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Math Help - 2 questions : Kernel/Image of a linear transformation and orthonormal basis

  1. #1
    MHF Contributor arbolis's Avatar
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    2 questions : Kernel/Image of a linear transformation and orthonormal basis

    Hello MHF,
    It is the continuation of the exercise I posted in my precedent thread.
    So I recall, I have a vectorial subspace of \mathbb{R}^4, W. W is defined to be \{ (x,y,z,w)\in \mathbb{R}^4 \text{such that }2x-z+w=0, x-y-z=0 \}.
    I've found a basis of W and a basis of W^{\perp}. They now ask me to find an orthonormal basis of \mathbb{R}^4 using the basis I found in the past items.
    For W I have that a basis is  \{ (0,1,-1,-1), (1,0,1,-1) \} . While for W^{\perp} I have that a basis is \{ (2,0,-1,1), (1,-1,0,-1) \}.
    Oh wait... doesn't that mean that a basis of  \mathbb{R}^4 is the union of the 2 basis I found? But it wouldn't necessarily be orthonormal especially because I have a factor 2 in the basis of W^{\perp}. So I don't see how I can answer the question.


    Now the next question seems even much harder. I must find a linear transformation T:\mathbb{R}^4 \to \mathbb{R}^4 such that \ker (T)=W and \Im (T)=W^{\perp}.
    My attempt : T(x,y,z,w)=0\Leftrightarrow T(x,y,x-y,-x-y)=0 and I'm at a loss.
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    Quote Originally Posted by arbolis View Post
    Hello MHF,
    It is the continuation of the exercise I posted in my precedent thread.
    So I recall, I have a vectorial subspace of \mathbb{R}^4, W. W is defined to be \{ (x,y,z,w)\in \mathbb{R}^4 \text{such that }2x-z+w=0, x-y-z=0 \}.
    I've found a basis of W and a basis of W^{\perp}. They now ask me to find an orthonormal basis of \mathbb{R}^4 using the basis I found in the past items.
    For W I have that a basis is  \{ (0,1,-1,-1), (1,0,1,-1) \} . While for W^{\perp} I have that a basis is \{ (2,0,-1,1), (1,-1,0,-1) \}.

    Oh wait... doesn't that mean that a basis of  \mathbb{R}^4 is the union of the 2 basis I found? yes!
    let u_1, u_2, u_3, u_4 be as i found in here. the set \{u_1,u_2,u_3,u_4 \} is an orthogonal basis for \mathbb{R}^4. so if you put v_j=\frac{u_j}{||u_j||}, \ 1 \leq j \leq 4, then \{v_1,v_2,v_3,v_4 \} would be an orthonormal basis for \mathbb{R}^4.

    Now the next question seems even much harder. I must find a linear transformation T:\mathbb{R}^4 \to \mathbb{R}^4 such that \ker (T)=W and \Im (T)=W^{\perp}.
    My attempt : T(x,y,z,w)=0\Leftrightarrow T(x,y,x-y,-x-y)=0 and I'm at a loss.
    let B_1=\{u_1,u_2,u_3,u_4 \} and B_2 be the standard basis for \mathbb{R}^4. the linear transformation you're looking for in basis B_1 simply sends u_1,u_2 to 0 and u_3,u_4 remain fixed, i.e.

    T(c_1u_1+c_2u_2+c_3u_3+c_4u_4)=c_3u_3+c_4u_4. thus [T]_{B_1}=\begin{bmatrix}0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}. now use what you've learned about "change of basis" to find S=[T]_{B_2}. then the answer to
    your question would be: T(\bold{x})=S \bold{x}, where \bold{x}=\begin{bmatrix}x \\ y \\ z \\ w \end{bmatrix}.
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    W basis \{\vec{v}_1= (0,1,-1,-1), \vec{v}_2=(1,0,1,-1) \} are orthonal each other. we only compute orthnormal (unit vector),
     \vec{u}_1=\frac{\vec{v}_1}{\|\vec{v}_1 \|}=\frac{1}{\sqrt{3}}\begin{bmatrix} 0 \\ 1 \\ -1 \\ -1 \end{bmatrix}<br />
,  \vec{u}_2=\frac{\vec{v}_2}{\|\vec{v}_2 \|}=\frac{1}{\sqrt{3}}\begin{bmatrix} 1 \\ 0 \\ 1 \\ -1 \end{bmatrix}<br />

    **********
     W^{\perp} basis are not orthnormal. We can compute by formular:
    \vec{u}_1=\frac{\vec{v}_1}{\|\vec{v}_1 \|}
    \vec{v}^{\perp}_2=\vec{v}_2 - \vec{v}_2^{\parallel}=\vec{v}_2-(\vec{u}_1.\vec{v}_2)\vec{u}_1
    \vec{u}_2=\frac{\vec{v}^{\perp}_2}{\|\vec{v}^{\per  p}_2 \|}

    \vec{u}_1=\frac{1}{\sqrt{6}}\begin{bmatrix} 2 \\ 0 \\ -1 \\ 1 \end{bmatrix} , \vec{u}_2=\frac{1}{\sqrt{6}}\begin{bmatrix} 0 \\ 2 \\  1 \\ 1 \end{bmatrix}

    **********
    In previous question "Give a basis of a subspace", Dim(ker(T))+Dim(Im(T))=2+2=4 =Dim(R^4), it mean T exist. And it is rotation matrix. It's more complex.
    Last edited by math2009; February 11th 2009 at 12:55 AM.
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    typewrite is very slow. I upload answer file
    Please check attach files.

    T is not only one solution. Dim(T)=2.
    Attached Thumbnails Attached Thumbnails 2 questions : Kernel/Image of a linear transformation and orthonormal basis-s3.jpg  
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    Last edited by math2009; February 11th 2009 at 04:27 PM.
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