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Current time:0:00Total duration:5:39

Example solving for the eigenvalues of a 2x2 matrix

Video transcript

in the last video we were able to show that any lambda that satisfies this equation for some nonzero vectors V then the determinant of lambda times I the identity matrix minus a must be equal to 0 or we could rewrite this as saying lambda is an eigen value eigen value of a if and only if all right it is if if and only if the determinant of lambda times the identity matrix minus a is equal to 0 now let's see if we can actually use this in any kind of concrete way to figure out I gain values so let's do a simple 2x2 let's deal in r2 let's say that a is equal to the matrix one two and four three and I want to find the eigen values of a so if lambda is an eigen value of a let's say lambda is eigen value of a then this right here tells us that the determinant of lambda times the identity matrix so it's going to be the identity matrix in r2 so lambda times 1 0 0 1 minus a 1 2 4 3 is going to be equal to 0 well what is this equal to this right here is the determinant of lambda times this is just lambda times all of these terms so it's lambda times 1 is lambda lambda times 0 is 0 lambda times 0 0 lambda times 1 is lambda and from that we'll subtract a so you get 1 2 4 3 and this has got to equal 0 and then this matrix or this difference of matrices this is just so keep the determinant this is the determinant of this first term is going to be lambda minus 1 lambda minus 1 the second term is 0 minus 2 so it's just minus 2 the third term is 0 minus 4 so it's just minus four and then the third the fourth term is lambda minus 3 lambda minus 3 just like that so kind of a shortcut to see what happened the terms along the diagonal well everything became a negative right we negated everything under the terms around the diagonal we've had a lambda out front that was essentially the byproduct of this expression right there so what's the determinant of this 2 by 2 matrix well the determinant of this is just this times that minus this times that so it's lambda minus 1 times lambda minus 3 minus minus these two guys multiplied by each other so minus 2 times minus 4 is plus 8 minus 8 this is the determinant of this matrix right here of this matrix right here or this matrix right here which simplify to that matrix and this has got to be equal that has got to be equal to 0 and the whole reason why that's got to be equal to 0 is because we saw earlier this matrix has a non-trivial null space because it has a non-trivial null space it can't be invertible and it's determinant has to be equal to 0 so now we have an interesting polynomial equation right here we can multiply it out we get what let's multiply it out we get lambda squared right minus 3 lambda minus 3 lambda minus lambda plus 3 minus 8 is equal to 0 or lambda squared lambda squared minus 4 lambda minus 4 lambda minus 5 minus 5 is equal to 0 and this in case you're want to know some terminology this this expression right here is known as the characteristic polynomial the characteristic polynomial just a little terminology polynomial but if we want to find the eigenvalues for a we just have to solve this right here this is just a basic quadratic problem and this is actually factorable to see two numbers and you take the product as minus five when you add them you get minus four by it's minus 5 and plus one so you get lambda minus five times lambda plus one is equal to zero right minus 5 times 1 is minus 5 and then minus 5 lambda plus 1 lambda is equal to minus 4 lambda so the two solutions of our characteristic equation being set to 0 our characteristic polynomial our lambda is equal to 5 or lambda is equal to minus 1 so just like that using the information that we prove to ourselves in the last video we're able to figure out that the two eigenvalues of a are lambda equals 5 and lambda equals negative 1 now that only just solves part of the problem right we know we're looking for AI ghen values and eigenvectors right we know that this equation can be satisfied with lambdas equaling 5 or minus 1 so we know the eigenvalues but we're yet to determine the actual eigenvectors so that's what we're going to do in the next video