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I have a question from a test I solved (without that question.. =) "If a matrix A s.t A is in M(C) (in the complex space) have only 1 eigenvalue than A is a diagonalizable matrix"

That is a false assumption since a (nXn matrix) a square matrix needs to have at least n different eigenvalues (to make eigenvectors from) - but doesn't the identity matrix have only 1 eigenvalue?...

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That is a false assumption since a (nXn matrix) a square matrix needs to have at least n different eigenvalues (to make eigenvectors from)

No, they need not be different, as you yourself have noticed (the identity matrix example) and others have also pointed it out.

You can write each matrix in the Jordan normal form:

$$A = S J S^{-1}, \quad J = \operatorname{diag}(J_1, J_2, \dots, J_m),$$

where

$$J_k = \begin{bmatrix} \lambda_k & 1 \\ & \ddots & \ddots \\ & & \lambda_k & 1 \\ & & & \lambda_k \end{bmatrix}.$$

A matrix $A$ is diagonalizable if and only if all $J_k$ are of order $1$, i.e., $J_k = \begin{bmatrix} \lambda_k \end{bmatrix}$. In other words, $m = n$ and

$$J = \operatorname{diag}(\lambda_1, \lambda_2, \dots, \lambda_n).$$

Now, if $A$ has only one eigenvalue, that means that $\lambda := \lambda_1 = \lambda_2 = \cdots = \lambda_n$, so

$$J = \operatorname{diag}(\lambda, \lambda, \dots, \lambda) = \lambda \operatorname{diag}(1, 1, \dots, 1) = \lambda {\rm I}.$$

Now, let us get back to $A$:

$$A = S J S^{-1} = S (\lambda {\rm I}) S^{-1} = \lambda S S^{-1} = \lambda {\rm I}.$$

So, $A$ is diagonalizable with only one eigenvalue if and only if it is a scalar matrix.

However, not every matrix with only one eigenvalue is diagonalizable. Just put $\lambda := \lambda_1 = \lambda_2 = \cdots = \lambda_m$, but let some of $J_k$ be of order strictly bigger than $1$. For example,

$$B = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$$

is not diagonalizable.

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Yes, it is possible for a matrix to be diagonalizable and to have only one eigenvalue; as you suggested, the identity matrix is proof of that. But if you know nothing else about the matrix, you cannot guarantee that it is diagonalizable if it has only one eigenvalue.

(The easiest way to see this is that by the Fundemental Theorem of Algebra, every complex matrix has at least one eigenvalue. But of course not every complex matrix is diagonalizable.)

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You need your $n\times n$ matrix to have n linearly-independent eigenvectors. And the identity matrix is already in diagonal form.

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It's pretty simple to see that:

A diagonalizable matrix has a unique eigenvalue if and only if it's a scalar matrix.

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Keep in mind that one eigenvalue can have multiple eigenvectors associated with them due to the possibility of having multiple parameters in the basis of the eigenspace

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