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Linear Algebra - Kernel and range of T

  1. Jun 17, 2017 #1
    1. The problem statement, all variables and given/known data

    Let ##T:M_2 \to M_2## a linear transformation defined by

    ##T \begin{bmatrix}
    a&b\\
    c&d
    \end{bmatrix} =
    \begin{bmatrix}
    a&0\\
    0&d
    \end{bmatrix}##

    Describe ##ker(T)## and ##range(T)##, and find their basis.

    2. Relevant equations

    For a linear transformation ##T:V\to W##

    ##range(T)={T(x) \epsilon W : x \epsilon V}##

    ##ker(T)= {x \epsilon V : T(x)= 0 \epsilon W}##


    3. The attempt at a solution


    Skipping the first part of the proof, I get to the part where I describe the range of the transformation and express it as a linear combination of two ##M_2## matrix

    ##T \begin{bmatrix}
    a&b\\
    c&d
    \end{bmatrix} =
    \begin{bmatrix}
    a&0\\
    0&d
    \end{bmatrix}##

    ##T \begin{bmatrix}
    a&b\\
    c&d
    \end{bmatrix} =
    a\begin{bmatrix}
    1&0\\
    0&0
    \end{bmatrix}+
    d\begin{bmatrix}
    0&0\\
    0&1
    \end{bmatrix}
    ##

    So $$\begin{bmatrix}
    1&0\\
    0&0
    \end{bmatrix} and \begin{bmatrix}
    0&0\\
    0&1
    \end{bmatrix}$$ span the range for ##T##, also they are linearly independent, thus forming a basis for the range.

    The kernel can be expressed as

    ##ker(T)={A \epsilon M_2 : Ax=0 \epsilon M_2}##

    ##ker(T)=A \epsilon M_2 :## ##A =
    \begin{bmatrix}
    a&b\\
    c&d
    \end{bmatrix} \forall a,b,c,d \epsilon ℝ, a=d=0##


    If I got it right up to this point then

    ##dim[range(T)]=2##

    But then according to the theorem that says

    ##dim[range(T)]+dim[ker(T)]=dim(V)##

    Being ##V## the space of ##2_x####2## square matrix, then ##dim(V)=2## but that would make ##dim[ker(T)]=0## which doesn't make sense to me, so I believe I got a concept wrong somewhere on my analysis.

    Any advise would be appreciated.
     
  2. jcsd
  3. Jun 17, 2017 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Why do you think the dimension of V is 2?
     
  4. Jun 17, 2017 #3
    I see, I just noticed I followed a wrong assumption

    ##V=
    a\begin{bmatrix}
    1&0\\
    0&0
    \end{bmatrix}+
    b\begin{bmatrix}
    0&1\\
    0&0
    \end{bmatrix}+
    c\begin{bmatrix}
    0&0\\
    1&0
    \end{bmatrix}+
    d\begin{bmatrix}
    0&0\\
    0&1
    \end{bmatrix}##

    Thus ##dim(V)=4##, right?
     
  5. Jun 17, 2017 #4

    Math_QED

    User Avatar
    Homework Helper

    Yes, in general, if you consider the vector space ##M_{m,n}(\mathbb{K})## (the ##m \times n## matrices), it has dimension ##mn##. This can easily be proven by showing that the set ##\{E_{ij}|1 \leq i \leq m, 1 \leq j \leq n\}## defines a basis for this space. Here, ##E_{ij}## is the matrix that is everywhere zero, except on place ##(i,j)## where it is ##1## (or another non zero number).
     
  6. Jun 17, 2017 #5

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes.
     
  7. Jun 17, 2017 #6
    Thank you both, now it all makes sense.
     
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