Question

text{Let A be an} n times n text{matrix and suppose that} L : M_{nn} rightarrow M_{nn} text{is defined by} L(x)=AX, text{for} X in M_{nn}. text{Show that L is a linear transformation.}

Transformation properties
ANSWERED
asked 2021-02-10
\(\displaystyle\text{Let A be an}\ {n}\ \times\ {n}\ \text{matrix and suppose that}\ {L}:{M}_{{\cap}}\ \rightarrow\ {M}_{{\cap}}\ \text{is defined by}{L}{\left({x}\right)}={A}{X},\text{for}\ {X}\in{M}_{{\cap}}.\text{Show that L is a linear transformation.}\)

Answers (1)

2021-02-11
It's enought to show that the below property got satisfied:
\(\displaystyle{\left({i}\right)}{L}{\left({x}\ +\ {y}\right)}={L}{\left({x}\right)}\ +\ {L}{\left({y}\right)}\)
\(\displaystyle{\left({i}{i}\right)}{L}{\left(\alpha{x}\right)}=\alpha{L}{\left({x}\right)}\)
Now Consider,
\(\displaystyle{L}{\left({x}\ +\ {y}\right)}={A}{\left({X}\ ++{Y}\right)}{\left(\text{use given definition}\ {L}{\left({x}\right)}={A}{X}\right)}\)
\(\displaystyle={A}{X}\ +\ {A}{Y}\)
\(\displaystyle={L}{\left({x}\right)}\ +\ {L}{\left({y}\right)}{\left(\text{By definition we get}\right)}\)
\(\displaystyle\text{Therefore},{L}{\left({x}\ +\ {y}\right)}={L}{\left({x}\right)}\ +\ {L}{\left({y}\right)}\)
Property (i) is satisfied.
Now consider,
\(\displaystyle{L}{\left({a}{x}\right)}={A}{\left({a}{X}\right)}{\left(\text{use given definition}\ {L}{\left({x}\right)}={A}{X}\right)}\)
\(\displaystyle={\left({A}\alpha\right)}{X}{\left(\text{Use associative property}\right)}\)
\(\displaystyle={\left(\alpha{A}\right)}{X}{\left(\text{Use Commutative property}\right)}\)
\(\displaystyle=\alpha{\left({A}{X}\right)}{\left(\text{Use associative property}\right)}\)
\(\displaystyle=\alpha{L}{\left({x}\right)}{\left(\text{By definition we get}\right)}\)
\(\displaystyle\text{Therefore},{L}{\left(\alpha{x}\right)}=\alpha{L}{\left({x}\right)}\)
Property (ii) is satisfied.
Therefore both properties got satisfied.
Therefore L is a linear transformation.
0
 
Best answer

expert advice

Need a better answer?

Relevant Questions

asked 2021-07-04

Let \(AX = B\) be a system of linear equations, where A is an \(m\times nm\times n\) matrix, X is an n-vector, and \(BB\) is an m-vector. Assume that there is one solution \(X=X0\). Show that every solution is of the form \(X0+Y\), where Y is a solution of the homogeneous system \(AY = 0\), and conversely any vector of the form \(X0+Y\) is a solution.

asked 2021-06-10
Determine whether the given set S is a subspace of the vector space V.
A. V=\(P_5\), and S is the subset of \(P_5\) consisting of those polynomials satisfying p(1)>p(0).
B. \(V=R_3\), and S is the set of vectors \((x_1,x_2,x_3)\) in V satisfying \(x_1-6x_2+x_3=5\).
C. \(V=R^n\), and S is the set of solutions to the homogeneous linear system Ax=0 where A is a fixed m×n matrix.
D. V=\(C^2(I)\), and S is the subset of V consisting of those functions satisfying the differential equation y″−4y′+3y=0.
E. V is the vector space of all real-valued functions defined on the interval [a,b], and S is the subset of V consisting of those functions satisfying f(a)=5.
F. V=\(P_n\), and S is the subset of \(P_n\) consisting of those polynomials satisfying p(0)=0.
G. \(V=M_n(R)\), and S is the subset of all symmetric matrices
asked 2020-10-20
To show:
The set \(\displaystyle{\left\lbrace{T}{\left({x}_{{1}}\right)},\ \ldots\ ,{T}{\left({x}_{{k}}\right)}\right\rbrace}\) is a linearly independent subset of \(\displaystyle{R}^{{{m}}}\)
Given:
Let \(\displaystyle{T}\ :\ {T}\ :\ {R}^{{{n}}}\rightarrow{R}^{{{m}}}\) be a linear transformation with nulity zero. If \(\displaystyle{S}={\left\lbrace{x}_{{{1}}},\ \cdots\ \ ,{x}_{{{k}}}\right\rbrace}\) is a linearly independent subset of \(\displaystyle{R}^{{{n}}}.\)
asked 2021-01-04
Whether the statement “ Let A be an \(m\ \times\ n\) matrix.
The system \(Ax = b\ \text{is consistent for all b in}\ R^{m}\ \text{if and only if the columns of A form a generating set for}\ R^{m}\) " is true or false.
...