thanks man

What a king

u the man.

38:01

Im only half through one lecture and I already love him. :'D

I was going through a headache, after 15 minutes of his lecture it got evaporated.

Sure

maybe he s not because he will be sad if his grandson s stupid and cannot inverse a matrix.... just kidding XD

@@NguyenAn-kf9ho lol

Wishing he was and isn't ? Better wishing he is.

I think you are right

Yes

@@vishalyadav2958 yes

U r doing phd or post grad?

U can follow horn n johnson and strang book... it's relatively easier to understand

yes when u get admitted to MIT u can take up the class and partake in assignments

He means difference between eigenvalues, |lambda1 - lambda2|, is big, then we have the case where "the bowl is long and thin" he mentions right before that.

@@nguyennguyenphuc5217, yes, it looks like it would make it easier to miss the point and bounce back and forth around the minimum

@@gabrielmachado5708 right. it the bowl is narrow and your descent is slightly off you'll start climbing again.... so we take baby steps.

The homework can be found in the Assignments section of the course on MIT OpenCourseWare at: ocw.mit.edu/18-065S18. Best wishes on your studies!

@@mitocw are the Julia language online asigmants mentioned also available somewhere? I see only problems from the textbook in the Assignments section of the OCW

julialang.org/

@@mitocw Where can we locate the programming assignments?

@@mitocw I have a question about the kzfaq.info/get/bejne/rtmAYLZo3rHFgnU.html Where can I find this lab work about convolution? On MIT OpenCourceWare at ocw.mit.edu/courses/mathematics/18-065-matrix-methods-in-data-analysis-signal-processing-and-machine-learning-spring-2018/assignments/ I can find only book assignments ocw.mit.edu/courses/mathematics/18-065-matrix-methods-in-data-analysis-signal-processing-and-machine-learning-spring-2018/assignments/MIT18_065S18PSets.pdf#page=7 Could you help me? Thanks!

The eigenvectors with non zero eigenvalues must be mapped to somewhere within the column space, in all other directions outside the column space it collapses to 0, bear in mind that the null space vectors are also solutions to Ax=\lambda x where \lambda is 0.

The answer is at 41:17 ... you notice how we can decompose the matrix into a weighted sum of its eigenvectors.. the weights being the eigenvalues obviously, and since Rank(A) is by definition the number of linearly independent vectors in the column space of A, i.e., it is the same as the number of non-zero terms in the decomposition, which is in turn the number of non-zero eigenvalues

@@fustilarian1 Thanks for your explanation. That's very helpful.

Regarding similarity you don't need the spectral theorem, just to remember that we say that A and B are similar if there exists an invertible matrix M such that A = M^(-1) * B * M You can immediately verify that if A = Q^(-1) * S* Q, B = S, and M=Q, then the equation is satisfied so A=Q^(-1) *S* Q and B=S are similar. Regarding the bowl statement, it should be pretty clear when the eigenvectors are [1,0] and [0,1]. In that case the energy function is given by: [x,y] * S * [x,y]^T = x^2 * lambda1 + y^2 * lambda2. So in the xz-plane it is just the quadratic function scaled by lambda1. In the yz-plane it is just the quadratic function scaled by lambda2 (and in general it is a linear combination of the two). If either eigenvalue is much larger than the other the scalings will be disproportionate and therefore we will get a bowl with a steep slope in the direction of the large eigenvalue, and pretty flat slope in the direction of the small eigenvalue. However the whole point of diagonalization is that basically we can treat any diagonalizable matrix like the diagonal matrix of its eigenvalues as long as we do the appropriate orthogonal base change (or equivalently work in the correct coordinate system), so really we already know that the general bowl will be an orthogonal transformation of the bowl described above and therefore itself be a narrow valley bowl. Concretely, if v1,v2 is an orthonormal basis of eigenvectors of S, with associated eigenvalues lambda1,lambda2, then the energy function is v^T QDQ^T v where Q is the orthonormal matrix whose columns are v1,v2. D is the diagonal matrix with elements lambda1,lambda2. We can write v as a unique linear combination of the eigenvectors (it is a basis after all): v= x * v1 + y * v2 Then the energy function evaluates to: v^T QDQ^T v = v^T QD [x,y]^T = v^T Q[lambda1 * x, lambda2 * y] = v^T (lambda1 * x * v1 + lambda2 * y * v2) = lambda1 * x^2 + lambda2 * y^2, so again it is a bowl which in the direction of v1 is a 1-dimensional quadratic scaled by lambda1, and in the direction of v2 is a 1-dimensional quadratic scaled by lambda2. So if lambda1 is huge the slope in the direction v1 will be steep. Same as before, just from the point of view of the coordinate system given by the eigenvectors (v1,v2).

@@ramman405 thanks

Are you asking why this quadratic form is called energy?

@@spoopedoop3142 yes exactly

@@CM-Gram Kinetic energy is 1/2mv^2, where v is the velocity vector, and potential energy is 1/2kx^2, where x is the position vector.

who can't understand that.

It will shift or tilt the bowl in X axis direction. You can try the vizualizer al-roomi.org/3DPlot/index.html

@@jeevanel44 Hey, sorry to bother you a year later - what expression would I input to receive the bowl shown here?

The course materials are available on MIT OpenCourseWare at: ocw.mit.edu/18-065S18. Best wishes on your studies!

Octave : -0.12311 , 8.12311 agrees with abc formula

Matlab too 😀

14:00 deep learning

24:00 gradient descent

27:00 eigenvalue tells the shape of the bowl

38:00 semi def pos

🤣

I guess not, unless the air gets personified and files a case.

Maybe no one raise their hand