I have a question regarding Linear Algebra chapter 5.3: If I am trying to determine if a matrix A= [ 0 1 1 2 1 2 3 3 2 ] is diagonizable, and the solution I got for the third row was 1 = 0, does this mean that the matrix is not diagonizable since the eigenvalue = 0 proves that the matrix is inconsistent? Or am I able to still get the basis vector for the eigenspace?
I have a question regarding Linear Algebra chapter 5.3: If I am trying to determine if a matrix A= [ 0 1 1 2 1 2 3 3 2 ] is diagonizable, and the solution I got for the third row was 1 = 0, does this mean that the matrix is not diagonizable since the eigenvalue = 0 proves that the matrix is inconsistent? Or am I able to still get the basis vector for the eigenspace?
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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I have a question regarding
If I am trying to determine if a matrix A= [ 0 1 1
2 1 2
3 3 2 ]
is diagonizable, and the solution I got for the third row was 1 = 0, does this mean that the matrix is not diagonizable since the eigenvalue = 0 proves that the matrix is inconsistent? Or am I able to still get the basis vector for the eigenspace?
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