- #1
rmiller70015
- 110
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Homework Statement
This is for a linear algebra class, but it's taught my mathematicians, for mathematicians and not physicists or engineers so we write pseudo-proofs to explain things.
In Exercises 37–40, let T be the linear transformation whose standard matrix is given. In Exercises 37 and 38, decide if T is a one-to-one mapping.
38. ##A = \begin{bmatrix}
7&5&4&-9\\
10&6&16&-4\\
12&8&12&7\\
-8&-6&-2&5\\
\end{bmatrix}##
Homework Equations
Theorem 9:
Theorem 12:
The Attempt at a Solution
I used mathematica to row reduce the matrix and I got:
##A = \begin{bmatrix} 1&0&7&0\\
0&1&-9&0\\
0&0&0&1\\
0&0&0&0\\
\end{bmatrix}##
Let T: ##\mathbb{R}^m \rightarrow \mathbb{R}^n## such that ##T(\bar{x}) = A\bar{x}##, where ##A=[\bar{a_1} \bar{a_2} \bar{a_3} \bar{a_4}]##
By Theorem 12, ##T(\bar{x})## is one to one if the columns of ##A## are linearly independent.
Let there be an indexed set of vectors ##S_A = \{\bar{a_1}, \bar{a_2}, \bar{a_3}, \bar{a_4}\}##, then by theorem 9 ##S_A## is linearly dependent if ##\bar{0} \in S_A##, but ##\bar{0} \notin S_A##, so ##S_A## is linearly independent.
Therefore, ##T(x)## is one to one.
TL;DR My answer is that the mapping is one to one, but I am not sure if that is true or not.