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I have a linear transformation $t: \mathbb{R}^3 \to \mathbb{R}^3$ and I need to determine the image of $t$. $t$ is given by $$t\left(\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} \right)= \begin{pmatrix} 1 & 1 & 0\\ 1 & 0 & 1\\ -2 & -1 & -1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} $$

and I know that the image of $t$ is the subspace $U = \{ x \in \mathbb{R}^3 \ | \ x_1 + x_2 = -x_3 \}$ from the exercise I'm doing. I also know that the image of the linear map is the span of the column vectors of the matrix. How would I go about showing that the span of the matrix is the given subspace?

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You wish to determine the image $U$ of the linear map $T:\Bbb R^3\to\Bbb R^3$ defined by $T(\vec x) = A\vec x$ where $$ A= \left[\begin{array}{rrr} 1 & 1 & 0 \\ 1 & 0 & 1 \\ -2 & -1 & -1 \end{array}\right] $$ You have correctly observed that $U$ is the column space of $A$. The nonzero rows of the reduced row echelon form of $A^\top$ form a basis for the column space of $A$. In our case, $$ \DeclareMathOperator{rref}{rref}\rref A^\top= \left[\begin{array}{rrr} 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{array}\right] $$ This proves that the two vectors \begin{align*} u_1 &= \langle 1,0,-1\rangle & u_2 &= \langle 0,1,-1\rangle \end{align*} form a basis for $U$. Now, note that $\DeclareMathOperator{Span}{Span}$ \begin{align*} U &= \Span\{u_1,u_2\} \\ &= \{\lambda_1 u_1+\lambda_2 u_2:\lambda_1,\lambda_2\in\Bbb R\} \\ &= \{\langle\lambda_1,\lambda_2,-(\lambda_1+\lambda_2)\rangle:\lambda_1,\lambda_2\in\Bbb R\} \\ &= \{\langle x,y,z\rangle:x+y=-z\} \end{align*} as desired.

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  • $\begingroup$ I like using rref of $A^T$ myself, but I’ll note that another common way to find a basis for the image is to row-reduce $A$ and then take the columns of $A$ that correspond to columns with pivots in the rref matrix. $\endgroup$ – amd Feb 28 '16 at 20:56
  • $\begingroup$ @amd This is of course correct, however the basis we get using this method is less useful as it is not "reduced" $\endgroup$ – Brian Fitzpatrick Feb 28 '16 at 20:58
  • $\begingroup$ That’s one of the reasons I prefer the method you gave. $\endgroup$ – amd Feb 28 '16 at 21:01

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