Show a Matrix is Symmetric Knowing its Eigenvectors I was looking at some past exams for a linear algebra course and came upon this question:
Let $A$ be a $3 \times 3$ real matrix with the following eigenvectors:
$$ \begin{pmatrix}
1 \\
1 \\
0 
\end{pmatrix}, \begin{pmatrix}
1 \\
2 \\
0 
\end{pmatrix}, \begin{pmatrix}
2 \\
1 \\
0 
\end{pmatrix}, \begin{pmatrix}
0 \\
0 \\
1 
\end{pmatrix}$$
Show that $A$ is symmetric.
My attempt: Since eigenvectors 2, 3, and 4 are linearly independent, they form a basis for $\mathbb{R}^3$. We can then diagonalize $A$. This gives that
$$A = \frac13 \begin{pmatrix}
-1 \lambda_2 + 4 \lambda_3& 2\lambda_2 - 2 \lambda_3& 0 \\
-2 \lambda_2 + 2 \lambda_3& 4\lambda_2 - 1 \lambda_3& 0 \\
0& 0& \frac13 \lambda_4 
\end{pmatrix}$$
From this I conclude $A$ is symmetric iff $\lambda_2 = \lambda_3$. But, I'm not sure where to go from here.
 A: It is easy to see that ${\bf v}_2$ and ${\bf v}_3$ are independent; also,
$$3{\bf v}_1={\bf v}_2+{\bf v}_3$$
as noted in comments.  Multiplying by $A$ gives
$$3\lambda_1{\bf v}_1=\lambda_2{\bf v}_2+\lambda_3{\bf v}_3$$
where $\lambda_k$ are the respective eigenvalues.  Therefore
$$\lambda_2{\bf v}_2+\lambda_3{\bf v}_3=\lambda_1{\bf v}_2+\lambda_1{\bf v}_3\ .$$
By independence, $\lambda_1=\lambda_2=\lambda_3$.
A: Eigenvectors for distinct eigenvalues are always linearly independent. Since the first three given vectors are linearly dependent, at least two of them are for the same eigenvalue$~\lambda$. But then the space spanned by those two (the subspace with third coordinate zero) is contained in the eigenspace for$~\lambda$, and since it contains the third of those three, the latter is also an eigenvector for$~\lambda$. Now either the fourth given vector is also an eigenvector for$~\lambda$, in which case that eigenspace is the whole space, or it is an eigenvector for a different eigenvalue$~\mu$, whose eigenspace it then spans (and which is orthogonal to the $2$-dimensional eigenspace for$~\lambda$). In either case, the eigenspaces span the whole space (so $A$ is diagonalisable), and are mutually orthogonal, so $A$ is symmetric.
In fact, since all standard basis vectors are eigenvectors in both cases, $A$ is actually diagonal.
A: Let :
$$A = \begin{pmatrix}  
A11 & A12 & A13 \\  
A21 & A22 & A23 \\  
A31 & A32 & A33 \\  
\end{pmatrix}$$
With the given Eigenvectors , $E_1 , E_2 , E_3 , E_4$ , we can try to get the Individual terms of $A$ , using $\lambda_1 , \lambda_2 , \lambda_3 , \lambda_4 $ , the 4 Eigenvalues , which must have at least 2 repeats.
With $A \times E_4 = \lambda_4 E_4$ , we see that
$A13=A23=0$ , & $A33=\lambda_4$
With $A \times E_1 = \lambda_1 E_1$ , we see that
{ $1A11+1A12=1\lambda_1$ } , [ $1A21+1A22=1\lambda_1$ ] , "$A31+A32=0$"
With $A \times E_2 = \lambda_2 E_2$ , we see that
{ $1A11+2A12=1\lambda_2$ } , [ $1A21+2A22=2\lambda_2$ ] , "$A31+2A32=0$"
With $A \times E_3 = \lambda_3 E_3$ , we see that
{ $2A11+1A12=2\lambda_3$ } , [ $2A21+1A22=1\lambda_3$ ] , "$2A31+A32=0$"
The "3 Equations in Quotes" show that $A31=A32=0$
We have , till now :
$$A = \begin{pmatrix}  
A11 & A12 &         0 \\  
A21 & A22 &         0 \\  
  0 &   0 & \lambda_4 \\  
\end{pmatrix}$$
With the "3 Equations in {}" , we get :
$A12=1\lambda_2-1\lambda_1$ & $A12=2\lambda_1-2\lambda_3$ & $3A12=2\lambda_2-2\lambda_3$
With the "3 Equations in []" , we get :
$A21=2\lambda_1-2\lambda_2$ & $A21=1\lambda_3-1\lambda_1$ & $3A21=2\lambda_3-2\lambda_2$
Even when $ \lambda_4 $ is the first repeat ( maybe not a repeat ) , at least 1 among $\lambda_1 , \lambda_2 , \lambda_3 $ must be the second repeat.
That will make $A12=A21=0$ , which shows that we have a Symmetric Matrix.
More-over , all three Eigenvalues are repeats , maybe $ \lambda_4 $ is not a repeat.
Conclusion :
We have in the end :
$$A = \begin{pmatrix}  
\lambda_1 &         0 &         0 \\  
        0 & \lambda_1 &         0 \\  
        0 &         0 & \lambda_4 \\  
\end{pmatrix}$$
