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Row and Column equivalent matrices

  1. Oct 26, 2014 #1
    Hey PF, I'm having trouble seeing the bigger picture here.

    Take matrix A and matrix B. If B can be obtained from A by elementary row operations then the two matrices are row equivalent. The only explanation my book gives is that since B was obtained by elementary row operations, (scalar multiplication and vector addition) that the row vectors of B can be written in terms of the row vectors in A. I follow this sort of, I at least see the 'intuitiveness' behind it but I feel like there is some piece of the puzzle missing in my brain that's not letting it click completely. Anyone want to drop some knowledge on me?

    And while you're at it answer me this: why aren't matrices A and B guaranteed to be column equivalent? What is the difference between rows and columns? If you get matrix A from matrix B by only elementary-column operations (is there such a thing?) then that would mean they are guaranteed to be column equivalent?
  2. jcsd
  3. Oct 26, 2014 #2


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    Two m×n-matrices A and B are "row equivalent" if for all n×1-matrices x, we have
    $$Ax=0\ \Leftrightarrow\ Bx=0.$$ (Either both equalities true or both equalities false). This is equivalent to saying that A and B have the same kernel.

    To say that A and B are "column equivalent" would be the same as saying that ##A^T## and ##B^T## are row equivalent, i.e. that ##A^T## and ##B^T## have the same kernel. ##\begin{pmatrix}1 & 0\\ 0 & 0\end{pmatrix}## and ##\begin{pmatrix}1 & 0\\ 1 & 0\end{pmatrix}## have the same kernel ("the y axis") , but their transposes ##\begin{pmatrix}1 & 0\\ 0 & 0\end{pmatrix}## and ##\begin{pmatrix}1 & 1\\ 0 & 0\end{pmatrix}## do not ("the y axis" and "the line y=-x" respectively).

    This simple example should be enough to see why elementary row operations are different from "elementary column operations".
    $$\begin{pmatrix}0\\ 0\end{pmatrix} =\begin{pmatrix} a & b\\ c & d\end{pmatrix}\begin{pmatrix} x\\ y\end{pmatrix} =\begin{pmatrix}ax+by\\ cx+dy\end{pmatrix}.$$ If you multiply the first row of the matrix by 3, you get an equivalent system of equations. (The entire first equation is multiplied by 3). If you multiply the first column of the matrix by 3, both equations change into something that's not equivalent to what they were before.

    If the original equation had been written as
    $$\begin{pmatrix}0 & 0\end{pmatrix} =\begin{pmatrix}x & y\end{pmatrix} \begin{pmatrix} a & b\\ c & d\end{pmatrix}$$ instead, it would have been appropriate to do column operations instead of row operations.
  4. Oct 26, 2014 #3
    So if we two matrices have the same kernel (the same basis for their null space?) then they will be row equivalent as well? Thanks btw, you pwn my textbook.
  5. Oct 27, 2014 #4


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    The terms "kernel" and "null space" mean exactly the same thing.

    Yes, A and B have the same kernel if and only if they're row equivalent.

    It follows from the definition of matrix multiplication that Ax=0 if and only if x is orthogonal to each column of A. This implies that x is in the kernel of A if and only if it's orthogonal to the row space of A. (The row space is the space spanned by the rows of A). So the kernel and the row space are each other's orthogonal complements. The Wikipedia page on row equivalence describes the key steps in a proof of the result that A and B have the same row space if and only if A and B are row equivalent. So "same kernel" ⇔ "same row space" ⇔ "row equivalent"
  6. Oct 28, 2014 #5


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    For vector spaces, "kernel" and "null space" have exactly the same meaning. For other algebraic structures that contain an additive identity, "0", we can define the kernel of a function as the set of all objects the function maps into 0. But that would not be called a "null space" since it is not a "space" in the sense of vector spaces.
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