W4 Eigenvalues to Eigenvectors example in lesson, is it accurate?

Course Q&A Math for Machine Learning Linear Algebra for Machine Learning and Data Sc...
1

I was able to calculate the eigenvectors for lamda value 4, but when I perform the steps for value 1, I get x1 = -2X2 and X2 = -X3. Which means the values in the less 0,1,1 for lamda value 1 isn’t possible. What am I missing?

EigenValues|627x452

2

Ah, yes, you’re right: that is a bug in the slide. They reversed the eigenvectors for \lambda = 1 and -3.

I’ll check and file a bug about this if it has not already been filed. Do you have a reference to which lecture and the time offset? Thanks!

3

W4: ‘Calculating Eigenvalues and EigenVectors’, time offset: 8:01. Thank you for a prompt response Paul!

4

image
image701×251 13 KB

I asked a q. on the common thread, but later realized its one of the graded quiz, so deleted the original post.
Is it alright to directly message you?

5

oops. Still getting used to the DeepLearning.ai site and how to post.

6

But for question 2) you’d need a way to prove that, right? How can you do that?

Starting with the determinant (option 1) is the easiest. That gives you a “yes/no” answer right away. If it is non-zero, then the matrix is non-singular and all the columns (or rows) are linearly independent. If you have 3 linearly independent vectors, then they are a basis for 3D space.

The other thing you could do would be to transpose the matrix and then reduce it to row echelon form. If all the rows are non-zero at the end, then it’s non-singular and they are all linearly independent.

7

I get a non-zero determinant value for the vectors which means its linearly independent. But when I insert the dimensions on the interactive tool, it shows up a plane. Not sure what I’m missing.

Let me try your proposed way of reducing it to row echelon form. Thanks

8

I don’t know what you mean by

What interactive tool? And what dimensions?

9

Apologies, Im referring to the Interactor Tool: Linear Span under week 4 for Coursera. As you can see for the vectors v1, v2, v3 for those values, the span type is a plane. But given, it seems, there is linear independence between v1, v2 and v3, I was expecting a ‘all space’ span type.
image

10

Oh, ok, sorry: the problem is that your evaluation of the determinant is incorrect. The determinant of that matrix is 0.

Now that I think a bit harder you can see that the first row + 2 * the second row equals the third row. So the rank of that matrix is 2.