 So the spectrum of a matrix sigma of A is the set of eigenvalues of A and again the spectral radius 2 of A is the largest magnitude eigenvalue which is equal to max of mod lambda where lambda belongs to sigma of A. So this is just writing the spectral radius in terms of sigma of A. Another small point to note is that if A x equals lambda x then A times C x is equal to lambda times C x for any 0 not equal to C belonging to some complex base and so basically if x is an eigenvector so is C x for any C not equal to 0. Okay so now the other thing is which will sort of lead us to the characteristic polynomial and studying it in is that given any polynomial so let us say p of t a k t power k plus a k minus 1 t power k minus 1 plus a 0 this is a polynomial in t and a in c to the n cross n okay so we can define p of a this is the polynomial evaluated at a matrix valued point a so we will define it as a k times a power k plus a k minus 1 a power k minus 1 plus a 0 identity matrix otherwise you cannot add it to these terms so this is how we define the polynomial evaluated at a matrix. Now this p of a it will always have the same eigenvectors as the matrix a itself and the polynomials the the eigenvalues of p of a p of a is an n cross n matrix because this is n cross n this is n cross n this is n cross n when you add up all these terms you will get an n cross n matrix and its eigenvalues are closely related to the eigenvalues of a itself and so that is the following theorem yes can somebody else confirm it is getting updated for me sir yeah maybe you know if you just wait for a second if it doesn't update then just drop off and join again you should see the new screen polynomial okay and if lambda is an eigenvalue a and x is a corresponding eigenvector then p of lambda is an eigenvalue of p of a x is an eigenvector of p of a associated with the eigenvalue p of lambda okay so how do we show this this is very simple so if I take p of a times x this is going to be equal to a k a power k x plus a k minus 1 a power k minus 1 x plus etcetera plus a 0 times the identity times x which is equal to x now of course a power j times x is equal to a power j minus 1 times a x which is equal to lambda a power j minus 1 times x and so we can keep going like this go to j minus 2 and so on and so this is going to be equal to lambda power j times x okay so that means that p of a times x is equal to so it becomes lambda power k times x here and lambda power k minus 1 times x here and so on so it becomes a k lambda power k x plus a k minus 1 lambda power k minus 1 x plus etcetera plus a 0 times x and x is multiplying all these terms so I can pull it out and write this as a k lambda power k plus a k minus 1 lambda power k minus 1 plus etcetera plus a 0 times x and this is nothing but p of lambda this polynomial evaluated at the number lambda times x so we see that p of a times x equals p of lambda times x this is an n cross n matrix this times x equals this is a scalar so p of lambda times x which means that p of lambda is an eigenvalue of a and the corresponding eigenvector is x p of lambda is an eigenvalue of p of a and the corresponding eigenvector is x so based on this so for example if sigma of a or the two values say minus 1 and 2 then what is sigma of a square a is a 2 cross 2 matrix which has two eigenvalues minus 1 and 2 so sigma of a square will be lambda 1 square and lambda 2 square which is equal to 1 and 4 and as I already mentioned to you earlier a is singular if and only if 0 is part of sigma of a because if a is singular if and only if ax equals 0 for some non-zero x which means which is true if and only if lambda equal to 0 is an eigenvalue sir yes so in the above example the polynomial is px equals x square right correct okay thank you okay so now let's discuss this characteristic polynomial in some more detail so this basically we answer questions about the eigenvalues of a and for example how many eigenvalues does an n cross n matrix have and when will it have and linearly independent eigenvectors okay so there are the type of questions we want to answer so the characteristic polynomial will define it in a slightly different way from what I said earlier it's just a change of sign we'll define it to be pa of t which is equal to determinant of ti minus a okay this is a polynomial of degree n and its roots are the eigenvectors eigenvalues so why is this a polynomial of degree n it's because if I think of it I mean just write it for the 2 cross 2 k so that you see what happens this matrix will look like t minus a11 minus a12 minus a21 and t minus a22 and if I take this determinant it will be equal to t minus a11 t minus a22 minus a21 a12 and so you can see that this is a quadratic polynomial so in the n cross n case by similarly writing it out you will see that it has to be a polynomial of degree n the coefficient of t is always 1 so it when I take the determinant of this I cannot suddenly get a polynomial whose degree is n minus 1 it will always be of degree n so there is a what is known as the fundamental theorem of algebra we will not prove this here we will take it on faith but it says that any polynomial of degree n and complex coefficients or width complex coefficients has exactly n zeros counting multiplicities okay so multiplicity just simply means how many times a particular lambda occurs as a zero of pa of lambda so if you count multiplicities then there are exactly n eigenvalues so for example the n cross n identity matrix has exactly n eigenvalues all are equal to 1 now because yes sir commerce coefficients means a real also is included yes okay now suppose we let lambda 1 through lambda n be the eigenvalues of a where repeated eigenvalues are okay they're just counted as lambda 1 lambda 2 to n so so let lambda 1 lambda n be the eigenvalues of a counting multiplicities the n eigenvalues of a counting multiplicities then these are the zeros of the characteristic polynomial so we can write pa of t equals t minus lambda 1 t minus lambda 2 t minus lambda n okay so this is a factorized form of the characteristic polynomial and this polynomial is identical to this determinant of ti minus a so in particular if I look at the coefficient of t to the n minus 1 in this polynomial that's just the sum of the eigenvalues and similarly if I look at the coefficient of t power 1 here you can see that it's just going to be t power 1 will be minus a 11 minus a 22 and the coefficient of t power n minus 1 here will be minus of lambda 1 plus lambda 2 plus etc up to lambda n so those two coefficients must match if the polynomials are if we say two polynomials are equal it means that the coefficients of t power n t power n minus 1 all way down to t power 0 must match t power n obviously matches because the coefficient of t squared is 1 and here also when I take the coefficient of t squared in the 2 m in the lambda 1 lambda 2 case that will also be equal to 1 but basically if I look at what happens to the coefficient of t to the n minus 1 here it's going to be equal to a 11 plus a 22 which is the coefficient which is equal to which must equal lambda 1 plus lambda 2 which is the coefficient of t to the n minus 1 here so basically we and similarly if I take the constant term here that's the product of these eigenvalues and the constant here here is going to be all I have to do is the constant term will be obtained by setting t equal to 0 in in this pa of t and so the constant term is just nothing but the determinant of the matrix a so the product of the eigenvalues equals the and the sum of the eigenvalues equals the trace of the matrix so that's the following theorem if lambda 1 through lambda n are the eigenvalues of a so the determinant of a is the product of the eigenvalues and the trace of a is the sum of the eigenvalues so just to maybe make it clear if you look at the if I set t equals 0 here what I get is minus lambda 1 into minus lambda 2 up to minus lambda n which is minus 1 power n times the product of the eigenvalues if I set t equals 0 here I will get determinant of minus a which is again going to be equal to minus 1 power n times determinant of a those two must be equal and therefore minus 1 power n cancels with minus 1 power n and what you're left with is determinant of a is the product of the eigenvalues and similarly if I look at the coefficient of t power n minus 1 here it's going to be minus lambda 1 minus lambda 2 up to minus lambda n and here if I look at the coefficient of t power n t power 1 I have a11 plus minus of a11 plus a22 which is nothing but the negative trace of the matrix so negative trace of the matrix is the same as the negative sum of the eigenvalues and so we have this part here so for the sake of completeness I just say here coefficient of t to the n minus 1 and this is obtained by looking at the coefficient of t power 0 or the constant term the other way is set t equals 0 in okay so there are a couple of other results before that maybe I'll just write this one thing out which is what is the procedure to find eigenvalues and eigenvectors