query: Are the eigenvectors of $A^2$ the same as the eigenvectors for $A$? Are the lambdas just squared?
positive_passages.docid: 54118
positive_passages.text: Proving Eigenvalue squared is Eigenvalue of $A^2$
negative_passages.docid: 104424
negative_passages.text: If each eigenvalueof $A$ is either $+1$ or $-1$ $ \Rightarrow$ $A$ is similar to ${A^{ - 1}}$
negative_passages.docid: 145396
negative_passages.text: Symmetric Matrix , Eigenvectors are not orthogonal to the same eigenvalue.
negative_passages.docid: 145342
negative_passages.text: Why do the eigenvectors for two similar matrixes differ by a large amount
negative_passages.docid: 39112
negative_passages.text: Does a correlation matrix of two variables always have the same eigenvectors?
negative_passages.docid: 110086
negative_passages.text: Let $A$ be an $m \times n$ matrix. Show $A^TA$ and $AA^T$ have the same eigenvalues
negative_passages.docid: 126711
negative_passages.text: What is the square matrix $A$ equal to if $A=A^2$
negative_passages.docid: 144097
negative_passages.text: eigenvalues and eigenvectors of $vv^T$
negative_passages.docid: 9250
negative_passages.text: Can the zero vector be an eigenvector for a matrix?
negative_passages.docid: 131611
negative_passages.text: $A^TA$ is always a symmetric matrix?
negative_passages.docid: 99585
negative_passages.text: Do the non-zero eigenvalues of AB and BA have the same algebraic multiplicity (for AB and BA not square)?
negative_passages.docid: 94424
negative_passages.text: Matrices and Eigenvalues of Norm
negative_passages.docid: 56670
negative_passages.text: The intuition behind generalized eigenvectors
negative_passages.docid: 126747
negative_passages.text: Null space for $AA^{T}$ is the same as Null space for $A^{T}$
negative_passages.docid: 122951
negative_passages.text: If the $\det(A)=0$ why does the matrix $A$ have an eigenvector?
negative_passages.docid: 149923
negative_passages.text: eigenvalues of A - aI in terms of eigenvalues of A
negative_passages.docid: 75536
negative_passages.text: every eigenvalue of $T$ has only one corresponding eigenvector up to a scalar multiplication