# The range of $T^*$ is the orthogonal complement of $\ker(T)$

How can I prove that, if $V$ is a finite-dimensional vector space with inner product and $T$ a linear operator in $V$, then the range of $T^*$ is the orthogonal complement of the null space of $T$?

I know what I must do (for a $v$ in the range of $T^*$, I have to show that $v\perp w$ for every $w$ in $\ker(T)$ and then do the opposite), but I don't know how to show that this inner product is zero.

• Is $T^*$ the traspose operator of $T$? Commented Mar 1, 2013 at 21:36
• @Loronegro it's the adjoint. Commented Mar 1, 2013 at 21:54

In order to show that the range of $T^*$ is the orthogonal complement of $\ker T$, we have to show that $\forall v \in \operatorname{Im}T^*$, $\forall w\in \ker T$: $\left<v,w\right>=0$.

Note that vectors in the range of $T^*$ are of the form $T^*v$ for $v\in V$. Now, let $w\in\ker T$. We have to show that $\left<T^*v,w\right>=0$. And, indeed, $\left<T^*v,w\right>=\left<v,Tw\right>=\left<v,0\right>=0$.

• You're using the fact that the range of $T^*$ is invariant under $T^*$ right? That is, if $\left\{ v_1,...,v_n\right\}$ is a basis for $V$ then $\left\{ T^* v_1,...,T^* v_n \right\}=\left\{ T^* v_1,...,T^* v_m\right\}$, with $m<n$, is a basis for the range, and $\left\{ T^* v_{m+1},...,T^* v_n\right\}$ a basis for the null space, right? Commented Mar 1, 2013 at 19:43
• I'm not sure I understand the "invariant" part of what you said... But, anyway, I am not working with bases at all. I mean, I showed this part of your original statement: for every vector $v$ in the range of $T^*$, and every $w\in \ker T$, $v$ is orthogonal to $w$. Commented Mar 1, 2013 at 19:50
• I meant invariant under T. Commented Mar 1, 2013 at 20:07
• @Ludolila I think your first paragraph claims that the range or $T^*$ is IN the orthogonal complement of the kernel of $T$. Commented Sep 18, 2019 at 15:15
• It seems you only demonstrated that range(T*) is orthogonal to Kernel(T) but not the fact the two are a complement to each other. Commented Sep 15, 2022 at 10:22

Let $$A=\operatorname{ran}(T^*), B=\ker(T)^\perp$$.

$$\boxed{A\subseteq B:}$$

For $$x\in A, \ x=T^*y\$$ for some $$y\in V$$. Then, for any $$z\in \ker(T),\ \langle x,z\rangle=\langle T^*y,z\rangle=\langle y,Tz\rangle=\langle y,0\rangle=0.$$ Hence $$x\in B.$$

$$\\ \\ \boxed{B\subseteq A:}$$

Because $$V$$ is finite dimensional and $$A,B$$ is subspace, it is equivalent to $$A^\perp \subseteq B^\perp= \ker(T)$$. If $$x\in A^\perp$$, for any $$y\in V$$, $$0=\langle x,T^*y\rangle=\langle Tx,y\rangle$$ Therofore, $$Tx=0$$ (see exercise 8.1.1 (b) or simply take $$y=Tx$$) , and thus $$x\in\ker(T)$$.

• Exercise 8.1.1 (b) is from what source? Commented May 30, 2016 at 16:20
• Linear algebra 2nd Edition, Hofmann&Kunze. I'm sorry about that. I thought the question is from this book's exercise, so i missed a quotation. Commented Jun 1, 2016 at 14:31
• Can you explain why it is "equivalent to"? Commented Apr 20, 2021 at 18:43
• @Hawk It is because $(A^{\perp})^{\perp}=A$ if $V$ is finite-dimensional and $A$ is a subspace of $V$ (see Exercise 8.2.13 in Hofmann&Kunze or math.stackexchange.com/questions/2319680/…). You can easily show that if $B\subset A$ then $A^{\perp}\subset B^{\perp}$. Apply this argument for $A^{\perp}$ and $B^{\perp}$. We then obtain $B=(B^{\perp})^{\perp} \subset (A^{\perp})^{\perp} = A$. Commented Apr 22, 2021 at 2:55
• Finite dimensional hypothesis is a quite important assumption. Otherwise, we would need to work with a complete Hilbert subspace of a Hilbert space. Actually, what is needed to derive is a type of Moreau Decomposition theorem: for each Hilbert space $W$ and every Hilbert subspace $V$, $W=V^{\perp} + V.$ Taking this into consideration, $B=A^{\perp} + A$.Hence,since$u\in B,$ then $u=v+w$, with $w\in A^{\perp}$ and $v\in A$. We need only to prove that $Tw=0$, but given $x$, $x^{T} T w = (T^{*} x)^{T} w = 0.$ Hence, $T w = 0$.Thus,$0=u^{T}w=v^{T}w+\|w\|^2 = \|w\|^2$,since $u \in$ker$(T)^{\perp}.$ Commented Nov 15, 2023 at 0:07