Department of Mathematics, K.T.H.M. College, Nashik

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Registered: 23-07-2015

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NET B Part (Jnue 2016 )

Given \(n \times n\) matrix B define \(e^B\)  by $$e^B= \sum_{j=0}^{\infty} \frac{B^j}{j !} $$ Let p be the characteristic polynomial of B. Then the matrix \(e^{p(B)}\) is???

A)\(  I_{n \times n}\)
B)\(  0_{n \times n}\)
C)\( e I_{n \times n} \)
D)\(  \pi I_{n \times n} \)

First look at  \(e^B\) , It gives \(e^B= \sum_{j=0}^{\infty} \frac{B^j}{j !} \)
\(e^B= \sum_{j=0}^{\infty} \frac{B^j}{j !}=I + \frac{B^1}{ 1!}+\frac{B^2+}{ 2!}+\frac{B^3}{ 3!}+\dots \)

Now p is characteristic polynomial of B \(\implies \) \(p(B)=0\) By cayleyHamilton theorem.
Now calculate \(e^{p(B)}\) .
\(e^{p(B)}=\sum_{j=0}^{\infty} \frac{p(B)^j}{j !}=I + \frac{p(B)^1}{ 1!}+\frac{p(B)^2+}{ 2!}+\frac{p(B)^3}{ 3!}+\dots \)
\(e^{p(B)}=\sum_{j=0}^{\infty} \frac{p(B)^j}{j !}=I + 0+0+0+\dots \)
\(e^{p(B)}=I_{n \times n} \)