Let $\langle \cdot, \cdot \rangle$ be Hermitian inner product and $A=(a_{ij})_{1\leqq i,j\leqq n}$ be $n\times n$ matrix of $\mathbb C.$

Prove this claim.

Claim :

If $\langle Ax,x\rangle=\langle x,Ax\rangle$ for all $x\in \mathbb C^n$, then $A$ is Hermitian matrix.

It suffices to show $a_{jk}=\overline{a_{kj}}$ for all $k,j=1,\cdots,n.$

Let $x\in \mathbb C^n$ and then \begin{align} &\langle Ax,x\rangle=\sum_{j,k=1}^n a_{jk}\overline{x_j} x_k\\ &\langle x,Ax\rangle=\sum_{j,k=1}^n \overline{a_{jk}} x_j \overline{x_k}=\sum_{j,k=1}^n \overline{a_{kj}} \ \overline{x_j}x_k \end{align}

From $\langle Ax,x\rangle=\langle x,Ax\rangle$, I get $\displaystyle\sum_{j,k=1}^n (a_{jk}-\overline{a_{kj}})\overline{x_j}x_k=0 \cdots (\ast)$

I want to derive $a_{jk}=\overline{a_{kj}}$ from $(\ast).$

The one way is here.

$(\ast)$ holds for all $x\in \mathbb C^n$, so consider

(i) $x=\begin{pmatrix}0\\ \vdots \\ 1 \\ \vdots \\ 0 \end{pmatrix}$ : the $m$-th component is $1$ and the others are $0$

(ii) $x=\begin{pmatrix}0\\ \vdots \\ i \\ \vdots \\ 1 \\ \vdots \\ 0 \end{pmatrix}$ : the $m$-th component is $i$, the $l$-th component is $1$, and the others are $0$.

Putting (i) into $(\ast)$, I get $a_{mm}=\overline{a_{mm}}$,

and putting (ii) into $(\ast)$ and considering $a_{mm}=\overline{a_{mm}}$, I get $a_{ml}=\overline{a_{lm}}$ for $l\neq m$.

Thus, $a_{ml}=\overline{a_{lm}}$ for all $m,l=1,\cdots,n.$

But this method needs some calculation. Of course I think this method makes sense, but I wonder there is a shorter (cleverer) way.

I think $(\ast)$ is similar to the form relating linear independence.

The definition of liner independence is $$\sum_{k=1}^n c_k v_k=0 \Rightarrow c_k=0 (\forall k)$$

Now, $\displaystyle\sum_{j,k=1}^n (a_{jk}-\overline{a_{kj}})\overline{x_j}x_k=0$ is similar to $\displaystyle\sum_{k=1}^n c_k v_k=0$, and $c_k=0(\forall k)$ seems to correspond to $a_{jk}=\overline{a_{kj}}(\forall k,j)$.

So, the question : Is there a way to prove the claim using the linear independence ?

Of course, another way to prove the claim is also welcome.

  • 1
    $\begingroup$ One of $$\langle Ax,x\rangle=\sum_{j,k=1}^n a_{jk}\overline{x_j} x_k\\ \langle x,Ax\rangle=\sum_{j,k=1}^n \overline{a_{jk}} x_j \overline{x_k} $$ is wrong. The same $x$ should have the bar both times. $\endgroup$
    – Arthur
    Jul 18, 2022 at 4:20
  • $\begingroup$ Write $A = S + T$ with $S^* = S$ and $T^* = -T$. Afye obtain $\endgroup$
    – Mason
    Jul 18, 2022 at 4:54

3 Answers 3


Here is a different argument than what you are trying to do. It takes full advantage of properties of the Hermitian product to avoid all coordinate calculations. (Of course the proofs of the standard properties of the Hermitian inner product on $\mathbf C^n$ need coordinate calculations.) What I do may look too slick or abstract at first, but it is second nature to people with enough experience with such things, and it is the kind of method you need to be able to use (over time) in order to prove properties of Hermitian inner products outside the setting of the concrete vector space $\mathbf C^n$.

The key point is how $A$ turns into its adjoint (conjugate transpose) $A^*$ on the other side of the Hermitian inner product: for all $v$ and $w$ in $\mathbf C^n$, $$ \langle Av,w\rangle = \langle v,A^*w\rangle. $$ Prove this if you are not familiar with it. Ultimately this equation for all $v$ and $w$ is the most important property of $A^*$, since in more abstract settings it provides us with a way to define $A^*$ without having to use coordinates.

Anyway, granting the above identity, we have the hypothesis $\langle Ax,x\rangle = \langle x,Ax\rangle$ for all $x$ in $\mathbf C^n$ and want to conclude $A=A^*$.

We will turn the hypothesis into a property of two variable vectors by writing $x=v+w$, so $$ \langle A(v+w),v+w\rangle = \langle v+w,A(v+w)\rangle, $$ for all $v$ and $w$. Expanding both sides and using biadditivity of the inner product, and canceling equal terms on each side, we get $$ \langle Aw,v\rangle + \langle Av,w\rangle = \langle v,Aw\rangle + \langle w,Av\rangle $$ for all $v$ and $w$. That is an identity involving two arbitrary vectors in $\mathbf C^n$! Setting $w=v$ turns this into the hypothesis, so we have turned the original single variable vector condition into an equivalent two-variable vector condition. If you learn anything important mathematically here, let it be that amazing trick.

What next?

The right side is $\langle A^*v,w\rangle + \langle A^*w,v\rangle$, so after moving some terms around and using biadditivity again, $$ \langle (A-A^*)v,w\rangle = \langle (A^*-A)w,v\rangle. $$ We can move the $A^*-A$ on the right side to the second component as its adjoint, which is $A-A^*$, so $$ \langle (A-A^*)v,w\rangle = \langle w,(A-A^*)v\rangle = \overline{\langle (A-A^*)v,w\rangle}. $$ A number is its own complex conjugate only when it is real, so $\langle (A-A^*)v,w\rangle$ is real for all $v$ and $w$. However, replacing $v$ with $iv$ in $\langle (A-A^*)v,w\rangle$ multiplies the inner product by $i$. The only real number equal to $i$ times itself is $0$, so we (finally) have $\langle (A-A^*)v,w\rangle = 0$ for all $v$ and $w$.

For an $n \times n$ complex matrix $B$, if $\langle Bv,w\rangle = 0$ for all $v$ and $w$, then setting $w = Bv$ shows $\langle Bv,Bv\rangle = 0$, or $||Bv||^2 = 0$ for all $v$. Thus $Bv = 0$ for all $v$, so $B = O$.

Thus $A-A^* = O$, so $A = A^*$.

Because I never used a single coordinate calculation, this argument works on arbitrary finite-dimensional complex vector spaces equipped with a Hermitian inner product: all I used were properties of Hermitian inner products and the meaning of the adjoint to a linear map with respect to a Hermitian inner product. In the infinite-dimensional setting this proof still works as long as the inner product space is complete (a Hilbert space) since for a Hilbert space $H$ you can define the adjoint of every linear map $A : H \to H$. (Finite-dimensional inner product spaces are always complete, and we can use dimension arguments on them to avoid having to bring up completeness.)


A more general result is that if $\langle x, A x \rangle = 0$ for all $x \in \mathbb{C}^n$ then $A = 0$ (not true for $\mathbb{R}^n$ in general).

To see this expand $\langle x+y, A (x+y) \rangle, \langle x+iy, A (x+iy) \rangle$ to show that $\langle x, A y \rangle = 0$ for all $x,y \in \mathbb{C}^n$.

Hence if $\langle x, A x \rangle = \langle Ax, x \rangle = \langle x, A^*x \rangle$, we have $\langle x, (A-A^*) x \rangle =0$ for all $x$ and hence $A=A^*$.

  • $\begingroup$ Excellent. This the summary of my answer in a clean way. $\endgroup$ Jul 18, 2022 at 7:10

Let $V, W$ be two finite dimensional vector spaces over $\mathbb{K}$.

$T\in\mathcal{L}(V,W) $

Then Adjoint of $T := T^{\star}: W\to V$ $$\langle Tv, w\rangle=\langle v, T^{\star}w\rangle$$

$T$ is called self-adjoint if $T=T^{\star}$ i.e $$\langle Tv, w\rangle=\langle v, Tw\rangle$$

For $A\in M_n(\Bbb{K}) $ , let us consider $T_A : \Bbb{K^n}\to \Bbb{K^n}$ defined by $$T_A(x) =Ax$$

Then $A$ is self-adjoint matrix i.e $(\overline{A})^T=A$ iff $T_A$ is an self-adjoint operator i.e ${T_A}^{\star}=T_A$

For the rest of the proof we will assume $T_A=A$ .

Goal : To show $A^{\star}=A$

Step-$1$ : For an operator $T$ , $\langle Tu, v\rangle= 0$ $\forall u, v$ implies $T=0$

Proof : Choose $v=Tu$ , then $\langle Tu, Tu\rangle= 0$ implies $Tu=0$ and this is true for all $u$ . Hence $T=0$ .

Step-$2$ : For an operator $T$ on complex vector space $\langle Tv, v\rangle= 0$ iff $T=0$

Sketch : Prove the identity

$$\langle Tu, v\rangle= \frac{1}{4}[\langle T(u+v),u+v\rangle -\langle T(u-v),u-v\rangle +i\langle T(u+iv),u+iv\rangle -i \langle T(u-iv),u-iv\rangle ]$$

Now by $\langle Tv,v\rangle=0$ , we have $\langle Tu,v\rangle =0$ for all $u, v$ . Hence $T=0$.

Note: This result in not true in Real vector space . For an example $T:\Bbb{R}^2\to\Bbb{R}^2$ defined by $T(x, y) =(-y, x) $ satisfy $\langle T(x, y),(x, y) \rangle =0$ forall $(x, y) \in\Bbb{R}^2$ but $T\neq 0$ .

Step-$3$ : What left? We are done!

$\langle ( T-T^{\star})v,v\rangle =\langle Tv,v\rangle -\langle T^{\star}v,v\rangle =\langle Tv,v\rangle-\langle v,Tv\rangle=0$

But step-$2$ , $T-T^{\star}=0$ implies $T=T^{\star}$.This proves that $T$ is Hermition.

I have learned this technique from "Linear Algebra Done Right by Sheldon Axler"

  • $\begingroup$ seft-adjoint --> self-adjoint. $\endgroup$
    – KCd
    Jul 18, 2022 at 6:45
  • $\begingroup$ I haven't understood your comment. $\endgroup$ Jul 18, 2022 at 6:46
  • $\begingroup$ Read the two phrases: one is misspelled (in English). $\endgroup$
    – KCd
    Jul 18, 2022 at 8:56
  • $\begingroup$ My bad. Thanks for point out fatal mistake. $\endgroup$ Jul 18, 2022 at 8:59

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