The question about diagonalization

[ftym2011]

I have the confusion that the one question is shown below:
Consider the following matrix:
A= [1 -1;1 1] which is 2x2 matrix, the column of that is [1 1] and [-1 1] respectively.

What happens when we apply A to vector v a large number of times?

Hoping someone can help me solve this question, thanks a lot!

 

[I like Serena]

I have the confusion that the one question is shown below:
Consider the following matrix:
A= [1 -1;1 1] which is 2x2 matrix, the column of that is [1 1] and [-1 1] respectively.

What happens when we apply A to vector v a large number of times?

Hoping someone can help me solve this question, thanks a lot!

What you need to know is that A defines a rotation combined with a scaling effect.
Do you know the angle and the scale?
What happens if you do this n times?

 

[HallsofIvy]

I like "I like Serena"'s answer, and I would further suggest seeing what A does to
$$\begin{bmatrix}1 \\ 0 \end{bmatrix}$$ and $$\begin{bmatrix}0 \\ 1\end{bmatrix}$$
the basis vectors, and imagining that occurring repeatedly.

However, ftym2011 titled this thread "The question about diagonalization" which makes me think a more general method is intended.

If the matrix, A, is "diagonalizable" then there exist an invertible matrix, P, and a diagonal matrix, D, such that $A= PDP^{-1}$. Then $A^2= (PDP^{-1})(PDP^{-1}= PD^2P^{-1}$ because the "$P^{-1}$" and "$P$" in the middle cancel. And, then $A^3= A(A^2)= (PDP^{-1})(PD^2P^{-1})= PD^3P^{-1}$. In general, $A^n= PD^nP^{-1}$ and it is easy to find the nth power of a diagonal matrix- it is the diagonal matrix with the nth powers on its main diagonal.

An n by n matrix is diagonalizable if and only if it has n independent eigenvectors. Specifically, D is the diagonal matrix with the eigenvalues of A on its diagonal and P is the matrix whose columns are the corresponding eigenvectors.

So let's put the question back to ftym2011: Can you find the eigenvalues and corresponding eigenvectors of
$$\begin{bmatrix}1 & -1 \\ 1 & 1\end{bmatrix}$$?

However, having said that, I note that, since A is anti-symmetric, its eigenvalues and eigenvectors are complex so, again, I think I like Serena's suggestion is best.