0
$\begingroup$

Find values of parameter t for which transformation is epimorphic: $\psi([x_1,x_2,x_3,x_4])=[x_1+x_2+x_3+2x_4,x_1+tx_2+x_3+3x_4,2x_1+x_2+tx_3+3x_4] $ When this transformation is epimorphic i.e. what should i look for in the reduced form of matrix of this linear transformation?

$\endgroup$
  • $\begingroup$ You should write up your definition of epimorphism first. $\endgroup$ – AlexR Dec 18 '14 at 14:15
0
$\begingroup$

The row-reduced form of the matrix must have 3 independent rows (i.e., all 3 rows must end up with "leading 1s" in them -- no rows of all zeroes).

Why? Because row-rank (= number of indep't rows) = column rank, and for the transformation to be an epimorphism, you need column rank (which is the dimension of the image) to be 3.

$\endgroup$
  • $\begingroup$ So to make evrything clear this transformation is an epiomorphism for $t\ne 1$? Right? $\endgroup$ – kurkowski Dec 18 '14 at 15:28
  • $\begingroup$ Nope. The problem occurs at a different value of $t$. Why not show us a sequence of matrices row-equivalent to the original, but simpler and simpler, so we can check your work? (Alternatively, you could use column-equivalence, if that seems easier...which it does, at least to me.) $\endgroup$ – John Hughes Dec 18 '14 at 15:37
  • $\begingroup$ My rref matrix is: \begin{pmatrix} 1 & 1 & 1 & 2 \\ 0 & t-1 & 0 & 1 \\ 0 & -1 & t-2 & -1 \end{pmatrix} thus i must have three independent rows thus the one with t-1 needs to be zero. So maybe it is an epimorhpism for t=1? That's all? $\endgroup$ – kurkowski Dec 18 '14 at 15:43
  • $\begingroup$ That's not row-reduced! Swap 2nd and 3rd rows to get $$\begin{pmatrix} 1 & 1 & 1 & 2 \\ 0 & -1 & t-2 & -1 \\0 & t-1 & 0 & 1 \end{pmatrix}$$ and add $(t-1)$ times the second row to the third to clear the (3, 2) entry. What do you get? For what values of $t$ does the resulting row-reduced matrix have only two independent rows? $\endgroup$ – John Hughes Dec 18 '14 at 16:34
  • $\begingroup$ thus i have $\begin{pmatrix} 1 & 1 & 1 & 2 \\ 0 & -1 & t-2 & -1 \\0 & 0 & t^2-3t+2 & -t+2 \end{pmatrix}$ thus t=2 and then it is epimorphic? but previously you wrote thata 3 independent rows are needed, i am lost $\endgroup$ – kurkowski Dec 18 '14 at 16:43
0
$\begingroup$

Hint
Epimorphisms must satisfy

  • $\phi$ is linear (this is the case for all $t$, since $\phi$ can be represented by a matrix-vector multiplication)
  • $\phi$ is surjective (that means that $\phi(\mathbb R^4) = \mathbb R^3$, i.e. $\mathrm{rank}(\phi) = 3$)

So you must check whether the matrix $$[\psi] = \pmatrix{1&1&1&2\\1&t&1&3\\2&1&t&3}$$ Has rank $3$ (in general: full row rank). This is the case iff the REF of $[\phi]$ has the form $$[\psi] \sim \pmatrix{1&0&0&\ast\\0&1&0&\ast\\0&0&1&\ast}$$ (in general: $[\psi]\sim\pmatrix{I_n &\ast}$ where $n$ is the number of rows)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.