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Yes, the two independent eigenvectors $(1,-1)$ and $(1,1)$ form an eigenbasis for this specific example. They span the entire 2D space, allowing any vector to be expressed as a linear combination of these eigenvectors.

Conditions: The matrix is the reflection about $y=-x$.; The eigenvectors are linearly independent.; The dimension of the space is 2.

Yes, the two independent eigenvectors $(1,-1)$ and $(1,1)$ form an eigenbasis for this specific example. They span the entire 2D space, allowing any vector to be expressed as a linear combination of these eigenvectors.

Conditions: The matrix is the reflection about $y=-x$.; The eigenvectors are linearly independent.; The dimension of the space is 2.

The formula $A^k = PD^kP^{-1}$ represents a sequence of three geometric operations: first, a change of basis using $P^{-1}$ to move to the eigenbasis; second, a diagonal scaling using $D^k$ which stretches or shrinks along the eigenvector directions; and third, a reverse change of basis using $P$ to return to the original coordinate system.

Conditions: $A = PDP^{-1}$ is a valid diagonalization.; $k$ is a positive integer.

The formula $A^k = PD^kP^{-1}$ represents a sequence of three geometric operations: first, a change of basis using $P^{-1}$ to move to the eigenbasis; second, a diagonal scaling using $D^k$ which stretches or shrinks along the eigenvector directions; and third, a reverse change of basis using $P$ to return to the original coordinate system.

Conditions: $A = PDP^{-1}$ is a valid diagonalization.; $k$ is a positive integer.