# Epimorphism of linear transformation

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?

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

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.

• So to make evrything clear this transformation is an epiomorphism for $t\ne 1$? Right? – kurkowski Dec 18 '14 at 15:28
• 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.) – John Hughes Dec 18 '14 at 15:37
• 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? – kurkowski Dec 18 '14 at 15:43
• 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? – John Hughes Dec 18 '14 at 16:34
• 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 – kurkowski Dec 18 '14 at 16:43

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)