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I have this $5 \times 5$ matrix : \begin{pmatrix} 1&1&1&1&1 \\ 2&2&2&2&2 \\ 3&3&3&3&3 \\ 4&4&4&4&4 \\ 5&5&5&5&5 \end{pmatrix} I need to find the eigenvalues and the eigenvectors. I found out that the eigenvalules are $15$ and $0$, $0$ with an algebraic multiplicity of $4$.

I am now calculating the eigenvectors and I was wondering is there a simple direct way of knowing the eigenvector of the eigenvalue of $15$, not by the usual calculation?

Is there a known eigenvector for an eigenvalue that is the sum of every column, as there is an eigenvector for an eigenvalue that is the sum of every row $[(1,1,1,1,...) ]$?

And what about cases in which the columns have the same sum, but are not identical?What is the eigenvector in that case? For example: \begin{pmatrix} 3&5&4 \\ 2&2&1 \\ 2&0&2 \end{pmatrix}

noam Azulay
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  • You’re making the same conceptual error that many do when first studying this subject: there’s no such thing as “the” eigenvector. Any nonzero scalar multiple of an eigenvector is also an eigenvector with the same eigenvalue. When all of the columns of the matrix are multiples of any one column, then all of the nonzero columns are eigenvectors, as are any other multiples of that column. – amd Oct 02 '19 at 18:12

3 Answers3

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The eigenvector to $15$ is the vector $(1,2,3,4,5)^T$, the eigenvectors to $0$ are vectors with sum of entries equal to one.

daw
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  • So , in general, for every matrix that has duplicate columns, the eigenvector of the eigenvalue that is the sum of each column is the column itself? – noam Azulay Oct 01 '19 at 13:12
  • And what about cases in which the columns have the same sum, but are not identical?What is the eigenvector in that case? – noam Azulay Oct 01 '19 at 13:14
  • 1st question: yes, 2nd question: you could ask this as a separate question – daw Oct 01 '19 at 13:49
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Just observe that given matrix is a rank one matrix and every rank one matrix have a nice form. Now for any $n \times n$ matrix $A$ with same columns can be written as $$ A = \begin{bmatrix} a_{1}\\ .\\ .\\ .\\ a_{n} \end{bmatrix} \begin{bmatrix} 1 ~~~.~~~.~~~.~~~1 \end{bmatrix} =u v^{t} $$ which has only one non zero eigen value which is trace of matrix and all other eigen values are zero. The eigen vectors corresponding to non zero eigen value is nothing but $u$, which can be proved easily.

For your second question, if each row sum is $s$, then $s$ will be eigen value of given matrix and corresponding eigen vector will be $\begin{bmatrix} 1\\ .\\ .\\ .\\ 1 \end{bmatrix} .$

Since $A$ and $A^{t}$ have same eigen values, $A^{t}$ will have column sum $s$ with eigen value $s$. But for same column sum I don't think, there is any direct technique to find eigen vector.

Manoj Kumar
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  • So that is why the eigenvector is actually the column itself, thank you very much! Very clear explanation.... Do you maybe know the answer to my second question-what about cases in which the columns have the same sum, but are not identical?What is the eigenvector in that case? – noam Azulay Oct 01 '19 at 16:01
  • @noamAzulay Edited my answer. – Manoj Kumar Oct 01 '19 at 17:34
  • But, I don't see how , for example, (1,1,1) is an eigenvector for the second matrix I posted, where the columns are not identical but each column's sum is 7. – noam Azulay Oct 02 '19 at 08:26
  • @noamAzulay you are right, I made a mistake. This eigen vector will be for same row sum. – Manoj Kumar Oct 02 '19 at 09:54
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This is obviously a rank-one matrix, which you’ve verified by finding that the algebraic multiplicity of $0$ is four. Its column space (image) is spanned by $v=(1,2,3,4,5)^T$, so the only possibility for an eigenvector with a nonzero eigenvalue is a multiple of $v$.

As for the second question, remember that there’s no such thing as “the” eigenvector: every nonzero scalar multiple of an eigenvector is also an eigenvector with the same eigenvalue. So, if the columns of a matrix are all multiples of some nonzero column, then the column space is again one-dimensional and every nonzero column is an eigenvector.

Algebraically, such a matrix can be decomposed into the outer product $uv^T$ of a pair of vectors. We then have $(uv^T)u=u(v^Tu)=(v^Tu)u=\lambda u$, therefore $u$ is an eigenvector with eigenvalue $v^Tu$.

amd
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