The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. \[\begin{aligned} \mathrm{im}(T) & = \{ p(1) ~|~ p(x)\in \mathbb{P}_1 \} \\ & = \{ a+b ~|~ ax+b\in \mathbb{P}_1 \} \\ & = \{ a+b ~|~ a,b\in\mathbb{R} \}\\ & = \mathbb{R}\end{aligned}\] Therefore a basis for \(\mathrm{im}(T)\) is \[\left\{ 1 \right\}\nonumber \] Notice that this is a subspace of \(\mathbb{R}\), and in fact is the space \(\mathbb{R}\) itself. Recall that because \(T\) can be expressed as matrix multiplication, we know that \(T\) is a linear transformation. Most modern geometrical concepts are based on linear algebra. We start with a very simple example. Therefore \(x_1\) and \(x_3\) are dependent variables; all other variables (in this case, \(x_2\) and \(x_4\)) are free variables. Each vector, \(\overrightarrow{0P}\) and \(\overrightarrow{AB}\) has the same length (or magnitude) and direction. Let \(V\) and \(W\) be vector spaces and let \(T:V\rightarrow W\) be a linear transformation. By Proposition \(\PageIndex{1}\) \(T\) is one to one if and only if \(T(\vec{x}) = \vec{0}\) implies that \(\vec{x} = \vec{0}\). In fact, they are both subspaces. You can verify that \(T\) represents a linear transformation. However its performance is still quite good (not extremely good though) and is used quite often; mostly because of its portability. By looking at the matrix given by \(\eqref{ontomatrix}\), you can see that there is a unique solution given by \(x=2a-b\) and \(y=b-a\). Accessibility StatementFor more information contact us atinfo@libretexts.org. Once this value is chosen, the value of \(x_1\) is determined. Find the position vector of a point in \(\mathbb{R}^n\). To prove that \(S \circ T\) is one to one, we need to show that if \(S(T (\vec{v})) = \vec{0}\) it follows that \(\vec{v} = \vec{0}\). Key Idea 1.4.1: Consistent Solution Types. We can describe \(\mathrm{ker}(T)\) as follows. Now consider the linear system \[\begin{align}\begin{aligned} x+y&=1\\2x+2y&=2.\end{aligned}\end{align} \nonumber \] It is clear that while we have two equations, they are essentially the same equation; the second is just a multiple of the first.
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