Let B be a fixed n x n invertible matrix. Define T:Mnn →Mnn by T (A) = B-¹AB. Find T (I) and T(B). i) ii) Show that T is a linear transformation. iii) Show that ker(T) = (0). What ia nullity(7)? Show that if CE Man, then CER(T). iv)

a) [tex]\(T(I) = B^{-1}IB = B^{-1}B = I\), \(T(B) = B^{-1}BB = B^{-1}B = I\)[/tex]

b) T is a linear transformation as it satisfies the properties of additivity and homogeneity.

c)[tex]\(\text{ker}(T) = \{0\}\), \(\text{nullity}(T) = 0\)\\[/tex], [tex]\(CE \in M_{an}\) implies \(C \in \text{ker}(T)\)[/tex]

a) [tex]\(T(I) = B^{-1}IB = B^{-1}B = I\)[/tex], [tex]\(T(B) = B^{-1}BB = B^{-1}B = I\)[/tex].

b) T is a linear transformation as it preserves addition and scalar multiplication:[tex]\(T(A+C) = T(A) + T(C)\)[/tex] and[tex]\(T(A+C) = T(A) + T(C)\)[/tex].

c) The kernel of T, [tex]\(T(A+C) = T(A) + T(C)\)[/tex], consists only of the zero matrix:[tex]\(T(A+C) = T(A) + T(C)\)[/tex].

d) The nullity of T, [tex]\(\text{nullity}(T)\)[/tex], is 0 since the kernel contains only the zero matrix.

e) If C is invertible and A is in [tex]\(\text{nullity}(T)\)[/tex], then CA is also in[tex]\(M_{an}\)[/tex].

Therefore, if [tex]\(\text{nullity}(T)\)[/tex] then[tex]\(\text{nullity}(T)\)[/tex]

T and TB both equal . T is a linear transformation preserving addition and scalar multiplication. The kernel of T is 0 with nullity 0. If [tex]\(\text{nullity}(T)\)[/tex], the[tex]\(\text{nullity}(T)\)[/tex]

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The linear transformation T(A) = B^(-1)AB, where B is a fixed n x n invertible matrix, is analyzed. It is found that T(I) = I and T(B) = B, demonstrating the behavior of T on the identity matrix and the matrix B itself. The linearity of T is proven by showing that it satisfies the properties of additivity and scalar multiplication. The kernel of T is shown to be {0}, indicating that the only matrix mapped to the zero matrix under T is the zero matrix itself. The nullity of T is determined to be 0, meaning T is injective or one-to-one. Lastly, it is demonstrated that if C is an eigenvector of B, then C is also an eigenvector of T.

For T(I), substituting I into the expression T(A) = B^(-1)AB yields T(I) = B^(-1)IB = B^(-1)B = I, where I is the identity matrix. This shows that the identity matrix remains unchanged under the transformation T.
Similarly, for T(B), substituting B into the expression T(A) = B^(-1)AB gives T(B) = B^(-1)BB = B^(-1)B^2 = B. This indicates that the matrix B itself is mapped to B under T.
To prove the linearity of T, let A and C be two matrices, and let k be a scalar. By substituting kA and A + C into the expression for T(A), we can demonstrate additivity and scalar multiplication. T(kA) = B^(-1)(kA)B = kB^(-1)AB = kT(A), showing scalar multiplication. T(A + C) = B^(-1)(A + C)B = B^(-1)AB + B^(-1)CB = T(A) + T(C), demonstrating additivity.
The kernel of T, denoted ker(T), is the set of matrices A for which T(A) = 0. Since T(A) = B^(-1)AB, the only matrix that satisfies this condition is A = 0, resulting in ker(T) = {0}.
The nullity of T, denoted nullity(T), is the dimension of the kernel of T. Since ker(T) = {0}, which contains only the zero matrix, nullity(T) = 0. This implies that T is injective or one-to-one.
Lastly, if C is an eigenvector of B, then B^(-1)CB = λC, where λ is the corresponding eigenvalue. Substituting this into the expression for T(A), we find T(C) = B^(-1)CB = λC. Thus, if C is an eigenvector of B, it is also an eigenvector of T with the same eigenvalue λ.


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the complete question is:

  Let B be a fixed n x n invertible matrix. Define T:Mnn →Mnn by T (A) = B-¹AB. Find T (I) and T(B). i) ii) Show that T is a linear transformation. iii) Show that ker(T) = (0). What ia nullity(7)? Show that if CE Man, then CER(T). iv)Show that if CE Man, then CE R(T).GENERATE THE ANSWER IN 150 WORDS IT WILL BE DIVIDED INTO TWO PARAGRAPHS THE FIRST PARA WILL BE THE SUMMARY OF THE ANSWER in 2 lines, AND SECOND PARA WILL BE THE EXPLANATION OF THE ANSWER

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