If a matrix-valued function converges element-wise (at the same rate) to a constant matrix, do the eigenvectors also converge element-wise?

Question

If a matrix valued function $\psi(\lambda)$ converges element-wise as to a constant matrix $A$(all elements converging at the same rate) as $\lambda\to\infty$, can we say that all the eigenvectors of the matrix $\psi(\lambda)$ converge element-wise to the eigenvectors of the matrix $A$?

PS at same rate, I mean all elements are Asymptotically as $\lambda\to\infty$ are of the form $A + B\Theta(\frac{1}{\lambda})$

PS2 : By eigen vectors I mean normalized eigen vectors which form an orthonormal basis. Assume $\psi(\lambda)$ is symmetric and positive definite always. $A$ is symmetric but may be singular.

PS3 : All matrices are real valued.

More clarity

1. Assume eigenvalues are also converging $\Theta(\frac{1}{\lambda})$ to eigenvalues of $A$.

2. Let $e_1(\lambda),e_2(\lambda),e_3(\lambda),...e_n(\lambda)$ be the normalized eigenvectors of $\psi(\lambda)$ in decreasing order of corresponding eigenvalues.

Let $E(\lambda) = [e_1(\lambda),e_2(\lambda),...e_n(\lambda)]$ be a matrix whose columns are normalized eigenvectors. let $D(\lambda)$ be a diagonal matrix whose elements eigenvalues of $\psi(\lambda)$ in the same order of eigenvectors.

Now $\psi(\lambda) = E(\lambda)D(\lambda)E(\lambda)^T$ and as eigenvalues $D$ is converging elementwise with same asymptotics, shouldn't $E(\lambda)$ converge element-wise?

Answer 1

Let $C$ be a symmetric positive-definite matrix. Let $R$ be a rotation by $\pi/2$ in some axis, and let $\psi(n)=B_n:=\frac{1}{n}R^nCR^{-n}$.

There are issues when an eigenspace has dimension more than 1 (as in the comments), but let us suppose that $C$ has a unique eigenvector $v$ for some specific eigenvalue $\lambda$. Then $B_n$ has eigenvector $R^nv$ with eigenvalue $\lambda/n$.

Thus $B_n\to A=0$ ($\|B_n\|=\|C\|/n\to0$), its eigenvalue $\lambda/n\to0$, but its corresponding eigenvectors $R^nv$ do not converge.

(Note: $E_n=R^n[v_1,\ldots,v_n]=[R^nv_1,\ldots,R^nv_n]$ does not converge component-wise.)

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Second Example: Let $\psi(n):=\begin{pmatrix}1&\tfrac{1}{n}\\\tfrac{1}{n}&1\end{pmatrix}$ when $n$ is odd, and $\begin{pmatrix}1+\tfrac{1}{n}&0\\0&1-\tfrac{1}{n}\end{pmatrix}$ for $n$ even. The eigenvalues of $\psi(n)$ are $1\pm\tfrac{1}{n}$ irrespective of parity, but with corresponding eigenvectors $(1,1)$, $(1,-1)$, when $n$ is odd, and $(1,0)$, $(0,1)$ when $n$ is even. Thus the diagonalizing matrix $E_n=\tfrac{1}{\sqrt2}\begin{pmatrix}1&1\\1&-1\end{pmatrix}$ for $n$ odd and $I$ for $n$ even. Clearly $E_n$ does not converge, even though their eigenvalues do, and $\psi(n)\to I$.