In this question, the OP asks about finding the matrix exponential of the matrix $$M=\begin{bmatrix} 1 & 1 & 1\\ 1 & 1 & 1\\ 1 & 1 & 1 \end{bmatrix}.$$ It works out quite nicely because $M^2 = 3M$ so $M^n = 3^{n-1}M$. The reason this occurs is that the vector $$v = \begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}$$ is an eigenvector for $M$ (with eigenvalue $3$). The same is true of the $n\times n$ matrix consisting of all ones and the corresponding vector. With this in mind, I ask the following:

  1. Can we find a standard form for an $n\times n$ matrix with the property that each of its columns are eigenvectors? (Added later: This may be easier to answer if we allow columns to be zero vectors. Thanks Adam W for the comment.)

  2. What about the case when we require all the eigenvectors to correspond to the same eigenvalue?

The matrices in the second question are precisely the ones for which the calculation of the matrix exponential would be analogous to that for $M$ as above.

Added later: In Hurkyl's answer, he/she shows that an invertible matrix satisfying 1 is diagonal. For the case $n=2$, it is fairly easy to see that any non-invertible matrix satisfies 1 (which is generalised by the situation in Seirios's answer).

However, for $n > 2$, not every non-invertible matrix satisfies this property (as one may expect). For example,

$$M = \begin{bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 1 & 0 & 1 \end{bmatrix}.$$


If each of the columns of $A$ are eigenvectors of $A$, then

$$ AA = AD $$

where $D$ is the diagonal matrix of the corresponding eigenvalues. So the matrix exponential is easy to compute:

$$ A^n = A D^{n-1}$$

As a partial answer to describing which matrices have this property, if $A$ is invertible, then we must actually have $A = D$. Also, every diagonal matrix has the stated property.


Another partial answer: One can notice that matrices of rank 1 are such examples.

Indeed, if $M$ is of rank 1, its columns are of the form $a_1C,a_2C,...,a_nC$ for a given vector $C \in \mathbb{R}^n$ ; if $L=(a_1 \ a_2 \ ... \ a_n)$, then $M=CL$.

So $MC=pC$ with the inner product $p=LC$. Also, $M^n=p^{n-1}M$.


The two answers so far (the diagonal matrix a la Hurkyl's answer and the rank one matrix a la Seirios's answer) may be combined as such: a block diagonal matrix with the blocks having rank one form. Call it the rank one blocks diagonal.

Following is a review of the spectral decomposition. Maybe someone can use it to find other possibilities or to prove that the rank-one blocks diagonal is the only possibility.

Let $\mathbf{r}_i$ be the row vectors that are left eigenvectors $$\mathbf{r}_i A= \lambda_i\mathbf{r}_i$$ Let $\mathbf{c}_i$ be the column vectors that are right eigenvectors $$A\mathbf{c}_i = \mathbf{c}_i \lambda_i$$ Then we have the spectral decomposition $$A = \sum_{i=0}^{d} \mathbf{c}_i \lambda_i \mathbf{r}_i$$ with $$ \mathbf{r}_n \mathbf{c}_m=\cases{ 0 & for $n \ne m$ \\ \lambda_n & for $n=m$ \\}$$

The last statement follows from the decomposition. Think of the $\mathbf{c}_i$ as "selectors" in the multiplication $\mathbf{r}_i A = \lambda_i \mathbf{r}_i$ since all terms in the sum for $(\mathbf{r}_i A) = \sum k_n \mathbf{r}_n$ are zero except the term of index $i$, ie $k_i = \lambda_i$.

For any column of $A$ to be an eigenvector, then the corresponding row eigenvector must containt the value $1$ for that column index. I believe that this observation would be the direction to go for a proof that the rank one blocks diagonal matrix would be the only matrix with the described properties.

  • $\begingroup$ Matrices with $A^2=0$ have the target property. Here's a rank 2 example that can't be written (nontrivially) as a block diagonal matrix: $$ \left( \begin{matrix} 0&0&0&0 \\ 0&0&1&0 \\ 0&0&0&0 \\ 1&0&0&0 \end{matrix} \right) $$ $\endgroup$ – Hurkyl Jan 13 '13 at 16:57
  • $\begingroup$ Nice example. So something about nilpotence would have to be addressed for sure, as the colums as eigenvectors here have zero eigenvalue. $\endgroup$ – adam W Jan 13 '13 at 17:04
  • $\begingroup$ Also, from the question, "each of the columns as eigenvectors" would necessarily require non-zero columns (unless zero vectors are considered eigenvectors?) $\endgroup$ – adam W Jan 13 '13 at 17:58

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