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Prove that if $A$ is a self adjoint operator in vector space $V$, then $\text{im}(A) = \text{ker}^{\perp}(A)$.

My approach:

By spectral theorem, $A$ is diagonalizable, hence, it's matrix in it's eigenvector basis $\beta$ is going to be $$ \begin{pmatrix} \lambda_1 & 0 & 0 \\ 0 & \ddots & 0 \\ 0 & 0 & \lambda_n \end{pmatrix} $$ and by that matrix we can conclude that the vector $\vec{w} \in V $ such that $w = 0$ is the only vector in $\text{ker}(A)$. $$ \text{ker}(A) = \left\{ \vec{w} \right\} = \left\{ \vec{0} \right\} $$ By our conclusion, we can affirm that for any $\vec{u} \in V$ such that $\vec{u} \neq 0 $ is going to be orthogonal to $\text{ker}(A)$: $$ <\vec{u}, \vec{0}> = 0 $$ Then, we can affirm that all $\vec{z} \in \text{im}(A)$ is going to be represented by some linear combination of the vectors in $\beta$. But we can affirm the same for any vector $\vec{u} \in V$. Since $\vec{u}\perp\text{ker}(A)$ and it's decomposed in the same basis $\beta$, we can conclude that the set of all possible $\vec{u}$'s = $im(A)$

I'm really new into proofs in math, that's why I'm writing more than trying to prove it with math. But I think that my line of thoughts is ok. Can someone please check if that's a valid proof? Thanks.

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    It does not make sense to say "If $A$ is selfadjoint, prove that blah": what you mean is "prove that if $A$ is selfajdjoint, then blah" – 2017-01-22
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    Ok @MarianoSuárez-Álvarez I'll edit it! But can you please check if what I've done is indeed correct? – 2017-01-22

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Suppose that $A$ is selfadjoint. If $x$ is in the image of $A$ and $y$ is in the kernel, then there exists a $z$ such that $x=Az$ and then $$(x,y)=(Az,y)=(z,Ay)=(z,0)=0.$$ It follows that every element of the image of $A$ is orthogonal to every element of the kernel of $A$.

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    Perfect! but shouldn't it be: $x = A(z)$ ? – 2017-01-22
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    I think you mean $x = Az$. – 2017-01-22
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    Awesome proof. But can you check if what I've done is correct too? – 2017-01-22
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    I stopped reading when you say that the only vector in the kernel is zero, which is of course false. The matrix A could well be the zero matrix, for example. – 2017-01-22
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    And why did you stop? – 2017-01-22
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    @MarianoSuárez-Álvarez hmmm that's true... if you assume that you have eigenvalues = 0. you're totally correct. Do you have any kind of tips to give me on math proofs? – 2017-01-22
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    The best tip is, make sure you do not claim false things. – 2017-01-22
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    Thank you. It's kinda obvious, but as you can see, I claimed a false statement in my proof. Thank you anyway. – 2017-01-22
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    Well, the point is that you have to check every thing you claim. Why exactly did you claim that the only vector in the kernel was 0? If you had actually bothered to check, you'd have had to solve an equaltion and you'd have seen that the solutions depend on the values of the lambdas. – 2017-01-22
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    When I said *make sure you do not claim false things* what I meant was *check*. – 2017-01-22
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    One way to check is to consider examples. Your "proof" breaks down with the zero matrix! That is one of the obvious examples to try... – 2017-01-22
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    You're kinda arrogant mate. I'm sorry if my proof is too dumb for you, I'm new in math ok? I'll try to be more rigorous and try... I just forgot that lambdas can be 0... Sorry about that I've ruined your day with my dumb proof. – 2017-01-22
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    Your proof is not dumb —for me or for anyone. It is just wrong. And you asked me what sort of things one can do to avoid being wrong, and I told you. It was not my intention, really, but if me observing that you have to consier at leeast the zero matrix as an example has piqued your pride and from now on you remember to do that, I would say that this was a quite fine day for me! – 2017-01-22