When is a Matrix Diagonalizable I: Results and Examples

when is a matrix diagonalizable
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In this video were going to consider the question of when a matrix is diagonalizable. Now most matrices are diagonalizable but there are some exceptions. Even if you allow complex eigenvalues and complex eigenvectors, there are some matrices that simply do not admit a basis of eigenvectors. And you know, the space of all n by n matrices, well thats n^2 dimensional. Turns out that the space of not diagonal n by n matrices is (n^2 – 1) dimensional. So if you pick a random matrix Its almost certain to be diagonalizable but if you pick a one parameter family of random matrices, somewhere along the way theres a good chance that youll run into a non-diagonalizable one. So lets see how a matrix can fail to be diagonalizable. The standard example is the matrix 1001. And lets go find its eigenvalues and eigenvectors. Now as always we compute λ (lambda) times the identity minus the matrix. λ minus 1, -1, 0, λ -1. And we take this determinate to get the characteristic polynomial. (λ – 1) (λ -1) – (-1) 0 just gives us (λ -1)^2 and that has one and only one root. So the only eigenvalue is λ = 1. Now only having one eigenvalue isnt necessarily a problem. If we have two eigenvectors with this eigenvalue, great. Wed have a basis of eigenvectors. So lets find out how many eigenvectors we actually have. You have to take A-λ * identity. Thats 0100 and row reduce it. Now in this case, theres not much the row reduction. It already is in reduced row echelon form. The pivot variable is the second variable. The free variable is the first. Our first equation is 0x_1 + x_2 is 0, in other words x_1 can be whatever it wants and x_2 has to be 0. And so our eigenvector has to be

a multiple of 1, 0. That means there is only one linearly independent eigenvector. If you take any two eigenvectors, they have to be multiples of each other and the matrix isnt diagonalizable. So only one eigenvector, thats not enough to form a basis for R2 or C2 and the problem is that one was a double root of the characteristic polynomial but it only gave us a single eigenvector. In general, we know that the eigenvalues are always the roots of the characteristic polynomial. This gives us two different ways to decide what the multiplicity of an eigenvalue is. You can say is it a regular root or a double root or a triple root or quadruple root of the polynomial? Thats called the algebraic multiplicity. So if you have a characteristic polynomial thats (λ – 1^2) (λ – 2^3) (λ – 7) we say the algebraic multiplicity of one is two, the algebraic multiplicity of two is three, the algebraic multiplicity of seven is one, thats just a regular root. And we denote algebraic multiplicity by m_a. Now, the geometric multiplicity describes how big the eigenspace is. Now for that, you have to actually figure out what E_{λ} is, and you do that by taking A minus λ times the identity and row reducing it. So the geometric multiplicity of one is the dimension of E_1 and thats going to be n-Rank(A-I) because thats the number of free variables. You have this many pivots, this many total variables, so the difference is the number of free variables. Likewise, m_g(2) is n-Rank(A-2I), m_g(7) is going to be n-Rank(A-7I). Now, you might ask how these two different multiplicities are related. And the answer is that the geometric multiplicity can never be bigger than the algebraic multiplicity, its always less than or equal to the algebraic multiplicity. On the other hand, its always

at least one. If {λ} is a root of the characteristic polynomial, then there exists an eigenvector, and so the geometric multiplicity is at least one. Now, in particular, that means that the algebraic multiplicity is one and the geometric multiplicity is at least one and at most one, so its just one. Theres nothing to check when the algebraic multiplicity is one. The only time things get interesting is when the algebraic multiplicity is bigger than one. Then you have to figure out the geometric multiplicity and figure out is one, two, three, or some number up to the algebraic. So the big theorem, this is the theorem that tells us when a matrix is digonalizable. Its diagonalizable if and only if the geometric multiplicities add up to n. So how do you get a basis of eigenvectors? We take a basis of eigenvectors for each eigenspace and you just concatenate them. And thats true if and only if for every eigenvalue, the geometric multiplicity is equal to the algebraic multiplicity. And you only need to check the cases where the algebraic is bigger than one because if the algebraic is one, then the geometric is one. So for example, look at this matrix here, 100 001 010. If you worked out what the characteristic polynomial is, the determinant of this matrix, winds up being (λ – 1)^2*(λ+1). So the eigenvalues are plus and minus one and the algebraic multiplicity of -1 is one, the algebraic multiplicity of +1 is two, its (λ-1)^2. So this one is a double root, so the algebraic multiplicity of one is two. Dont have to worry about the single root, we do have to worry about the double root, so lets figure out what the dimension of the eigenspace is. We have to take A-I, and there it is. You row reduce and you see that theres

If theres only one pivot there are two free variables, x_1 and x_3, and so that means that we have a two dimensional eigenspace. So in this case, m_g(1) is also two, and we win, its diagonalizable. In our next example, its just like the first example except over here in this corner, instead of putting a zero here I put a one there. That doesnt change the characteristic polynomial, the characteristic polynomial still winds up being exactly the same as before. So the algebraic multiplicity of one is two, and now if we go about figuring out the geometric multiplicity, take A-I and thats this matrix. And to row reduce it, well swap the first and third row. Then well add the first row to the second, and Ill swap the second and third rows, add the second row to the first and you see now weve got two pivots. There are two pivots, so theres only one free variable, so the geometric multiplicity is one. The geometric multiplicity did not equal the algebraic, this was one, this was two, so A is not diagonalizable. And remember, when I say the geometric multiplicity, I mean the geometric multiplicity of one. Theres also a geometric multiplicity of the other eigenvalue. Every eigenvalue has a geometric and an algebraic multiplicity. Last example is just like the second example, except Ive changed this one to a two. Now if you compute the characteristic polynomial, its (λ-2)({λ}^2-1), so the roots are one, negative one and two. They all have algebraic multiplicity 1, so they all must have geometric multiplicity 1. So its diagonalizable. By the way, all three examples were invertible matrices. People often confuse invertible matrices with diagonalizable matrices. They have nothing to do with each other. You can have invertible matrices that arent diagonalizable. You can have non-invertible matrices that are diagonalizable. Two completely different concepts.

tags:
linear algebra, matrix, diagonalization, diagonalizable, algebraic multiplicity, geometric multiplicity, eigenvalue, Diagonalizable Matrix, M346, math
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