With the linear maps $T_1$ and $T_2$ be linear endomorphisms on a vector space $V$. If $T_2$ is diagonalizable, is $T_1$$(T_2)$ also diagonalizable? Also if $f(T)$ = $T_1(T_2)$ is $f$ diagonalizable? How do I show this is the case?
Determining if a linear transformation composition is diagonalizable.
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linear-algebra
eigenvalues-eigenvectors
linear-transformations
diagonalization
function-and-relation-composition
1 Answers
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What if $T_1$ is not diagonalizable, and $T_2$ is the identity map?
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0How would $T_2$ being identity map make the composition diagonalizable? – 2017-02-23
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0@BBest That’s exactly the point. – 2017-02-23
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0@amd but wouldn't $T_2$ map all **v** to itself? Then I don't understand how that has an effect of diagonalizablilty for $T_1$ – 2017-02-23
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0@BBest In your question, you didn’t ask about the diagonalizability of $T_1$. You asked about the diagonalizability of the composition of the two maps. – 2017-02-23
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0Right, I guess I should of more specifically asked how do I show this – 2017-02-23
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0@B Best: My answer shows that it's possible to have $T2$ diagonalizable, but $T1(T2)$ _not_ diagonalizable. – 2017-02-23