Difference between revisions of "031 Review Part 3, Problem 3"

Revision as of 13:08, 13 October 2017

Let $A={\begin{bmatrix}5&1\\0&5\end{bmatrix}}.$

(a) Find a basis for the eigenspace(s) of $A.$

(b) Is the matrix $A$ diagonalizable? Explain.

Foundations:
Recall:
1. The eigenvalues of a triangular matrix are the entries on the diagonal.
2. By the Diagonalization Theorem, an $n\times n$ matrix $A$ is diagonalizable
if and only if $A$ has $n$ linearly independent eigenvectors.

Solution:

(a)

Step 1:

Step 2:

(b)

Step 1:

Step 2:

Final Answer:
(a)
(b)

@@ Line 13: / Line 13: @@
 !Foundations: &nbsp;
 |-
-|
+|Recall:
+|-
+|'''1.''' The eigenvalues of a triangular matrix are the entries on the diagonal.
+|-
+|'''2.''' By the Diagonalization Theorem, an &nbsp;<math style="vertical-align: 0px">n\times n</math>&nbsp; matrix &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; is diagonalizable
+|-
+|if and only if &nbsp;<math style="vertical-align: 0px">A</math>&nbsp; has &nbsp;<math style="vertical-align: 0px">n</math>&nbsp; linearly independent eigenvectors.
 |}