Do the eigenvectors of a reflection about $y=-x$ form an eigenbasis?

A matrix is diagonalizable if and only if it possesses a full set of linearly independent eigenvectors (an eigenbasis). The primary benefit is computational efficiency: calculating high powers of the matrix (e.g., $A^{100}$) becomes trivial because you simply raise the diagonal entries (eigenvalues) to that power, rather than performing repeated matrix multiplication.

Conditions: Matrix has $n$ linearly independent eigenvectors in $n$-dimensional space; Change of basis matrix $P$ formed by eigenvectors is invertible

No, a general $n \times n$ matrix need not have $n$ distinct real eigenvalues. Some matrices may have repeated eigenvalues, complex eigenvalues, or fewer than $n$ linearly independent eigenvectors.

Conditions: The matrix is $n \times n$.; The field of scalars is typically $\mathbb{R}$ unless specified otherwise.

For the reflection about the line $y=-x$, the vector $(1,-1)$ lies along the reflection line and remains unchanged. Therefore, $(1,-1)$ is an eigenvector with an eigenvalue of $1$.

Conditions: The transformation is a reflection about the line $y=-x$.; The vector is nonzero.; The vector lies on the axis of reflection.

A 90-degree rotation moves every non-zero vector in the plane off its original line (span). Since eigenvectors must remain on their span, no real vector qualifies.

Conditions: Matrix represents a pure 90-degree rotation in 2D; Searching for real-valued eigenvectors

For the reflection about $y=-x$, the vector $(1,1)$ is mapped to $(-1,-1)$. Since $(-1,-1) = -1 \cdot (1,1)$, the vector $(1,1)$ is an eigenvector with an eigenvalue of $-1$.

Conditions: The transformation is a reflection about the line $y=-x$.; The vector is nonzero.; The vector is perpendicular to the reflection line.