# Let $T:V → V$ be a linear transformation where $V$ is finite dimensional. [closed]

Let $T:V → V$ be a linear transformation where $V$ is finite dimensional. Show that exactly one of (i) and (ii) holds:
(i) $T(v)=0$ for some $v ≠ 0$ in $V$ ;
(ii) $T(x)=v$ has a solution $x$ in $V$ for every $v$ in $V$

Not sure how to answer this, I'll get any help I can get. Thanks in advance!

## closed as off-topic by Xander Henderson, Paramanand Singh, GNUSupporter 8964民主女神 地下教會, Strants, Carl MummertMar 14 '18 at 18:39

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• You have $A \operatorname{adj} A = (\det A) I$. – copper.hat Mar 13 '18 at 21:28
• Compare this to the Fredholm alternative as shown for finite dimensional vector space $V$. – hardmath Mar 14 '18 at 16:31

## 3 Answers

If $(i)$ doesn't hold, then $T$ is injective. Therefore, it is surjective, which means that (ii) holds.

And if (i) holds, then $T$ is not injective. Therefore, it is not surjective, and so (ii) doesn't hold.

• In the second part I guess you mean $T$ is not surjective. – InsideOut Mar 13 '18 at 21:32
• @Legoman I've edited my answer. Thank you. – José Carlos Santos Mar 13 '18 at 21:35

Note that the two statements are mutually exclusive indeed

• (i) $\iff$ T not invertible
• (ii) $\iff$ T invertible

Consider the dimension of the null space for T. The dimension of nullspace is either $0$ or non-zero.

If it is non-zero. By def, i) is satisfied. Then we know the rank of $T$ is smaller than $dim(V)$. So $T(x)$ has smaller dimension than $V$, this means ii) is not satisfied.

If nullity is zero. Then by def, i) is not satisfied and $T$ is invertible. Hence, it means the dimension of $T(x)$ is equal to $dim(V)$ and this implies ii) is satisfied.

Hence in every case, exactly one condition is satisfied.