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Math Help - More Confusion, (linear algebra)

  1. #1
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    More Confusion, (linear algebra)

    Handing in date is drawing very close and im losing my mind over this.

    \begin{pmatrix}1 & 4 & 6\\3 & 9 & 1\\2 & 6 & 5\\4 & 4 & 9\end{pmatrix}

    ive been asked to find the rref(M) which is confusing me because I just end up with:

    \begin{pmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\\0 & 0 & 0\end{pmatrix}

    which kind of makes sence to me as there are 3 variables so one of the rows has to be a linear combination of the others? but im then asked to find the kernal(M) which seems to me to be the zero vector????? how confusing...
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  2. #2
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    Quote Originally Posted by SilenceInShadows View Post
    Handing in date is drawing very close and im losing my mind over this.

    \begin{pmatrix}1 & 4 & 6\\3 & 9 & 1\\2 & 6 & 5\\4 & 4 & 9\end{pmatrix}

    ive been asked to find the rref(M) which is confusing me because I just end up with:

    \begin{pmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\\0 & 0 & 0\end{pmatrix}

    which kind of makes sence to me as there are 3 variables so one of the rows has to be a linear combination of the others? but im then asked to find the kernal(M) which seems to me to be the zero vector????? how confusing...
    Your transform is T:\mathbb{R}^3 \to \mathbb{R}^4

    Remember that the Kernal of a linear transformation is the set of vectors v \in \mathbb{R}^3 such that T(v)=0 where 0 is the 0 vector in \mathbb{R}^4

    Solving this gives

    <br />
\begin{pmatrix}1 & 4 & 6 & 0\\3 & 9 & 1 & 0\\2 & 6 & 5 & 0\\4 & 4 & 9 & 0\end{pmatrix}<br />

    \begin{pmatrix}1 & 0 & 0 & 0\\0 & 1 & 0 & 0\\0 & 0 & 1 & 0\\0 & 0 & 0 & 0\end{pmatrix}

    So the only solution to this system as you stated is the trivial solution. So the only vector in the kernel is (0,0,0)

    Note: this tells us that the Linear transformation is 1-1.

    Good luck.
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  3. #3
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    Thankyou,

    I wanted the answer before I added this bit; I need a basis for the kernal.

    I need to find a basis for the trival zero vector? hence my confusion...
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  4. #4
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    also what would be the dimension of the null space? the rank + nullity gives the number of columns right? the rank is given by the numbr of linearly independent rows or columns, giving 3.

    3 + nullity = 3?

    So my matrix with kernal the zero vector that I have to find a basis for has dimension zero?



    fun!
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  5. #5
    Behold, the power of SARDINES!
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    Quote Originally Posted by SilenceInShadows View Post
    also what would be the dimension of the null space? the rank + nullity gives the number of columns right? the rank is given by the numbr of linearly independent rows or columns, giving 3.

    3 + nullity = 3?

    So my matrix with kernal the zero vector that I have to find a basis for has dimension zero?



    fun!
    The Dimenstion of a finite-dimensional vector space V is defined to be the number of vectors in a basis for V. In addition the zero vector space is defined to have dimension zero.

    See examples in this wiki

    Dimension (vector space) - Wikipedia, the free encyclopedia

    Good luck.

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