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Current time:0:00Total duration:14:34

in the last video we started with the 2x2 matrix a is equal to 1 2 4 3 and we use the fact that lambda is an eigenvalue lambda is an eigenvalue eigenvalue of a if and only if the determinant of lambda times the identity matrix in this case it's the 2 by 2 identity matrix minus a is equal to 0 this gave us a characteristic polynomial and we solved for that and we said well the eigenvalues for a are lambda is equal to 5 and lambda is equal to negative 1 that's what we saw in the last in the last video we said that if you were trying to solve a times some eigenvector is equal to lambda times that eigenvector the two lambdas which this equation can be solved for are the lambdas 5 and minus 1 if we assume assuming assuming a non zero eigen vectors assuming nonzero non zero eigen vectors eigen vectors so we have our eigen values but that's you know then I don't even call that half the battle what we really want is our eigen vectors and our eigen values so let's see if we can do that so if we manipulate if we manipulate this equation a little bit and we manipulated in the past actually even come up with this statement over here we can you know rewrite this over here as the zero vector is equal to lambda times my eigen vectors minus a times my eigen vector just subtracted a V from both sides and we know lambda times some eigen vectors the same thing as lambda times the identity matrix times that I ghen vector so all I'm doing is rewriting this like that you multiply the identity matrix times an eigen vector or times any vector you just going to get that vector so these two things are equivalent - a V a V that's still going to be equal to the zero vector so far all I've done is manipulated this thing and this is really how we got to that thing up there then you take the you factor out the V so to speak because we know that matrix vector products exhibit the distributive property and we get lambda times the identity matrix minus a times times my eigenvector have got to be equal to zero or another way to say it is for any for any for any for any lambda eigenvalue let me write it for any eigen value lambda eigenvalue lambda the eigenvectors that correspond to that lambda we can call that the eigen space the eigenspace for lambda so that's a new word eigen space again space eigen space just means all of the eigenvectors that correspond to some eigenvalue the eigen space for that for some particular eigenvalue is going to be equal to the set of vectors that satisfy this equation well the set of vectors that satisfy this equation is just the null space it's just the null space of that right there so it's equal to the null space of this matrix right there the null space tricks of lambda times the identity matrix n by n identity matrix minus a and so everything I've done here this is true this is the general case but now we can apply this notion to this matrix a right here so we know that v is an eigen value so for let's say for lambda is equal to 5 the eigenspace that corresponds to 5 is equal to the null space of what what is 5 times the identity matrix 5 it's going to be the 2 by 2 identity matrix right 5 times the identity matrix is just it's just 5 0 0 5 minus a that's just 1 2 4 3 so that is equal to the null space the null space of the matrix 5 minus 1 is 4 0 minus 2 is minus 2 0 minus 4 is minus 4 and then 5 minus 3 is 2 so the null space of this matrix right here this matrix is just a actual number of numerical representation of this matrix right here the null space of this matrix is the set of all of the vectors that satisfy this or all of the eigenvectors that correspond to this eigenvalue or the eigenspace that corresponds to the eigenvalue 5 these are all equivalent statements so we just need to figure out the null space of this guy is all the vectors that satisfy the equation 4 minus 2 minus 4 2 times some eigenvector is equal to the 0 vector and the null space of a matrix is equal to the null space of the reduced row echelon form of a matrix so what's the reduced row echelon form of this guy well let's I guess a good starting point let me keep my first row the same 4 minus 2 and let me replace my second row with my second row plus my first row so minus 4 plus 4 is 0 2 plus minus 2 is 0 now let me divide my first row by 4 and I get 1 minus 1/2 and then I get 0 0 so what's the null space of this this corresponds to if V this times V 1 V 2 that's just another way of writing my eigenvector v has got to be equal to the 0 vector or another way to say it is that my first entry V 1 which corresponds to this pivot column plus or minus 1/2 times my second entry has got to be equal to that 0 right there or V 1 is equal to 1/2 is equal to 1/2 V 2 and so if I wanted to write all of the eigen vectors that satisfy this so I could write it this way my eigen space my eigenspace that corresponds to lambda equals 5 that corresponds to the eigenvalue 5 is equal to the set of all of the vectors v1 v2 that are equal to some scaling factor real estate say it's equal to T times what if we say that v2 is equal to T so v2 is going to be equal to T times 1 and then v1 is he going to be equal to 1/2 times v2 or 1/2 times T one-half times T just like that for any for any T is a member for any T is a member of the real numbers and if we wanted to we could scale this up because any real number times you know 1/2 that would also be a you know a the span let me do that actually it'll make it a little bit cleaner actually I don't have to do that so we could write the eigenspace the eigenspace for the eigenvalue v is equal to the span of the vector of the vector one half and one so it's a line in r2 those are all of the eigen vectors that satisfy that that work for the equation where the eigen value is equal to 5 now what about when the eigen value is equal to minus 1 and the eigen value is equal to minus 1 so let's do that case when lambda is equal to minus 1 then we have then the we're going to have the it's going to be the null space so the eigenspace for lambda is equal to minus 1 is going to be the null space of lambda times our identity matrix which is going to be minus 1 0 0 minus 1 all right it's going to be minus 1 times 1 0 0 1 which is just minus 1 there minus a so minus 1 2 4 3 and this is equal to the null space of minus 1 minus 1 is minus 2 0 minus 2 is minus 2 0 minus 4 is minus 4 and minus 1 minus 3 is minus 4 and that's going to be equal to the null space of the reduced row echelon form of that so the we can perform some row operations right here let me just put it in reduced row-echelon form so if I replace my second row with my second row plus two times my first row so I'll keep the first row the same minus two minus two and then my second row I'll replace it with two times I'll replace it with it plus two times the first or even better I'm going to replace it with it plus minus two times the first so minus four plus four is zero minus 4 plus 4 is zero and then if I divide the top row by minus two the reduced row echelon form of this matrix right here or this matrix right here is going to be one one zero so the eigenspace that corresponds to the eigenvalue minus one eigen space corresponds to the eigenvalue minus one is equal to the null space of this guy right here or it equals the it's the set of vectors that satisfy this equation 1 1 0 0 and then you have V 1 V 2 is equal to 0 or you get V 1 plus not these aren't vectors these are just values V 1 plus V 2 is equal to 0 sorry is equal to 0 because 0 is just equal to that thing right there so 1 times V 1 plus 1 times V 2 is going to be equal to that 0 right there or I could write V 1 is equal to minus V 2 or if we say that V 2 is equal to T we could say V 1 is equal to minus D or we could say that the eigenspace for the eigenvalue minus 1 is equal to all of the vectors all of the vectors V 1 V 2 that are equal to some scalar T times V 1 is minus T and V 2 is plus T or you could say this is equal to the span of the vector minus 1 and 1 so let's just graph this a little bit just to understand what we just did we were able to find two eigen values for this five and minus one and we're able to find all of the factors that are essentially the or we were able to find a set of vectors that are the eigenvectors that correspond to each of these eigenvalues so let's graph them so if we go to our two let me draw my axis it's my vertical axis that's my horizontal axis so all of the vectors that correspond to lambda equal five are aligned the line one-half one so or the span of one-half one so that is one that is one so you go one half and one just like that so that's that vector spanning vector but anything along the span of this all the multiples of this are going to be valid eigenvectors or anything along that line all of the vectors when you draw them in standard position point to a point on that line all of these vectors any vector on there is going to be a valid eigenvector and the corresponding eigenvalue is going to be equal to five so you give me you give me this guy right here when you apply the transformation it's going to be five times this guy you give me you give me if this guy is X T of X is going to be five times this guy whatever vector you give along this line the transformation of that guy and the transformation is literally multiplying it by the matrix a where did I have the matrix a the matrix a right up there you're essentially just scaling this guy by five in either direction this is for lambda equal 5 and for lambda equals one is the span of this vector which is minus 1 minus 1 1 which looks like this it will look so this vector looks like that we care about the span of it so anything any vector in this set any vector that when you draw in standard position lies or points two points on this line will be an eigenvector for the eigenvalue minus one so lambda equals minus one so if you take well let's say you take the spanning vector here you apply the transformation you're going to get minus one times it so this is X the transformation of X is going to be that right same length just in the opposite direction if you have this guy right here you apply the transformation it's going to be in the same spanning line just like that so the two eigen spaces for the matrix where did I write it I think it was the matrix 1 2 3 1 2 4 3 the two eigenvalues were 5 and minus 1 5 and minus 1 and then it has an infinite number of eigenvectors so they actually create two eigenspaces each of them correspond to one of the eigenvalues and these lines represent those two eigenspaces you give me any vector in either of these sets and they're going to be an eigenvector or you can be any vet I'm using the word vector too much give me any vector in either of these sets and they will be an eigenvector for our matrix a and then depending which line it is we know what their transformation is going to be it's going to be on this guy you take the transformation the resulting vector is going to be five times the vector if you take one of these eigenvectors and you transform it the resulting transformation of the vector is going to be minus one times that vector anyway we now know what eigenvalues eigenvectors eigenspaces are and even better we know how to actually find them