Finding Eigenvectors for 4x4 Matrix A = 4 2, 0 1 | Homework Help

[Dissonance in E]

Homework Statement

I have matrix
A = 4 2
0 1

Whose eigenvalues I found to be 4 & 1

I need to find the eigenvectors for the same matrix

Homework Equations

(A-lambdaI)V=0

The Attempt at a Solution

Lambda = 4 gives
0 2 x V1 = 0
0 -3 V2

0v1 + 2v2 = 0
0v1 - 3v2 = 0

Eigenvector = 0 , 0
How does that give me any choice? Either there's some fundamental rule I am missing or my maple skills need a workout as according to maple the vector is 1, 0
which would give 1+0 = 0 ? No?

Lambda = 1 gives

3 2 x V1 = 0
0 0 V2

3v1 + 2v2 = 0
0v1 + 0 v2 = 0

eigenvector = -2, 3
Ok I get that this works, why couldn't I say that the eigenvector is 2, -3?
wouldnt this give 3(2) + 2(-3) = 0
Whats the difference?

Clarifications greatly appreciated.

P.s: Sorry for the painful format, couldn't figure out how to draw matrices with latex.

 

[Donaldos]

How did you find those eigenvalues?

 

[Dissonance in E]

Using det(a - lambdaI) = 0

 

[Donaldos]

\lambda =4 \Rightarrow V_2=0 so V_1 can take any value, including 1.

\lambda =1 \Rightarrow 3V_1=-2V_2. Therefore the eigenvector you suggested is just as valid as the one you found using Maple.

 

[HallsofIvy]

How did you find those eigenvalues?

Using det(a - lambdaI) = 0

The eigenvalues of a trianguar (or diagonal) matrix are just the numbers on the main diagonal.

Lambda = 4 gives
0 2 x V1 = 0
0 -3 V2

0v1 + 2v2 = 0
0v1 - 3v2 = 0

Eigenvector = 0 , 0

No, you only have 2v2= 0 and -3v2= 0 which are satisfied as long as v2= 0. v1 can be anything. An eigenvector for eigenvalue 4 is any vector of the form <x, 0>.

The definition of "eigenvalue" is that there exist a non-zero vector such that Av= \lambda v
Here you must have
\begin{bmatrix}4 &amp; 2 \\ 0 &amp; 1\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}
= \begin{bmatrix}4x+ 2y \\ y\end{bmatrix}= \begin{bmatrix}4x \\ 4y\end{bmatrix}
which gives you the two equations 4x+ 2y= 4x and 4y= 4y. The first equation reduces to 2y= 0 which gives y= 0 and the second equation is automatically satisfied. Again, x can be anything.