Vtyjkyuk

The algebraic multiplicity of the eigenvalues of a real symmetric matrix is equal to the geometric multiplicity.

(The instructor explained the proof in the classroom. I have tried to reproduce it below. But at certain parts, I am getting some difficulty in understanding. Hope the scholars of this mathematical community would assist me in this regard.)

The Proof as explained in the class:-

Let $A$ be an $ntimes n$ real symmetric matrix and $lambda$ be an eigenvalue of $A$. If $r$ is the geometric multiplicity of $lambda$, then the dimension of the eigenspace $E_lambda=v in mathbbR^n:A v=lambda v$ is $r$. Suppose we find an orthonormal set of vectors $v_1,v_2,dots,v_r$ which constitute a basis of $E_lambda$. Now the additional $n-r$ vectors $v_r+1,v_r+2,dots,v_n$ are found such that $v_1,v_2,dots,v_r,v_r+1,dots,v_n$ form an orthonormal basis of $mathbbR^n$. Isn't it due to Gram Schmidt Orthogonalization?

We have, $Av_i=lambda v_i$ for $1le i le r$.

Let us denote, $A v_i=sum_j=1^n c_ijv_j quad forall quad r+1 le i le n$

We can represent these $n$ equations in matrix form
beginequation*
beginaligned
& Abeginbmatrix
&v_1&dots&v_r&v_r+1dots&v_n
endbmatrix=beginbmatrix
lambda v_1&dotslambda v_r&sum_j=1^n c_r+1,jv_j dots&sum_j=1^n c_nj v_j
endbmatrix\
Rightarrow & Abeginbmatrix
& v_1 &dots& v_r &v_r+1 dots& v_n
endbmatrix=beginbmatrix
& v_1 &dots& v_r & v_r+1 dots& v_n
endbmatrix
beginbmatrix
lambda& & & & c_r+1,1&dots&c_n1\
&lambda& & & c_r+1,2&dots&c_n2\
&& &&vdots&ddots&vdots\
& & & lambda& c_r+1,r&dots&c_nr\
& & & & c_r+1,r+1&dots&c_n,r+1\
& & & &vdots&ddots&vdots\
& & & & c_r+1,n&dots&c_nn\
endbmatrix \
Rightarrow & AS=SQ qquad left(mboxSayright)
endaligned
endequation*
It is easy to see that $S$ is invertible and orthogonal.
Thus, beginequation
beginaligned
& AS=SQ\
Rightarrow & A=SQS^-1\
Rightarrow & A^T=left(S^-1right)^T Q^T S^T\
Rightarrow & A= left(S^Tright)^T Q^T S^T qquad left(mbox$A=A^T$right)\
Rightarrow & A= S Q^T S^-1 qquad hspace7mm left(mbox$S^T=S^-1$right)\
Rightarrow & A=S beginbmatrix
lambda & & & & & & \
&lambda& & & & & \
& & ddots & & & &\
& & & lambda & & & \
c_r+1,1 &c_r+1,2 &dots &c_r+1,r&c_r+1,r+1&dots & c_r+1,n\
vdots& vdots & ddots & vdots & vdots&ddots & vdots \
c_n1& c_n2& dots & c_nr& c_n,r+1&dots& c_nn
endbmatrix S^-1
endaligned
endequation
(Except the last equation, the above part was not done in the class. Am correct in getting the last equation?)

Now, we obtain,
beginequation
beginaligned
& Q= S^-1AS \
Rightarrow & Q=S^T A S\
Rightarrow & Q^T=S^T A^T left(S^Tright)^T\
Rightarrow & Q^T=S^T A S =Q\
endaligned
endequation
which indicates that $Q$ is a symmetric matrix. So, it should look like

$$Q=left[
beginarrayc
lambda \
& lambda\
& & ddots& \
& & & lambda\
hline
& & & &Large textbfC
endarrayright]$$

where $C$ is a symmetric matrix of order $left(n-rright)timesleft(n-rright)$.
Thus, it can be easily shown that $det(A-kI)=det(Q-kI)=left(lambda-kright)^r det(C-kI)$.

Now, the proof tries to show that $lambda$ can't be an eigenvalue of $C$. It uses contradiction to reach the conclusion. Suppose there exist an $(n-r)times 1$ non zero vector $u$ such that $Cu=lambda u$. Therefore, the above form of $Q$ allow us to get the following result:-
beginequation
beginaligned
Q beginbmatrix
0\
0\
u
endbmatrix
=beginbmatrix
0\
0\
lambda u
endbmatrix=lambda beginbmatrix
0\
0\
u
endbmatrix
endaligned
endequation
I have clearly understood till this point without much difficulty. But now it is claimed that $beginbmatrix
0\
0\
u
endbmatrix$ is an eigenvector of $A$, corresponding to the eigenvalue $lambda$. I am not able to understand it.

The above relation says that $beginbmatrix
0\
0\
u
endbmatrix$ is an $lambda$ eigenvector of $Q$. And, $Q$ is related to $A$ by $Q=S^-1A S$. From this relation we can just say that $Sbeginbmatrix
0\
0\
u
endbmatrix$ is an $lambda$ eigenvector of $A$. (Because, I know that if $Q=S^-1AS$ and if $v$ is a $lambda$ eigenvector of $Q$, then $Sv$ is a $lambda$ eigenvector of $A$. It doesn't tells that $v$ is an eigenvector of $A$. But here, I think, it may be true due to the fact that $S$ is orthogonal or $A$ is symmetric. I am not sure how it is possible in our case. )

The last part of the proof tells that $beginbmatrix
0\
0\
u
endbmatrix$ can't be an $lambda$ eigenvector of $A$.

Here also I didn't get the point. One of my fellow mates told me that $beginbmatrix
0\
0\
u
endbmatrix$ is in the span of $e_r+1,e_r+2,dots,e_n$, because the first $r$ components are zero. So, it means that it is in the span of $v_r+1,v_r+2,dots,v_n$. It did not make any sense for me. How does it imply an contradiction? If I suppose that he is correct, then could I give any contradiction using the fact that the eigenvectors corresponding to distinct eigenvalues must be independent?