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Suppose $T$ and $U$ are linear transformations from $\mathbb{R}^{n}$ to $\mathbb{R}^{n}$ such that $T(U \mathbf{x})=\mathbf{x}$ for all $\mathbf{x}$ in $\mathbb{R}^{n} .$ Is it true that $U(T \mathbf{x})=\mathbf{x}$ for all $\mathbf{x}$ in $\mathbb{R}^{n}$ ? Why or why not?

true

Algebra

Chapter 2

Matrix Algebra

Section 3

Characterizations of Invertible Matrices

Introduction to Matrices

Campbell University

Oregon State University

Harvey Mudd College

Lectures

01:32

In mathematics, the absolu…

01:11

02:53

Suppose a linear transform…

04:11

Let $f : \mathbb{R}^{n} \r…

05:13

Suppose $T$ and $S$ satisf…

01:34

Show that the function $D:…

02:02

Let $T_{1}: M_{n}(\mathbb{…

02:40

Suppose vectors $\mathbf{v…

03:38

Let $T_{1}: V \rightarrow …

03:37

Let $T : \mathbb{R}^{2} \r…

04:17

In the following exercises…

02:27

Let $S : \mathbb{R}^{p} \r…

in problem 37. Given that that he transformation off that transfer, that you transformation for the victor X equals the Victory X for all X in our in and supporting that he and you our linear transformations from our end to our end. We want to see that if that the U transformation off the tee transformation of the victor Victor X equals X. This is the question we can conclude that from this fall we can rewrite it in the algebraic for any transformation off a function. For example, why equals the transformation off any victor? Why, for example, equals the standard metrics for that transformation multiplied by the victory, Then tee off any function see off any victor equals the standard metrics off the transformation t while deployed by the victim, New X equals X and using the same concept. That transformation off any vector Z, for example, equals the standard metrics for the U transformation. You can assume it. It's to be want to blow by the victories that then we can trump make it in algebraic full metrics a multiplied by the metrics be multiply by the victory X equals X. We can see that a B multiplied by the Victory X equals the Victory X, which means that a B equals the identity metrics. And now we can see that the metrics A has an inverse, which is B because when we multiply each other, we get the identity matrix. This is the definition off the embers, which makes which means that A and B are convertible but receives well and be or in veritable. But this is then A and B are a vertebral mattresses, and they are in veritable mattresses because they both are square mattresses square but receives because the transformation is from our end to our end, which means that the standard metrics is end by any and here be is MBA in. Then we have the identity matrix in This is the conclusion from the given. Let's see what's required toe be proven. We want to to see if this is true. Let's take the left hand side. The left hand side means that we have the standard metrics be multiplied by the victor off the transformation TX equals X, and from this assumption from this form, we will transform toe a job for which makes be multiplied by A which is the standard metrics for the transformation function. T the driver Victor X equals X not it's not equal X because we want toe. See that if it will give us X, then we have b equals p Multiply by t equals B multiply by a X and we know that e b r in vertebral mattresses and they are the inverse of each other. B is the inverse of a and A is in front of me which makes b equals oy equals I and multiplied by X equals X and it equals the right hand side, which means that this statement is true and this is the final answer of our property.

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