Can one eigenvalue have multiple independent eigenvectors?

(It’s possible to have two or more independent eigenvectors with the same eigenvalue. In which case, the eigenvectors form a 2 or more dimensional subspace of eigenvectors.

Are there infinitely many eigenvectors for an eigenvalue?

Since a nonzero subspace is infinite, every eigenvalue has infinitely many eigenvectors. (For example, multiplying an eigenvector by a nonzero scalar gives another eigenvector.) On the other hand, there can be at most n linearly independent eigenvectors of an n × n matrix, since R n has dimension n .

Can eigenvalues be more than one?

Using eigenvalues > 1 is only one indication of how many factors to retain. Other reasons include the scree test, getting a reasonable proportion of variance explained and (most importantly) substantive sense. That said, the rule came about because the average eigenvalue will be 1, so > 1 is “higher than average”.

Can there be more than N eigenvectors?

So a square matrix A of order n will not have more than n eigenvalues. So the eigenvalues of D are a, b, c, and d, i.e. the entries on the diagonal. So depending on the values you have on the diagonal, you may have one eigenvalue, two eigenvalues, or more. Anything is possible.

What does a repeated eigenvalue mean?

We say an eigenvalue A1 of A is repeated if it is a multiple root of the char acteristic equation of A; in our case, as this is a quadratic equation, the only possible case is when A1 is a double real root. We need to find two linearly independent solutions to the system (1). We can get one solution in the usual way.

How do you find the eigenvector of a repeated eigenvalue?

= (λ + 1)2. Setting this equal to zero we get that λ = −1 is a (repeated) eigenvalue. To find any associated eigenvectors we must solve for x = (x1,x2) so that (A + I)x = 0; that is, [ 0 2 0 0 ][ x1 x2 ] = [ 2×2 0 ] = [ 0 0 ] ⇒ x2 = 0.

How many eigenvectors can an eigenvalue have?

Since A is the identity matrix, Av=v for any vector v, i.e. any vector is an eigenvector of A. We can thus find two linearly independent eigenvectors (say <-2,1> and <3,-2>) one for each eigenvalue.

How many eigenvectors can you have?

EDIT: Of course every matrix with at least one eigenvalue λ has infinitely many eigenvectors (as pointed out in the comments), since the eigenspace corresponding to λ is at least one-dimensional.

Can you have two of the same eigenvalues?

Two similar matrices have the same eigenvalues, even though they will usually have different eigenvectors. Said more precisely, if B = Ai’AJ. I and x is an eigenvector of A, then M’x is an eigenvector of B = M’AM. So, A1’x is an eigenvector for B, with eigenvalue ).

What does an eigenvalue less than 1 mean?

An eigenvalue less than 1 means that the PC explains less than a single original variable explained, i.e. it has no value, the original variable was better than the new variable PC2. This would fit with factor rotation producing a second factor that is related to a single variable.

Does each eigenvalue have one eigenvector?

The converse statement, that an eigenvector can have more than one eigenvalue, is not true, which you can see from the definition of an eigenvector. However, there’s nothing in the definition that stops us having multiple eigenvectors with the same eigenvalue.