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# Eigenvectors and eigenspaces for a 3x3 matrix

Eigenvectors and eigenspaces for a 3x3 matrix. Created by Sal Khan.

## Want to join the conversation?

• First of all, amazing video once again. They're helping me a lot. I study economics in Belgium, and we've been getting a lot of these problems (matrices, also integrals, geometry...) but I think our book (it's written by the uni itself) isn't sufficient. Is there anyone that could reccommend maybe a good, general book on mathematics on this level (college)?thank you!
• try Linear Algebra and its applications by David C. Lays
• I am unable to find vidios on diagonalization
• 3 was a repeated root and the eigenspace resulted in a plane. Can the number of repeated roots be used to predict the dimension of the eigenspace?
• If I recall, you can't use the number of repeated roots to find the dimension of the eigenspace, because it completely depends on the matrix A that you are finding eigenvalues for. Sometimes, after obtaining an eigenvalue of multiplicity >1, and then row reducing A-lambda(IdentityMatrix), the amount of free variables in that matrix matches the multiplicity. Other times, the amount of free variables is less than the multiplicity.
All you can know, is that if an eigenvalue K has a multiplicity of n, then at most, the dimension of the eigenspace of the eigenvalue is n.
If your dimensions of your eigenspaces match with your roots of the characteristic equation (including repeated ones), then you can diagonalize the matrix.
• At , why v2 = a and v3 = b?
• Sal is setting the free variables v2 and v3 to be arbitrary real numbers, so he can write v1, v2, and v3 as linear combinations of vectors and the arbitrary real numbers a, b.
• In my book, the characteristic equation is A-lamda I, is that matters to switch the place of A and lamba I?
• Good question,

Both "A- lambda I=0" and "lambda I - A=0" work, in fact they are the same equation! To see why, look at the start of the video how Sal derives the formula. He starts with the definition of an eigenvalue, which is "Av=lambda v". He then subtracts one side of the equation from the other. It doesn't matter which side we subtract so we either end up with "Av- lambda I v=0" or "lambda I v - Av=0".
• When is row reduce a matrix, it ends up being the identity matrix, hence the vector is [1,1,1] with no parameters.. Is this still considered a eigenvector?
• Actually, if the row-reduced matrix is the identity matrix, then you have v1 = 0, v2 = 0, and v3 = 0. You get the zero vector. But eigenvectors can't be the zero vector, so this tells you that this matrix doesn't have any eigenvectors.
To get an eigenvector you have to have (at least) one row of zeroes, giving (at least) one parameter. It's an important feature of eigenvectors that they have a parameter, so you can lengthen and shorten the vector as much as you like and it will still be an eigenvector.
• Is the intersection of any two eigenspaces of the same matrix equal to the zero vector?