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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.

PinkyWay
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user62182
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2 Answers2

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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, $\left<T^*v,w\right>=\left<v,Tw\right>=\left<v,0\right>=0$.

Ludolila
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  • 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? – user62182 Mar 01 '13 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 $w$. – Ludolila Mar 01 '13 at 19:50
  • I meant invariant under T. – user62182 Mar 01 '13 at 20:07
  • Denote the range of $T^$ by $ImT^$. To say that $ImT^* $ is invariant under $T$, is to say that $\forall v \in ImT^$ we have $Tv \in ImT^$. And this is not something I want to claim/use. – Ludolila Mar 01 '13 at 20:12
  • Regarding what you said about the basis: note that if ${v_1,...,v_n}$ is a basis for $V$, it doesn't mean that ${T^v_1,...,T^v_m }$ form a basis for $ImT^*$ (although these vectors do span the range, they are not necessarily linearly independent). – Ludolila Mar 01 '13 at 20:15
  • Yes you are right... and I don't need to work with basis to show it like you did. Thanks! :) – user62182 Mar 01 '13 at 20:16
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    @Ludolila I think your first paragraph claims that the range or $T^*$ is IN the orthogonal complement of the kernel of $T$. – inquisitor Sep 18 '19 at 15:15
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    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. – Rafael Sep 15 '22 at 10:22
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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 $x\in\ker(T)$.

rudgns55
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    Exercise 8.1.1 (b) is from what source? – Leucippus May 30 '16 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. – rudgns55 Jun 01 '16 at 14:31
  • Can you explain why it is "equivalent to"? – Lemon Apr 20 '21 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 $B=(B^{\perp})^{\perp} \subset (A^{\perp})^{\perp} = A$. – rudgns55 Apr 22 '21 at 02:55
  • 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$(T)^{\perp}.$ – R. W. Prado Nov 15 '23 at 00:07
  • In the comment above, $u^T v=\langle u, v \rangle$. I was without space. – R. W. Prado Nov 15 '23 at 00:12