# Question about the quiz in video: Eigenvalues and eigenvectors

The quiz in the video Eigenvalues and eigenvectors, asks us to find the eigenvalues and eigenvectors of the matrix [[9,4][4,3]].

I understand that the eigenvalues are 11 and 1 after solving the quadratic equation.

However, when I try to find the eigenvectors using the method describe in the video:
For λ=11
9x + 4y = 11x => 4y = 2x => 2y = x
4x + 3y = 11y => 4x = 8y => x = 2y
So I cannot find a unique point.

For λ=1
9x + 4y = x => 4y = -8x => y = -2x
4x + 3y = y => 4x = -2y => 2x = -y
So again I cannot find a unique point.

Also, the solution in the video says that the eigenvectors are (2,1) and (-1,2). When I input the matrix to an online calculator (for example here: Eigenvalues and Eigenvectors), then I get eigenvectors (2,1) and (-0.5,1).

What am I doing wrong? Thank you!

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@Evripidis_Christodou

You are not wrong. It is not necessary that there must be a unique point. All of the points that satisfy these equations can be treated as eigenvector.

I totally agree with you, the computation of Eigenvectors has not been presented well. Just finding multiples does not make one grasp the concept rather concrete mathematical solutions should exist. Can anyone please help solve these equations by derivation/substitution?

I.e.
i. x + 2y = 3y
ii. 2x + y = 3x
How can we find the solution in this case?

Thank you, I understand this know.

@Muhammad-Kalim-Ullah to solve these equations you get:
x = y and
x = y

To find the eigenvectors, you just need to find a solution which will give you the direction of the eigenvector. For example, (1,1), or (2,2) or (3,3) etc. The eigenvectors just express a direction.

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Since the determinant of A - \lambda I is required to be zero, this implies that (A - \lambda I)X=0 has infinitely many solutions. Consequently, when we substitute an eigenvalue into \lambda, we won’t find an unique X. In other words, accept that it is never going to be unique

Let me take this as an example

If I remember correctly, the text-book approach would be to assume x = t, such that we will get y = 2t. In other words, the eigenvector is going to be \begin{bmatrix} 1\\ 2 \end{bmatrix} t, or \frac{\sqrt{5}}{5}\begin{bmatrix} 1\\ 2 \end{bmatrix} which is the unit vector.

I havn’t gone through the M4ML courses, so I am not sure how it approaches this.

Good luck.

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This is very well said. The magnitude is left to my t.

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