Prove that if matrix $A$ is similar to matrix $B$, then matrix $A^2 + A + I$ is similar to $B^2+B+I$
If matrix $A$ is similar to matrix $B$, then matrix $A^2 + A + I$ is similar to $B^2+B+I$
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linear-algebra
matrices
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0what is matrix $E$? – 2017-01-27
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0Identity matrix – 2017-01-27
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3I downvoted this question because it shows no effort and the solution is very basic – 2017-01-27
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0If $A$ and $B$ are similar, then $p(A)$ and $p(B)$ are similar for any polynomial $p$. In fact, it is true in more generality, for example, take $f$ to be entire, then $f(A)$ is similar to $f(B)$. – 2017-01-27
3 Answers
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$A$ is similar to matrix $B$ implies there exists a matrix $P$ such that $A = P^{-1}BP$
Now $A^2 + A + E = P^{-1}[B^2 +B+E]P$.
Hence matrix $A^2 + A + E$ is similar to $B^2+B+E$.
Used Facts: $A^2 = P^{-1}BPP^{-1}BP = P^{-1}B^2P$ and $E = P^{-1}P$.
0
Hint: By assumption we have $B=SAS^{-1}$ with invertible $S$. Now compute $B^2+B^1+B^0$. Note that $B^k=SA^kS^{-1}$ for all $k\ge 0$.
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If $A$ is similar to $B$, then there is an invertible matrix $P$, such that
$P^{-1}AP=B$
By computing $B^2=P^{-1}APP^{-1}AP=P^{-1}A^2P$
Which means $A^2$ and $B^2$ are similar.
Now apply $P$ and $P^{-1}$ to $A^2+A+E$ to get
$P^{-1}(A^2+A+E)P=P^{-1}A^2P+P^{-1}AP+E=B^2+B+E$