# Help Showing that the Adjoint Operator $T^*$ is Surjective if and only if $T$ is Injective

Let $T\in L(V,W)$,where $L(V,W)$ denotes a linear map from a vector space $V$ to vector space $W$. I want to prove that $T$ is injective iff $T^*$ is surjective, where $T^*$ is the adjoint of $T$. I start with the definition of adjoint: $\langle w,Tv \rangle= \langle T^*w,v \rangle$ for all $w \in W$, $v\in V$. What should I do next? Take $v=0$?

• Do you know about the matrix representation of the adjoint map? Commented Nov 27, 2017 at 11:58
• do you mean conjugate tranpose?
– user227158
Commented Nov 27, 2017 at 12:05
• Right. (In the real case, it's just the transpose.) Can you see how to proceed from here? Commented Nov 27, 2017 at 12:07
• how it relates to inner product in this case?
– user227158
Commented Nov 27, 2017 at 12:09
• Well, one way to see this is to analyze the ranks of the matrix representations of $T$ and $T^*$, noting that the rank is preserved under transposition. Commented Nov 27, 2017 at 12:11

Suppose $$T^*$$ exists and is surjective.

Let $$\mathbf{v} \in \ker(T)$$. Since $$T^*$$ is surjective, there exists $$\mathbf{w} \in W$$ such that $$T^*(\mathbf{w}) = \mathbf{v}$$. Then

$$\langle \mathbf{v}, \mathbf{v} \rangle = \langle \mathbf{v}, T^*(\mathbf{w}) \rangle = \langle T(\mathbf{v}), \mathbf{w} \rangle = \langle \mathbf{0}_W, \mathbf{w} \rangle = 0$$

This only happens when $$\mathbf{v} = \mathbf{0}_V$$. Hence, $$\ker(T) = \{\mathbf{0}_V\}$$ and $$T$$ is injective.

The converse is not true in general, here is a counter example. Let $$V = W = l^2$$, the space of square-summable sequences. Define a linear operator $$T$$ such that

$$T((a_n)) = (a_n - a_{n+1}),$$ for all $$(a_n) \in V$$. (Identity minus left-shift)

1. $$T$$ is injective. (The only constant, square-summable sequence is the zero-sequence.)
2. $$T^*$$ exists (Identity minus right-shift)
3. $$T^*$$ is not surjective. (The sequence $$(1,0,0,\ldots) \in V$$ is not in $$\mathcal{R}(T^*)$$)

Let us note for the sake of clarity that given $T:V\to W$, $T^*:W\to V$, we have $\text{im}(T^*)\subset V$ and $\text{ker}(T)\subset V$. Now, $$\text{im}(T^*)^\perp=\{x\in V: \langle x,T^*y\rangle=0,\forall y\in W\}=\{x\in V:\langle Tx,y\rangle=0, \forall y\in W\}$$ $$=\{x\in V: Tx=0\}=\text{ker}(T).$$ Therefore, we have that $$\text{im}(T^*)^\perp=\text{ker}(T)=\{0\}.$$ This implies that
$$\text{im}(T^*)=(\text{im}(T^*)^\perp)^\perp=\{0\}^\perp=V.$$ So, we observe that $T^*$ is surjective. Suppose, conversely, that $T^*$ is surjective. Then, we see that $\text{im}(T^*)=V$. From before, we know that $\text{im}(T^*)^\perp=\text{ker}(T)$. This implies that $$V=(\text{im}(T^*)^\perp)^\perp=\text{ker}(T)^\perp.$$ Thus, we see that $\{0\}=\text{ker}(T)$, so that $T$ is injective. Do be sure that you can justify each of these steps.

• Should it be $im(T*)=W$ in order to show it is surjective?
– user227158
Commented Nov 27, 2017 at 12:39
• No, by definition $T^*:W\to V$, so that the image of $T^*$ lives in the target space, $V$. Commented Nov 27, 2017 at 12:40
• oh I see.last time I tried to use $rangeT*=(ker T)^\perp$.I ran into this problem and then I tried to think of another way to prove and stuck.Thank you
– user227158
Commented Nov 27, 2017 at 12:43
• It is of immeasurable importance to be sure you understand the domains/codomains of the objects you are discussing. Maybe try writing all of these data out at the top of the page before attempting the problem. It tends to help alleviate these sorts of frustrations. Commented Nov 27, 2017 at 12:45
• $W = (W^\perp)^\perp$ may not hold when $W$ is not finite dimensional.
– Bio
Commented Dec 2, 2019 at 14:55