Let{x11X 12M2: xij E RX21X22be the vector space of 2 × 2 real matrices with entrywise addition and scalar multiplication. Let[1A =and consider the linear transformation T' : M2 → M2 defined byT(X)= AXХА.(a) Find a basis of Ker(T). what is the dimension of Ker(T)?(b) Find an eigenvector of T' with strictly negative eigenvalue, (i.e., a nonzero X E M2 suchthat T(X) = \X for some < 0) or show that no such eigenvector exists.

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{ X11 X12 M2 : Xij E R X 22 X21 be the vector space of 2 × 2 real matrices with entrywise addition and scalar multiplication. Let [1 A 0 2 and consider the linear transformation T : M2 → M2 defined by T(X) — АХ —ХА. (a) Find a basis of Ker(T). what is the dimension of Ker(T)? (b) Find an eigenvector of T with strictly negative eigenvalue, (i.e., a nonzero X E M2 such that T(X) = XX for somel< 0) or show that no such eigenvector exists.

{ X11 X12 M2 : Xij E R X 22 X21 be the vector space of 2 × 2 real matrices with entrywise addition and scalar multiplication. Let [1 A 0 2 and consider the linear transformation T : M2 → M2 defined by T(X) — АХ —ХА. (a) Find a basis of Ker(T). what is the dimension of Ker(T)? (b) Find an eigenvector of T with strictly negative eigenvalue, (i.e., a nonzero X E M2 such that T(X) = XX for somel< 0) or show that no such eigenvector exists.

Algebra and Trigonometry (6th Edition)

6th Edition

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Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

Section: Chapter Questions

Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

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Question

Let
{
x11
X 12
M2
: xij E R
X21
X22
be the vector space of 2 × 2 real matrices with entrywise addition and scalar multiplication. Let
[1
A =
and consider the linear transformation T' : M2 → M2 defined by
T(X)= AX
ХА.
(a) Find a basis of Ker(T). what is the dimension of Ker(T)?
(b) Find an eigenvector of T' with strictly negative eigenvalue, (i.e., a nonzero X E M2 such
that T(X) = \X for some < 0) or show that no such eigenvector exists.

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