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In Exercises $25-28$ , determine if the specified linear transformation is (a) one-to-one and (b) onto. Justify each answer. The transformation in Exercise 19

a. $A=\left[\begin{array}{ccc}{1} & {-5} & {4} \\ {0} & {1} & {-6}\end{array}\right]$b. onto

Algebra

Chapter 1

Linear Equations in Linear Algebra

Section 9

The Matrix of a Linear Transformation

Introduction to Matrices

Oregon State University

McMaster University

University of Michigan - Ann Arbor

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for this example, we're going to be analyzing the transformation. T, which maps the vector space, are three into the vector space are, too. It goes from our three the domain because we have three entries here, and it goes to our to because the co domain contains just two entries. Next, let's rewrite this in a slightly friendlier format where Row one of the SPECTRE is x one minus five x two plus four x three, then the second row of this factor is x two minus six x three. With this format, we can express T as mapping a vector X where X is a column vector containing x one x two x three into eight times X where the standard matrix A of the transformation T can be found as follows. For column one. Pull the coefficients we see, which are 10 for X one. Calm to pull coefficients again. They are negative 51 and coefficients and column three R four and negative six. So this is our standard matrix A for the transformation. T. Let's analyze this matrix a little bit further. If we look at the pivots, there's a pivot here and here and this tells us that a has a pivot in every row. Well, when we have a pivot in every row, that implies immediately that T maps are three on to are to let's now discuss whether or not this transfer teach transformation T is going to be one toe one. The first noticed. Something is interesting going on here. We have something that's considered three dimensional mapping into something that's considered two dimensional, where the transfer tick mation t is linear. If this happens, we're going to have a lot of vectors mapping into the zero vector and so t cannot be Oneto one. Well, that's the intuition behind it. Let's also express why T cannot be Oneto one. Using appropriate dirhams. Let's go back to our Matrix A and analyze the pivots once again, where this time we're more concerned about the columns than Rose. We have a pivot here here, and the fourth column is not ah, pivot column so we can state next A does not have a pivot in every column. What this then tells us that the columns of a are linearly dependent and whenever the columns of a matrix, the standard matrix from a transformation are linearly dependent. It follows immediately that t is not 1 to 1. So it's recap on what we've shown here for the standard matrix A. We found that there is a pivot in every row. That means it's on on to transformation. But we found if we think of columns, there's not a pivot in every column. So it is not 1 to 1. So for on to think rose for one toe, one Think columns.

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