r^ 4 and not a 2x1 basis column matrix? I thought your first explAnation made sense. Thanks for your help.
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  7. #7
    MHF Contributor FernandoRevilla's Avatar
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    There is a nice generalization:

    $\displaystyle T:\mathbb{R}^{n\times n}\rightarrow \mathbb{R}^{n\times n},\quad T(A)=A+A^t$

    (i) $\displaystyle T(A)=0 \Leftrightarrow A+A^t=0 \Leftrightarrow A^t=-A \Leftrightarrow{\;A\textrm{\;is\;skew\;symmetric}} $

    So, $\displaystyle \lambda_1=0$ is an eigenvalue of $\displaystyle f$ and $\displaystyle \ker (T-0I)=F_1$ (subspace of skew symmetric matrices).

    (ii) $\displaystyle B$ symmetric implies $\displaystyle B=(1/2)(B+B^t)$ , as a consequence $\displaystyle T(B)=2B$. Reciprocally,

    $\displaystyle T(B)=2B\Rightarrow B+B^t=2B \Rightarrow B^t=B\Rightarrow B\textrm{\;is\;symmetric} $

    That is, $\displaystyle \lambda_2=2$ is an eigenvalue of $\displaystyle T$ and $\displaystyle \ker (T-2I)=F_2$ (subspace of symmetric matrices). But $\displaystyle \mathbb{R}^{n\times n}=F_1\oplus{F_2}$. This implies:

    (a) $\displaystyle T$ is diagonalizable on $\displaystyle \mathbb{R}$.

    (b) $\displaystyle D=\textrm{diag}(0,\ldots,0,2,\ldots,2)$ with $\displaystyle 0$, $\displaystyle n(n-1)/2$ times and $\displaystyle 2$, $\displaystyle n(n+1)/2$ times.

    As a particular case for $\displaystyle n=2$:

    $\displaystyle V_0=\left\{{\begin{bmatrix}{0}&{-1}\\{1}&{0}\end{bmatrix}}\right\},\quad V_2=\left\{{ \begin{bmatrix}{1}&{0}\\{0}&{0}\end{bmatrix},\begi n{bmatrix}{0}&{1}\\{1}&{0}\end{bmatrix},\begin{bma trix}{0}&{0}\\{0}&{1}\end{bmatrix} }\right\}$

    and $\displaystyle D=\textrm{diag}(0,2,2,2)$

    Regards.

    Fernando Revilla
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  8. #8
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    But i haven learnt abt symmetry. Is there any other mtd?
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  9. #9
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by alexandrabel90 View Post
    But i haven learnt abt symmetry. Is there any other mtd?
    My last post was only a generalization, for your particular case follow DrSteve's outline. The canonical basis for $\displaystyle \mathbb{R}^{2\times 2}$ is:

    $\displaystyle B=\left\{{e_1=\begin{bmatrix}{1}&{0}\\{0}&{0}\end{ bmatrix},e_2=\begin{bmatrix}{0}&{1}\\{0}&{0}\end{b matrix},e_3=\begin{bmatrix}{0}&{0}\\{1}&{0}\end{bm atrix},e_4=\begin{bmatrix}{0}&{0}\\{0}&{1}\end{bma trix}}\right\}$

    and:

    $\displaystyle \begin{Bmatrix}T(e_1)=2e_1\\T(e_2)=e_2+e_3\\T(e_3) =e_2+e_3\\T(e_4)=2e_4\end{matrix} $

    So, the matrix of $\displaystyle T$ in respect to $\displaystyle B$ is:

    $\displaystyle A=\begin{bmatrix}{2}&{0}&{0}&{0}\\{0}&{1}&{1}&{0}\ \{0}&{1}&{1}&{0}\\{0}&{0}&{0}&{2}\end{bmatrix}$

    as DrSteve pointed out. You'll find as eigenvalues $\displaystyle \lambda_1=0$ (simple) and $\displaystyle \lambda_2=2$ (triple). And now, if you obtain for example $\displaystyle (0,-1,1,0)$ as eigenvector for $\displaystyle \lambda_1=0$, these are coordinates on $\displaystyle B$, so this vector is exactly:

    $\displaystyle 0e_1-1e_2+1e_3+0e_4=\begin{bmatrix}{0}&{-1}\\{1}&{0}\end{bmatrix}$

    Regards.

    Fernando Revilla
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