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

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

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!

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

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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?

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

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.

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

I don’t know what you mean by

What interactive tool? And what dimensions?

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.

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.

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