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Row/column operation on matrices and determinants

  1. Feb 9, 2015 #1
    How we cannot apply row and column operation simultaneously on matrix when finding its inverse by elementary transformation but can apply it in determinant?
    I think kernel and image gets disturbed in a matrix, though I don't know what it actually is.
    Why not in determinant case?
     
  2. jcsd
  3. Feb 10, 2015 #2

    Svein

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    Science Advisor

    You can - but only when you do it on the right hand side simultaneously. Assume you want to find [itex]B [/itex] such that [itex]A \cdot B = I [/itex]. You do the same row operations on [itex] A[/itex] and [itex]I [/itex]. After a while (excluding singularities etc.) you end up with [itex] U \cdot B = R[/itex], where [itex] U[/itex] is upper triangular. Now substitute back until you end up with [itex]I \cdot B = H [/itex] and you are done.
     
  4. Feb 10, 2015 #3
    But doesn't the kernel or image gets disturbed?
     
  5. Feb 10, 2015 #4

    Svein

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    Science Advisor

    I am not an expert on matrices so, I cannot answer that. What I know, is how to find the inverse (as detailed above). Useful facts:
    • The matrix can be inverted if its determinant is different from 0.
    • The determinant of the inverse matrix is the inverse of the determinant of the original matrix.
    See also http://en.wikipedia.org/wiki/Invertible_matrix .
     
  6. Feb 10, 2015 #5
    Row operations are equivalent to left multiplication by corresponding elementary matrices, column operations are equivalent to right multiplication. So, when you perform row operations on a (square) matrix ##A##, you get a matrix ##E## (the product of corresponding elementary matrices) such that ##EA=I##. For square matrices that means that ##E=A^{-1}##; performing the same row operations on the right hand side ##I## you get ##EI=E=A^{-1}##.

    Again, you can find the inverse performing only column operations: you get a matrix ##E## (the product of corresponding elementary matrices) such that ##AE=I##, which again means that ##E=A^{-1}##. Performing the same column operations on the right hand side ##I## you get there ##IE=E=A^{-1}##.

    If you perform both row and column operations you get two matrices ##E_1## and ##E_2## such that ##E_1AE_2=I##, so in the right hand side you get ##E_1IE_2=E_1E_2##. For ##E_1E_2## to be the inverse of ##A## you need ##AE_1E_2=I## (or equivalently ##E_1E_2A =I##), and you have ##E_1AE_2=I##. Matrix multiplication is not commutative, so there is no reason for the latter identity imply the former.

    "No reason" is not a formal proof that the method does not work, for the formal proof you need a counterexample. You can find it just by playing with applying row and column operations to ##2\times 2## matrices. There are also more scientific method of constructing a counterexample.

    The reason that applying both row and column operations is that while ##E_1AE_2=I## does not imply that ##E_1E_2=A^{-1}## it implies that ##\operatorname{det} (E_1E_2)=\operatorname{det} A^{-1}##. Namely, $$1=\operatorname{det}(E_1AE_1) = \operatorname{det} E_1 \operatorname{det} A \operatorname{det}E_2,$$ so $$\operatorname{det}(E_1E_2) \operatorname{det}E_1\operatorname{det}E_2= (\operatorname{det}A)^{-1} =\operatorname{det}A^{-1}.$$

    Finally, for invertible matrices the image is all ##\mathbb R^n## and the kernel is always trivial (i.e. ##\{\mathbf 0\}##), they cannot be "disturbed" by row/column operations, they will remain the same.
     
  7. Feb 10, 2015 #6

    Mark44

    Staff: Mentor

    If you're talking about a matrix you get by applying a row operation to another matrix, the answer is no. If you start with a matrix A, and apply one of the three row operations to it to get A1, the matrices A and A1 are equivalent. They have exactly the same kernel and image.
    Edit: The image can change. See my later post in this thread.

    If you're talking about applying column operations, I don't know -- I have never needed to apply column operations to reduce a matrix. However, if you swap the columns of a matrix, you are swapping the roles of the variables these columns represent.
     
    Last edited: Feb 10, 2015
  8. Feb 10, 2015 #7
    That is not true. Row operations preserve kernel, column operations preserve image (column space).

    In the case of invertible matrices, however, the image is always all ##\mathbb R^n## and the kernel is always ##\{\mathbf 0\}##, so we can say that in this case the image and kernel are "preserved" user row and column operations.
     
  9. Feb 10, 2015 #8
    A bit of specification. Row operations preserve kernel (but generally not image), column operations preserve image (but generally not kernel).
     
  10. Feb 10, 2015 #9

    Mark44

    Staff: Mentor

    I mispoke. Row operations don't necessarily preserve the range, as I said. A simple example shows this:
    $$A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 1 & 1\end{bmatrix}$$
    Using row reduction, we get an equivalent matrix.
    $$B = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0\end{bmatrix}$$
    Although the dimensions of the column spaces of A and B are equal (2), they span different subspaces of R3.

    The columns of A represent a plane in R3 that is perpendicular to <-1, -1, 1>. The columns of B represent a different plane in R3 that is perpendicular to the z-axis.
     
  11. Feb 11, 2015 #10
    I see you have applied R3--------> R3 - R1
    But there should be 1 in row 3 column 2 or more simply B32 should be equal to 1?
     
  12. Feb 11, 2015 #11

    Mark44

    Staff: Mentor

    No. Here's what I did: -R1 + R3 --> R3 and then -R2 + R3 --> R3.

    IOW, add -R1 to R3, and then add -R2 to R3.
     
  13. Feb 11, 2015 #12
    Got it. Thanks to all of you - HawkEye 18, Mark44 and Svein.
     
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