S10 Flashcards
The sum of two eigenvectors of a matrix A is also an eigenvector of A.
False: (1,0) and (0,1) are both eigenvectors of A, but (1, 1) is not.
(It is true if the two eigenvectors belong to the same eigenvalue, though.)
![](https://s3.amazonaws.com/brainscape-prod/system/cm/120/961/925/a_image_thumb.png?1659440378)
A is singular if and only if 0 is an eigenvalue of A.
True
0 is an eigenvalue if and only if 0 = det(A−0·I) = det(A) if and only if A is singular.
If v1 and v2 are linearly independent eigenvectors of A, then they correspond to distinct eigenvalues.
False
We have seen two-dimensional eigenspaces, which contain two linearly independent eigenvectors that correspond to the same eigenvalue.
If A is invertible, then it is diagonalizable.
False
A is invertible, but not diagonalizable.
![](https://s3.amazonaws.com/brainscape-prod/system/cm/120/961/974/a_image_thumb.png?1659440378)
If A is singular, then it is not diagonalizable.
False
A is singular, and diagonalizable because it is already diagonal.
![](https://s3.amazonaws.com/brainscape-prod/system/cm/120/962/021/a_image_thumb.png?1659440378)
If A has fewer than n distinct eigenvalues, then A is not diagonalizable.
False
Identity Martix (Shortest answer ever.)
Each eigenvector of an invertible matrix A is also an eigenvector of A−1.
True
If Av = λv, then (1/λ) v = vA−1, so v is an eigenvector of A−1 with eigenvalue (1/λ).