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.

  • $\begingroup$ Is $T^*$ the traspose operator of $T$? $\endgroup$
    – Loronegro
    Commented Mar 1, 2013 at 21:36
  • 1
    $\begingroup$ @Loronegro it's the adjoint. $\endgroup$
    – user62182
    Commented Mar 1, 2013 at 21:54

2 Answers 2


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

  • $\begingroup$ 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? $\endgroup$
    – user62182
    Commented Mar 1, 2013 at 19:43
  • $\begingroup$ 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 $w$. $\endgroup$
    – Ludolila
    Commented Mar 1, 2013 at 19:50
  • $\begingroup$ I meant invariant under T. $\endgroup$
    – user62182
    Commented Mar 1, 2013 at 20:07
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    $\begingroup$ @Ludolila I think your first paragraph claims that the range or $T^*$ is IN the orthogonal complement of the kernel of $T$. $\endgroup$
    – inquisitor
    Commented Sep 18, 2019 at 15:15
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    $\begingroup$ 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. $\endgroup$
    – Rafael
    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 $x\in\ker(T)$.

  • 1
    $\begingroup$ Exercise 8.1.1 (b) is from what source? $\endgroup$
    – Leucippus
    Commented May 30, 2016 at 16:20
  • $\begingroup$ 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. $\endgroup$
    – rudgns55
    Commented Jun 1, 2016 at 14:31
  • $\begingroup$ Can you explain why it is "equivalent to"? $\endgroup$
    – Lemon
    Commented Apr 20, 2021 at 18:43
  • $\begingroup$ @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 $B=(B^{\perp})^{\perp} \subset (A^{\perp})^{\perp} = A$. $\endgroup$
    – rudgns55
    Commented Apr 22, 2021 at 2:55
  • $\begingroup$ 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$(T)^{\perp}.$ $\endgroup$ Commented Nov 15, 2023 at 0:07

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