It is currently 3 am as I study for my linear algebra midterm usually I'd be falling asleep at my desk but you do such a great job of explaining and engaging I somehow manage to plow through your videos so easily thank you so much
Way to specific just finished watching all these vids at 3am & never got bored once, I understood everything after being confused all semester can say im pretty confident going into this exam (the colourful notes & short vids help a lot) since my notes usually look like this
For those that don't get 13:00. After getting RREF, we decompose the vector giving the general solution into linear combination of vectors where the weights are the free variables. That is, x1 |2 * x2 + x4 - 3*x5 | x2 | x2 | x3 = | - 2*x4 + 2*x5 | = to the final solution she had at 14:35 x4 | x4 | x5 | x5 |
13:06 This makes sense up to here. You just rearranged the augmented matrix for x -2.x2 -1x4 +3x5 = 0 (ect) but where does the x2 [2 1 0 0 0] come from? That seemed to appear out of nowhere. The pattern seems to be: (1) find the pivots for each row. If a column (x1...xn) doesn't have a pivot it is free (2) for each pivot, rearrange the algebraic equation to isolate the dependent variable x1, x3 ect (PS: not sure why you couldn't continue with row operations to remove x4 from x1 or x3 equations) (3) for each free variable set it to 1, the other free variables to 0 This will give you the coefficients for the dependent variables. Arrange them in the appropriate row Is this a new algorithmic approach is new or have I missed a video?
Don't you mean always linearly dependent on 16:12? because you are essentially stating that they are free variables so the scalars don't have to equal to zero, so the trivial solution has more than one possible scalar numbers
Without these types of TH-cam videos, I would have no shot of making A's in any of my classes. Unlike my other classes that I understand most stuff in class but just need refreshers from videos to help me study.. For this class on the other hand, I do not understand anything in the class 😂😂😂 I really need these videos. You're gonna carry me to my D. A C would be the Lord's blessing 😂I do not want to graduate late or have to retake this class!!!
Yea it stumped me for a bit but I think I figured it out. The free variables are only really there for the solutions for x, not for the solutions for 0. In the video, she wrote x = x2 [ 2 1 0 0 0 ] + x4 [ 1 0 -2 1 0] + x5 [ -3 0 2 0 1]. The free variables only affect the value of x (hence the x = ...). In order for us to determine whether the spanning set is linearly independent, you need to see if Ax = 0 has only the trivial solution (A is now the 3 vectors merged together into one matrix). If you do the process for this, you will see that Ax = 0 for this version of A is only true for the trivial solution (there are no free variables in this solution because there are only 3 vectors, not 5. It doesn't matter if we get the last two rows full of 0s because there is no column 4 or 5).
Thank you for the videos but we seem to have some missing steps we haven't seen before...The confusion with the 1 is that in previous videos we used 0 for the free variables. Have I missed a video where we go through that re-arrangement algorithm from 13:16 onwards? @@SawFinMath
I missed this part too. Here's what chat gpt said: Find Vectors for the Null Space: For 𝑥2=1x2 =1, set 𝑥4=0x4=0, and 𝑥5=0x5 =0. Plug these into the equations for 𝑥1x1 and 𝑥3x3 to find their values. You end up with a vector where the position corresponding to 𝑥2x2 is 1, and the other free variables are 0. Repeat this for x4=1, x2=0, x5=0 and for x5=1, x2=0, x4=0 @@SawFinMath
At around 5:00 what you're writing for conditions 2 and 3 isn't correct. Condition 2 should say: "When u and v are in Nul(A), is u+v in Nul(A)?" Condition 3 should be: "When u is in Nul(A), is cu is in Nul(A)?"
x can't be multiplied the matrix A if it is not nx1 since A is mxn. Remember that Ax only makes sense if the amount of columns of A matches with the amount of rows of x. Since x must be nx1, then it must exist in R^n (a 2D vector exists in R2, a 3D vector exists in 3D, and an n-D vector exists in R^n). So, null spaces only make sense if x belongs to R^n.
It must have zeros below ot we don't know if it's a pivot as it isn't reduced enough. To know the value of the solution we continue reducing to gets pivots above
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It is currently 3 am as I study for my linear algebra midterm usually I'd be falling asleep at my desk but you do such a great job of explaining and engaging I somehow manage to plow through your videos so easily thank you so much
Way to specific just finished watching all these vids at 3am & never got bored once, I understood everything after being confused all semester can say im pretty confident going into this exam (the colourful notes & short vids help a lot) since my notes usually look like this
@@Jayjay-eh4hf
Watching at 3:45 am and can still keep up :)
@@khaledsalah92488am day of test, step up your game
These are the best linear algebra videos I've found so far. Great Job!
For those that don't get 13:00.
After getting RREF, we decompose the vector giving the general solution into linear combination of vectors where the weights are the free variables. That is,
x1 |2 * x2 + x4 - 3*x5 |
x2 | x2 |
x3 = | - 2*x4 + 2*x5 | = to the final solution she had at 14:35
x4 | x4 |
x5 | x5 |
THANK CHRIST FOR TEACHERS LIKE YOU ON TH-cam!!!! WOW! ENJOYING MATHS FOR THE FIRST TIME!
13:06 This makes sense up to here. You just rearranged the augmented matrix for x -2.x2 -1x4 +3x5 = 0 (ect) but where does the x2 [2 1 0 0 0] come from? That seemed to appear out of nowhere.
The pattern seems to be:
(1) find the pivots for each row. If a column (x1...xn) doesn't have a pivot it is free
(2) for each pivot, rearrange the algebraic equation to isolate the dependent variable x1, x3 ect (PS: not sure why you couldn't continue with row operations to remove x4 from x1 or x3 equations)
(3) for each free variable set it to 1, the other free variables to 0
This will give you the coefficients for the dependent variables. Arrange them in the appropriate row
Is this a new algorithmic approach is new or have I missed a video?
I don't get that missing step either.
Where all my engineers at?
what up homie!
Don't you mean always linearly dependent on 16:12? because you are essentially stating that they are free variables so the scalars don't have to equal to zero, so the trivial solution has more than one possible scalar numbers
Thank you for proposing this question as I also wonder the same thing. @kimberly Brehm
I agreed with you on this although I'm not sure what Prof Brehm meant,
Prof Brehm.. would you help clarify on this .. Please! Thank you!
Without these types of TH-cam videos, I would have no shot of making A's in any of my classes. Unlike my other classes that I understand most stuff in class but just need refreshers from videos to help me study.. For this class on the other hand, I do not understand anything in the class 😂😂😂 I really need these videos. You're gonna carry me to my D. A C would be the Lord's blessing 😂I do not want to graduate late or have to retake this class!!!
You’ve got this!
@@SawFinMath Thank you! ☺️✊🏼
this is so so so awesome oh my god thank you so much
I do not understand why the set is linearly independent if there are free variables in the spanning set. Please explain!
@Ronnie W whar does mean by free variables are the weights on the spanning vectors??
Yea it stumped me for a bit but I think I figured it out. The free variables are only really there for the solutions for x, not for the solutions for 0. In the video, she wrote x = x2 [ 2 1 0 0 0 ] + x4 [ 1 0 -2 1 0] + x5 [ -3 0 2 0 1]. The free variables only affect the value of x (hence the x = ...). In order for us to determine whether the spanning set is linearly independent, you need to see if Ax = 0 has only the trivial solution (A is now the 3 vectors merged together into one matrix). If you do the process for this, you will see that Ax = 0 for this version of A is only true for the trivial solution (there are no free variables in this solution because there are only 3 vectors, not 5. It doesn't matter if we get the last two rows full of 0s because there is no column 4 or 5).
late to the party, but thank you for uploading these. Helping me big time late in the semester!
i didn't understand at 14:22 of how you come up with x2(2,1,0,0) ...
If someone knows plz tell me
I also don't know, I think she might've made a mistake?? Because the other two make sense but I could also be wrong.
yeah where did that rearrangement algorithm come from? This is the first time that we have seen free variables written as 1 in the video series too.
Hi, I don't understand why you put 1 for free variables inside the nx1 matrices.
We need to account for the free variable. If we put a zero, then that implies there is no variable there.
Thank you for the videos but we seem to have some missing steps we haven't seen before...The confusion with the 1 is that in previous videos we used 0 for the free variables. Have I missed a video where we go through that re-arrangement algorithm from 13:16 onwards? @@SawFinMath
I missed this part too. Here's what chat gpt said:
Find Vectors for the Null Space:
For 𝑥2=1x2 =1, set 𝑥4=0x4=0, and 𝑥5=0x5 =0.
Plug these into the equations for 𝑥1x1 and 𝑥3x3 to find their values. You end up with a vector where the position corresponding to 𝑥2x2 is 1, and the other free variables are 0.
Repeat this for x4=1, x2=0, x5=0
and for x5=1, x2=0, x4=0
@@SawFinMath
thank you
At around 5:00 what you're writing for conditions 2 and 3 isn't correct. Condition 2 should say: "When u and v are in Nul(A), is u+v in Nul(A)?" Condition 3 should be: "When u is in Nul(A), is cu is in Nul(A)?"
I was just about to comment this and saw your comment 😄
If this is not the best then I don't know what is...
Why does x belong to Rn for null space? I always see that in notes but I am unsure of why and I cant seem to visualize it.
x can't be multiplied the matrix A if it is not nx1 since A is mxn. Remember that Ax only makes sense if the amount of columns of A matches with the amount of rows of x. Since x must be nx1, then it must exist in R^n (a 2D vector exists in R2, a 3D vector exists in 3D, and an n-D vector exists in R^n). So, null spaces only make sense if x belongs to R^n.
thanks maam
why must the pivots have 0s above and below?
It must have zeros below ot we don't know if it's a pivot as it isn't reduced enough. To know the value of the solution we continue reducing to gets pivots above
@@SawFinMath is it because we cant have a variable we defined like x3 in the equation for x1 if x1 and x3 are pivots?
I hope you don't mind but I'm playing the video in 1.25 :)
I'm not offended! There are only so many hours in the day :)
thank you