Here in "linear algebra hoffman kunze" book it is given that if $T$ is a linear transformation on vector space $V$ then $T^n = T \ldots T $ ($n$ times) and they define $T^0=I$ if $T\neq0$ where $I$ is the identity transformation. I could not understand what is meant by $T \neq 0 $.
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1$T=0$ if it maps every element to $0$. – Mar 15 '16 at 13:30
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So, if it is given that $T=0$ then why $T^0\neq I$? – uuuuuuuuuu Mar 15 '16 at 13:36
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2Probably just personal preference on the authors' part. – Mar 15 '16 at 13:38
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2@Saikat I see no particular reason to make $T = 0$ an exception here, but it doesn't really make a difference. The only thing worth remembering is that it's often convenient to define $T^0 = I$ (especially in power series of matrices). – Ben Grossmann Mar 15 '16 at 13:39
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2They are not defining the $0^{\text{th}}$ power of the zero matrix. – Lee Mosher Mar 15 '16 at 13:39
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@Lee Mosher How is zero matrix coming into picture? – uuuuuuuuuu Mar 15 '16 at 13:43
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1@Saikat: How? The authors must have chosen, for their own ineffable reasons, to explicitly exclude it from their definition of $T^0$. We can't know why they chose that without being them. – hmakholm left over Monica Mar 15 '16 at 13:45
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You usually make the same exception when you define $x^0=1$ for real numbers, and the same reasoning (almost) applies, so it should not be surprising. – Arthur Mar 15 '16 at 13:58
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@HenningMakholm Well, there might be a good (and understandable by us) reason for that. However, if this is the case, we probably need more context (and then, the question of Saikat makes fully sense to me). – Surb Mar 15 '16 at 13:59
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1@Omnomnomnom there must a case where if we define "$T^o=I$" and $T=0$ ,then there is some problem. – uuuuuuuuuu Mar 15 '16 at 14:12
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1@Saikat there isn't one that I know of – Ben Grossmann Mar 15 '16 at 14:14
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@Omnomnomnom me neither, but isn't a proof that it doesn't exist :). – Surb Mar 15 '16 at 14:14
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1The zero matrix is the matrix representing the linear transformation $T=0$. – Lee Mosher Mar 15 '16 at 14:55
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If $T:V \to V$, then $T = 0$ means that $T$ satisfies $T(v) = 0$ for every $v \in V$.
Ben Grossmann
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@HM fair point. I might delete this answer at any rate, given the ensuing discussion above. – Ben Grossmann Mar 15 '16 at 13:49
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In my opinion, the authors define $T^{0}=I$ if $T \neq 0$, they want to express $T^{1}=T^{0}T=T$ and $T^{n}=T^{n-1}T$.
bing
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2What is the problem in defining $T^{n}=T^{n-1}T$ if $T=0$ and $T^{0}=I$? Then, we have $T= 0\cdot I=0$ and $T^{n}=0$ for any $n>1$. – Surb Mar 15 '16 at 13:54
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Suppose that a linear transformation on vector space $V$ is a matrix $A$. Then $T^n=TT ...T$ (n times) is equivalent to be $A^n$. – bing Mar 15 '16 at 14:05
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@bing the question is why did the author exclude $T=0$ from the definition. – Ben Grossmann Mar 15 '16 at 14:13
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@bing If it would be trivial, there wouldn't be a need to exclude this case I guess. – Surb Mar 15 '16 at 14:19
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1@bing Why is it ridiculous? If $T\colon V \to V$ and $V=\Bbb R^1$, then $T^0=0^0=1$ is a commonly accepted convention. – Surb Mar 15 '16 at 14:28
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